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"Watir" (pronounced water) stands for "Web Application Testing in Ruby". It is a toolkit for automated tests to be developed and run in a web browser that is simple and can be easily upgraded. Though it is an open-source family of Ruby libraries, it will support your app regardless of the technology you have used. It also supports multiple browsers on various platforms.
Since you can conduct the test online & the features like online proctoring, windows violation records make the screening secure, you can ask the candidate to appear for this online test on Watir testing from their any desired place as per their convenience. It will help you to eliminate the needs of infrastructure & travel reimbursement.
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Identify potential hires using Watir testing assessment test by evaluating working skills and job readiness of the applicants. For this reason, emphasis is laid upon evaluating the knowledge of applied skills gained through real work experience, rather than theoretical knowledge.
Watir skill test contains may contain MCQ's (Multiple Choice Questions), MAQ's (Multiple Answer Questions), Fill in the Blanks, Descriptive, Whiteboard Questions, Audio / Video Questions, LogicBox (AI-based Pseudo-Coding Platform), Coding Simulations, True or False Questions, etc.
If you want to open this link from a Watir test script then, which of the following regular expressions can replace this link given that the GC_1_7P7_ part doesn't affect the location of the application?
Q 2. If the following link is used in string format in a Watir test case then, which of the following option shows the correctly escaped form of the same?
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Lightweight champion Khabib Nurmagomedov showed he is ready to face Conor McGregor at UFC 229 by being the first on the scale Friday morning. He weighed in at the 155-pound limit Friday.
An hour later, Ireland's McGregor weighed in at 154.5 pounds and emphasized his own preparedness for Saturday's match by roaring out a primal scream accompanied by a look of rage while atop the scale.
Russia's Nurmagomedov (26-0) left Thursday's news conference early before former two-division champion McGregor (21-3) arrived, saying he needed to start his weight-cutting process. In March 2017, Nurmagomedov fell ill during his weight cut and had to withdraw versus Tony Ferguson.
"I want him to make weight," McGregor said upon his later arrival.
Friday marked the third consecutive fight that Nurmagomedov has arrived in the first group of the morning weigh-in, and he's dominated in those clashes, defeating then-top-five-ranked Edson Barboza and most recently claiming the belt stripped from McGregor in April by overwhelming replacement foe Al Iaquinta.
McGregor previously stood as 145-pound featherweight champion in 2015 by knocking out Jose Aldo, and he fought two bouts against Nate Diaz at 170 pounds in 2016.
The ceremonial weigh-in, where both fighters will meet in a face-off at T-Mobile Arena, is scheduled for Friday afternoon.


FPS drop in latest build?
I've been using Dolphin 3.5-367 to play all my games full speed with no problem at all ( monster hunter tri, zelda twilight princess, tales of symphonia dawn of the new world, among other games), and I decided to try the newest release (3.5-1655) and my fps lowers a lot even with the same settings, for example in monster hunter tri my fps with the 367 build is locked at 30, while in the 1655 build it lowers to 22-25.
Yes, this is normal. Either overclock your cpu, keep using old buggy versions and don't ask for help when things don't work, or deal with it being slow.
If I'm correct the development versions let me use the latest changes in Dolphin for the cost of less stability correct?
Then what could be causing these fps drops?
Think I'll stick with the superior old builds.
Much more accurate in unnoticeable ways.
Why try to fix it? Well, because it didn't work. Our goal here isn't to make an emulator that best suits your desires to play games. We want to make a good emulator that properly emulates the console. Sometimes that means emulating new things that were ignored previously, or removing old hacks that were fast but inaccurate. Sometimes this means slowing down the emulator for no noticable improvement. But just because you didn't notice a problem before doesn't mean there wasn't one. If you want to use 367, go for it, but don't ask us for help when something doesn't work.
I'm not trying to start a discussion and sorry if I was misunderstood, what I meant was that I Know there are still many things to improve and to add so that the emulator can be as close as possible to the real thing and I am very grateful for making Dolphin happen. What I was trying to say is that if the older build gives me a better experience by not having fps drops nor giving me trouble with the interface or the controls then I feel I should stick with it. If I do have a problem later on and can't get around it I will try the new builds and be grateful that all of the developers have put time and effort for making us have a better experience. Again I really apologize if I was misunderstood I do not want to discuss with you (or anyone for that matter) because I think the Dolphin community is filled with some really awesome people that help one another for a greater good.


text:
Super-star team places have restrictions?
xiaosachong2010-02-01 11:01:08 +0000 #1
For example, Paul and Wade and James, the three players in a team do? " On each title that does not it?
95463577142010-02-01 11:10:04 +0000 #2
Of course, no restrictions, Kevin special people known as the Big Three not to do, but the level of each team's salary is limited, players wages are generally high
lxzhi2010-02-01 11:40:17 +0000 #3
No, but there are salary restrictions ah ( salary cap)
zfx12332010-02-01 12:21:43 +0000 #4
not heard of this super-star, said open-ended, but because there are salary restrictions on NBA luxury tax provisions, to a certain extent, limit the maximum salary the team recruited players. There is no check data, but like you said, Paul, Dwyane Wade, James such a level of players, a single person's salary is absolutely top NBA column, these three add up to enough to choke a salary, but also with other locations is strong salary of the players, as well as low-cost substitute, this is impossible. Furthermore put forward a team like the other teams still playing some kind, NBA to develop such a system is to a certain level, the League has always been as far as possible in a competitive state. There is suspense is not only interesting? Hehe
steven1008372010-02-01 11:16:11 +0000 #5
No, but there are salary cap, that is, the team's salary cap is a definite! Beyond this limit have to pay the excess the same money, also known as the luxury tax! For example more than 2 million, they pay 2 million, and this is what the team in order to prevent the overwhelming majority, to dominate the alliance, of course, are willing to pay the luxury tax is a great team! Lakers ah, the Celtics ah!
Arsenal League Cup has been eliminated it?
Arsenal this week, to whom, when will Hou?
Why is the standard of football in China is not high but also engage Super, when kicked in A is not
Why Asennaken Campbell accepted Vieira is also not acceptable because of money factors?
Franck Ribery went close, will be back?
Arsenal League Cup is not all play with three lines teenager ah?
Brazil's World Cup, then let rest time, in line with China's request?
South Africa's World Cup broadcasting time work schedule in line with the Chinese people do?
Is to determine the 2014 World Cup held in Brazil on the do?
South Africa World Cup tickets, where to buy

redpajama_set_name: RedPajamaCommonCrawl


The Jamestown is a great home for entertaining friends and family. A large great room opens to the spacious kitchen with center island and a spacious dining area. A private den is tucked away in the back of the home. An option is available to convert the den into a fifth bedroom with full bath. A spacious owner suite includes a large walk-in closet and private bathroom.


Bernie Worrell (1944-2016) needs little introduction. He was an American keyboardist, arranger and composer best known as a founding member of Parliament/Funkadelic and for his work with Talking Heads. The list of his musical collaborations is extensive and includes renowned artists and acts from the likes of Manu Dibango, Fela Kuti, Jack Bruce, Mike Watt and Les Claypool. Bernie Worrell's list of achievements is equally impressive...This album was the iconic keyboardist's fourth solo album, recorded and released in 1993. The album has an all star line up that includes several renowned artists from the likes of Fred Wesley, Buckethead, Bill Laswell and Umar Bin Hassan (The Last Poets).'The Other Side' is absolutely uncategorizable. Modal strains of jazz, avant-noise, poetry readings, heavy funk overtones and Hammond organ-drenched experimental atmospheres give a lengthy and unique flavor to each song. The result of this visionary approach is downright fascinating & Worrell-s attitudes go way beyond the expected boundaries flanking free-form jazz and funk?he transforms these recordings into a jaw-dropping, delightful adventure of unexplored and unexpected sonic realms.Available as a deluxe 180 gram 2LP set with silkscreened D-side, limited to 500 copies worldwide. First time on vinyl!


Tuition includes morning snack, lunch, and afternoon snack after transitioning to table food.
Tuition includes morning snack, lunch, and afternoon snack.
Tuition includes morning snack and afternoon snack. Lunch is available for purchase for $3.00/day.
Sibling Discount: A 5% discount is available to the 2nd child enrolled in our Lower and Upper School programs, and a 10% discount is available for the 3rd child as well as any additional children enrolled in our Lower and Upper School programs. Full tuition is paid for the oldest student and the discount applies to each additional enrolled sibling (Grades K–Upper School).
Application Fee: A fee of $130 per child applies to each applicant at the time the application is submitted.
Kindergarten Fee: A $250 one-time fee applies to each student entering Kindergarten. This fee includes the initial set up of each Kindergartener entering the Lower School Program. This fee must be paid upon contract signing to secure enrollment.
New Student Admission Fee (Excluding Kindergarten): A $250 one-time fee applies upon the acceptance of each new student enrolling (excluding Kindergarten). This fee must be paid upon contract signing to secure enrollment.
Curriculum and Technology Fee: A fee of $195 per student is charged to help subsidize the increasing dollars being spent on curriculum and technology. This fee must be paid upon contract signing to secure enrollment.
Facilities Fee: A fee of $200 is charged to help subsidize the increasing dollars being spent on facilities and capital improvements. This fee must be paid upon contract signing and enrollment.
Enrollment in FACTS Management: If utilizing for tuition payments and incidental charges, FACTS assesses a yearly fee of approximately $20–$50 to use their services.
Upon acceptance, a $100 non-refundable deposit is required to reserve placement. This deposit will be applied equally towards the family tuition balance.
State scholarships are available. Please visit www.stepupforstudents.org for more information.
Tuition is due on the first day of each month.
Withdrawals after June 1, but prior to the start of school, are subject to the following financial obligation: After June 1, one-month's tuition is due; after July 1, two-months' tuition is due; and after August 1 and prior to beginning of school, three-months' tuition is due.
The semester's tuition is due if a student withdraws after the beginning of the school year because the hiring of teachers and staff and the ordering of textbooks and other classroom materials are based upon anticipated enrollment. Student records will not be released until the account is paid in full. Parents will be expected to honor their full year's commitment to the school. Requests for exceptions must be addressed in writing and will be considered on a case-by-case basis.
Tuition that becomes 30 days past due will subject the student to being withheld from attending school until such amounts due to First Baptist Academy are made current.
Monthly tuition payments are made to a tuition management system (FACTS). The annual fee is approximately $45 per family.
Payment of the fees and deposits are due upon enrollment. Parents or guardians may choose to pay the tuition in full or in semi-annual payments (August 2019 & December 2019). FBA also offers a 12-month (July 2019–June 2020) and a 10-month payment plan (August 2019–May 2020).


Sometimes it is a chore to feed database with all data we need for given database view to return results we are after within specs. Fortunately we can utilize temporary tables to mock such database views!
Then it is just a matter of creating simulated view results, i.e. using factories.


text:
\section{Introduction}
Algebras where the identities defining the structure are twisted by a homomorphism are called Hom-algebras. The notion of Hom-Lie algebras was introduced by Hartwig, Larsson and Silvestrov to describe the structures on certain deformations of the Witt algebras and the Virasoro algebras in \cite{HLS}. Hom-Lie algebras are also related to deformed vector fields, the various versions of the Yang-Baxter equations, braid group representations, and quantum groups in \cite{HLS, Y1, Y2}. More research on Hom-Lie algebras see references \cite{G1, S2}.

The notion of ring of quotients was introduced by Utumi in \cite{U1}, which played an important role in the development of the theories of associative and commutative rings. He showed that every associative ring without right zero divisors has a maximal left quotients ring and constructed it. Inspired by \cite{U1}, Siles Molina studied the algebras of quotients of Lie algebras in \cite{S1}. Martindale rings of quotients were introduced by Martindale in 1969 for prime rings in \cite{M1}, which was designed for applications to rings satisfying a generalized polynomial identity. In \cite{GG}, Garc\'{i}a and G\'{o}mez defined Martindale-like quotients for Lie triple systems with respect to power filters of sturdy ideals and constructed the maximal system of quotients in the nondegenerate cases. More research about quotients of Lie systems refer in references \cite{GC, MCL}. In \cite{M2}, Mart\'{i}nez derived a necessary and sufficient Ore type condition for a Jordan algebra to have a ring of fractions, which is the origin of algebras of quotients of Jordan systems. In \cite{AGG, AM, GG2, M3}, authors introduced and researched algebras of quotients and Martindale-like quotients of Jordan systems respectively. Inspired by \cite{S1}, we will introduce the notion of algebras of quotients of Hom-Lie algebras and prove the properties, such as semiprimeness and primeness,  can be lifted from a Hom-Lie algebra to its algebra of quotients in this paper.

As we know, one popular topic of research is the relationship between algebras of quotients of non-associative algebras and associative algebras in recent years. In \cite{PS}, authors examine the relationship between associative and Lie algebras of quotients. Inspired by them, we'll explore whether there exists a relationship between associative and Hom-Lie algebras of quotients. The answer is affirmative and we will describe in detail in the following.

The paper is organised as follows: in Section \ref{se:2}, we'll introduce some basic definitions and prove some elementary propositions which will be used in the following. In Section \ref{se:3}, we'll give the definition of algebra of quotients of a Hom-Lie algebra at first. Next, we'll introduce the concept of ideally absorbed of a Hom-Lie algebra and show that a Hom-Lie algebra $(Q, [\cdot, \cdot], \alpha)$ is an algebra of quotients of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ if and only if $(Q, [\cdot, \cdot], \alpha)$ is ideally absorbed into $(L, [\cdot, \cdot], \alpha)$, presented as Theorem \ref{thm:3.10}. In Section \ref{se:4}, we focus on semiprime Hom-Lie algebras and construct the maximal algebras of quotients of them and characterize them. In Section \ref{se:5}, we mainly study the relationship between algebras of quotients of Hom-Lie algebras and associative algebras generated by inner derivations of Hom-Lie algebras, and get the main result that $A(Q)$, the associative algebra generated by inner derivations of $(Q, [\cdot, \cdot], \alpha)$, is a left quotients of $A_{0}$ if $(Q, [\cdot, \cdot], \alpha)$ is an algebra of quotients of $(L, [\cdot, \cdot], \alpha)$, where $A_{0}$ denotes the subalgebra of $A(Q)$ satisfying $A_{0}(L) \subseteq L$, written as Theorem \ref{thm:5.11}. In Section \ref{se:6}, we introduce the concept of dense extension first of all. Next, we get a proposition about dense extension, Proposition \ref{prop:6.7}, that every $I \subseteq Q$ is also dense for every essential Hom-ideal $I$ of $(L, [\cdot, \cdot], \alpha)$ if $L \subseteq Q$ is a dense extension and $(Q, [\cdot, \cdot], \alpha)$ is an algebra of quotients of $(L, [\cdot, \cdot], \alpha)$.
\section{Preliminaries}\label{se:2}
\begin{defn}\cite{HLS}
Let $L$ be a vector space over a field $\mathbb{F}$. Define a bilinear map $[\cdot, \cdot] : L \times L \rightarrow L$ and a linear map $\alpha : L \rightarrow L$. If the triple $(L, [\cdot, \cdot], \alpha)$ satisfies the following conditions:
\begin{enumerate}[(1)]
\item $[x, y] = -[y, x]$, $\forall x, y \in L$;
\item $[\alpha(x), [y, z]] + [\alpha(y), [z, x]] + [\alpha(z), [x, y]] = 0$, $\forall x, y, z \in L$.
\end{enumerate}
then $(L, [\cdot, \cdot], \alpha)$ is called a Hom-Lie algebra.
\end{defn}
\begin{defn}\cite{HLS}
A subspace $W$ of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ is called a Hom-subalgebra of $L$ if it satisfies $\alpha(W) \subseteq W$ and $[W, W] \subseteq W$.
\end{defn}
\begin{defn}\cite{HLS}
A subspace $I$ of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ is called a Hom-ideal of $L$ if it satisfies $\alpha(I) \subseteq I$ and $[I, L] \subseteq I$.
\end{defn}
\begin{re}\label{re:2.4}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-Lie algebra. For any $x \in L$, define $\rm{ad}_{\it{x}} : \it{L \rightarrow L}$ to be a linear map by $\rm{ad}_{\it{x}}\it{(y)} = [\alpha(x), y]$ for any $y \in L$. Then $\rm{ad}_{\it{x}}$ is a derivation if $\alpha$ satisfies
\begin{equation}\label{eq:2.1}
\alpha([x, y]) = [\alpha(x), y] = [x, \alpha(y)],\quad\forall x, y \in L,\tag{2.1}
\end{equation}
which is to say that $\alpha$ belongs to $\alpha^{0}$-centroid. In this sense, $\rm{ad}_{\it{x}}$ is called an inner derivation.
\end{re}
\begin{ex}
Suppose that $L$ is a $4$-dimensional vector over $\mathbb{F}$ with a basis $\{e_{1}, e_{2}, e_{3}, e_{4}\}$. The multiplication is
\[[e_{2}, e_{3}] = [e_{2}, e_{4}] = e_{2},\; [e_{3}, e_{4}]= e_{3},\; \rm{others\; are\; all\; zero}.\]
One can see that $(L, [\cdot, \cdot])$ isn't a Lie algebra since
\[[[e_{2}, e_{3}], e_{4}] + [[e_{3}, e_{4}], e_{2}] + [[e_{4}, e_{2}], e_{3}] = -e_{2} \neq 0.\]
Define $\alpha : L \rightarrow L$ to be a linear map by
\[\alpha(e_{1}) = ae_{1}, \alpha(e_{2}) = \alpha(e_{3}) = 0, \alpha(e_{4}) = be_{1},\]
where $a, b \in \mathbb{F}$. Then $(L, [\cdot, \cdot], \alpha)$ is a non-trivial Hom-Lie algebra satisfying (\ref{eq:2.1}).
\end{ex}
\begin{defn}
For a Hom-subalgebra $H$ of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$, define
\[\rm{Ann}\it{_{L}(H) = \{x \in L | [x, \alpha(y)] = \rm{0},\;\forall \it{y} \in H\}}\]
to be the $\alpha$-annihilator of $H$ in $L$. If $H$ is a Hom-ideal of $L$ and $\alpha$ satisfies (\ref{eq:2.1}), then so is $\rm{Ann}\it{_{L}(H)}$. In particular, if $H = L$, write $\rm{Ann}\it{(L)} = \rm{Ann}\it{_{L}(L)}$.
\end{defn}
\begin{defn}
Let $(L, [\cdot, \cdot], \alpha)$ be a Hom-Lie algebra. An element $x \in L$ is an absolute zero divisor if $(\rm{ad}_{\it{x}})^{2} = \rm{0}$. The Hom-Lie algebra $L$ is said to be nondegenerate if it doesn't contain any nonzero absolute zero divisor.
\end{defn}
\begin{defn}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-Lie algebra. Then $L$ is said to be semiprime if for any nonzero Hom-ideal $I$ of $L$, $[I, \alpha(I)] \neq \{0\}$ and prime if for any two nonzero Hom-ideals $I$, $J$ of $L$, $[I, \alpha(J)] \neq \{0\}$.
\end{defn}

In the following, we suppose that all Hom-Lie algebras satisfy $(\ref{eq:2.1})$.
\begin{re}\label{re:2.9}
For a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$, we have

$L$ nondegenerate $\Rightarrow$ $L$ semiprime $\Rightarrow$ $\rm{Ann}\it{(L)} = \{\rm{0}\}$.
\end{re}
\begin{proof}
(1). Suppose that $L$ is nondegenerate. If $L$ isn't semiprime, there exists a nonzero Hom-ideal $I$ of $L$ satisfying $[I, \alpha(I)] = \{0\}$. Take $0 \neq x \in I$, then we have $(\rm{ad}_{\it{x}})^{2}(\it{L}) = [\alpha(x), [\alpha(x), L]] \subseteq [\alpha(I), I] = \{\rm{0}\}$, which implies that $x$ is a nonzero absolute zero divisor. Contradict with $L$ is nondegenerate.

(2). Suppose that $L$ is semiprime. If $\rm{Ann}\it{(L)} \neq \{\rm{0}\}$, then $\rm{Ann}\it{(L)}$ is a nonzero Hom-ideal of $L$ satisfying $[\rm{Ann}\it{(L)}, \alpha(\rm{Ann}\it{(L)})]\subseteq [\rm{Ann}\it{(L)}, \alpha(L)] = \{\rm{0}\}$. Contradiction.
\end{proof}

A Hom-ideal $I$ of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ is said to be essential if every nonzero Hom-ideal of $L$ hits $I$($I \cap J \neq \{0\}$ for every nonzero Hom-ideal $J$ of $L$).
\begin{prop}\label{prop:2.10}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-Lie algebra and $I$ a Hom-ideal of $L$.

(1) If $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$, then $I$ is essential.

(2) If $L$ is semiprime, then $I \cap \rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$ and $I$ is essential if and only if $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$.

\end{prop}
\begin{proof}
(1). Otherwise, there exists a nonzero Hom-ideal $J$ of $L$ such that $I \cap J = \{0\}$. Take $0 \neq v_{0} \in J$, we have $[v_{0}, \alpha(I)] \subseteq I \cap J = \{0\}$. So $v_{0} \in \rm{Ann}\it{_{L}(I)}$, which is a contradiction. Hence $I$ is essential.

(2). If not, $[I \cap \rm{Ann}\it{_{L}(I)}, \alpha(I \cap \rm{Ann}\it{_{L}(I)})] \subseteq [\rm{Ann}\it{_{L}(I)}, \alpha(I)] = \{\rm{0}\}$. Note that $I \cap \rm{Ann}\it{_{L}(I)}$ is a nonzero Hom-ideal of $L$, this contradicts with $L$ being semiprime.

Suppose that $I$ is essential. If $\rm{Ann}\it{_{L}(I)} \neq \{\rm{0}\}$, then $I \cap \rm{Ann}\it{_{L}(I)} \neq \{\rm{0}\}$. Contradiction. Hence, $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$. Combining with (1), we get that $I$ is essential if and only if $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$.
\end{proof}
\begin{prop}\label{prop:2.11}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-Lie algebra. Then $L$ is prime if and only if for every nonzero Hom-ideal $I$ of $L$, $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$.
\end{prop}
\begin{proof}
Suppose that $L$ is prime. If there exists a nonzero Hom-ideal $J$ of $L$ such that $\rm{Ann}\it{_{L}(J)} \neq \{\rm{0}\}$. Then $[\rm{Ann}\it{_{L}(J)}, \alpha(J)] = \{\rm{0}\}$ according to the definition of $\rm{Ann}\it{_{L}(J)}$, which contradicts with $L$ being prime.

On the other hand, suppose that for every nonzero Hom-ideal $I$ of $L$, $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$. If $L$ isn't prime, there exist two nonzero Hom-ideals $I$ and $J$ of $L$ such that $[I, \alpha(J)] = \{0\}$, which implies that $I \subseteq \rm{Ann}\it{_{L}(J)}$. Contradiction.
\end{proof}
\section{Algebras of quotients of Hom-Lie algebras}\label{se:3}
Inspired by the notion of algebras of quotients of Lie algebras in \cite{S1}, we introduce the notion of algebras of quotients of Hom-Lie algebras.

Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of Hom-Lie algebra $(Q, [\cdot, \cdot], \alpha)$. For every $q \in Q$, set
\[_{L}(q) = \mathbb{F}q + \left\{\sum_{i = 1}^{n}[[[[q, x_{1}^{i}], x_{2}^{i}], \cdots], x_{k_{i}}^{i}] \mid x_{j}^{i} \in L\; \rm{with}\; \it{n}, k_{i} \in \mathbb{N}\right\},\]

That is, $_{L}(q)$ is the linear span in $Q$ of the elements of the form $[[[[q, x_{1}], x_{2}], \cdots], x_{n}]$ and $q$, where $n \in \mathbb{N}$ and $x_{1}, \cdots, x_{n} \in L$. It is clear that $[_{L}(q), L] \subseteq _{L}(q)$. Now we define
\[(L : q) = \{x \in L \mid [x, \alpha(_{L}(q))] \subseteq L\}.\]
Obviously, if $q \in L$, then $(L : q) = L$.
\begin{prop}\label{prop:3.1}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Take $q \in Q$. Then $(L : q)$ is a Hom-ideal of $L$. Moreover, it's maximal among the Hom-ideals $I$ of $L$ such that $[I, \alpha(q)] \subseteq L$.
\end{prop}
\begin{proof}
For any $x \in (L : q)$, $[\alpha(x), \alpha(_{L}(q))] = \alpha([x, \alpha(_{L}(q))]) \subseteq \alpha(L) \subseteq L$, which implies that $\alpha(x) \in (L : q)$, i.e., $\alpha((L : q)) \subseteq (L : q)$.

For any $x \in (L : q)$, $y \in L$, $[[x, y], \alpha(_{L}(q))] = [\alpha(x), [y, _{L}(q)]] + [[x, _{L}(q)], \alpha(y)] = [x, \alpha([y, _{L}(q)])] + [[x, \alpha(_{L}(q))], y] \subseteq [x, \alpha(_{L}(q))] + [L, y] \subseteq L$, which implies that $[x, y] \in (L : q)$, i.e., $[(L : q), L] \subseteq (L : q)$. Hence, $(L : q)$ is a Hom-ideal of $L$.

Suppose that $I$ is a Hom-ideal of $L$ such that $[I, \alpha(q)] \subseteq L$. We'll show that for any $x_{1}, \cdots, x_{n} \in L$ and $n \in \mathbb{N}$, $[I, \alpha([[[[q, x_{1}], x_{2}], \cdots], x_{n}])] \subseteq L$. We'll prove it by induction on $n$.

When $n = 1$, for any $x_{1} \in L$, $[I, \alpha([q, x_{1}])] = [\alpha(I), [q, x_{1}]] = [\alpha(q), [I, x_{1}]] + [\alpha(x_{1}), [q, I]] = [\alpha(q), [I, x_{1}]] + [x_{1}, [\alpha(q), I]] \subseteq [\alpha(q), I] + [x_{1}, L] \subseteq L$.

Suppose that when $n = k$, $[I, \alpha([[[[q, x_{1}], x_{2}], \cdots], x_{k}])] \subseteq L$ for any $x_{1}, \cdots, x_{k} \in L$.

When $n = k + 1$, for any $x_{1}, \cdots, x_{k}, x_{k + 1} \in L$, we denote $[[[[q, x_{1}], x_{2}], \cdots], x_{k}]$ by $z$. By hypothesis, we have $[I, \alpha(z)] \subseteq L$.

Therefore, $[I, \alpha([[[[[q, x_{1}], x_{2}], \cdots], x_{k}], x_{k + 1}])] = [I, \alpha([z, x_{k + 1}])] = [\alpha(I), [z, x_{k + 1}]] = [\alpha(z), [I, x_{k + 1}]] + [\alpha(x_{k + 1}), [z, I]] = [\alpha(z), [I, x_{k + 1}]] + [x_{k + 1}, [\alpha(z), I]] \subseteq [\alpha(z), I] + [x_{k + 1}, L] \subseteq L$.

Hence, we have $[I, \alpha(_{L}(q))] \subseteq L$, i.e., $I \subseteq (L : q)$.
\end{proof}
\begin{defn}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Then $Q$ is called an algebra of quotients of $L$, if given $p, q \in Q$ with $p \neq 0$, there exists $x \in L$ such that $[x, \alpha(p)] \neq 0$ and $x \in (L : q)$.
\end{defn}
\begin{prop}\label{prop:3.3}
Let $(L, [\cdot, \cdot], \alpha)$ be a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$.
\begin{enumerate}[(1)]
\item If $\rm{Ann}\it{(L)} = \{\rm{0}\}$, then $L$ is an algebra of quotients of itself.
\item If $Q$ is an algebra of quotients of $L$, then $\rm{Ann}\it{_{Q}(L)} = \rm{Ann}\it{(L)} = \{\rm{0}\}$.
\end{enumerate}
\end{prop}
\begin{proof}
(1). For any $p, q \in L$ with $p \neq 0$, we have $p \notin \rm{Ann}\it{(L)}$. There exists $x \in L$ such that $[p, \alpha(x)] \neq 0$, so $[x, \alpha(p)] \neq 0$. It's obvious that $x \in (L : q)$. Therefore $L$ is an algebra of quotients of itself.

(2). For any $0 \neq q \in Q$, there exists $x \in L$ such that $[x, \alpha(q)] \neq 0$, so $[q, \alpha(x)] \neq 0$. Hence $q \notin \rm{Ann}\it{_{Q}(L)}$. Therefore $\rm{Ann}\it{_{Q}(L)} = \{\rm{0}\}$. Similarly, we have $\rm{Ann}\it{(L)} = \{\rm{0}\}$.
\end{proof}

The above proposition says for a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$, that $\rm{Ann}\it{(L)} = \{\rm{0}\}$ is a sufficient and necessary condition such that $L$ has algebras of quotients.

We will prove that some properties of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ can be inherited by its algebras of quotients $Q$. Actually, $Q$ just needs a weaker condition.
\begin{defn}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Then $Q$ is called a weak algebra of quotients of $L$, if given $0 \neq q \in Q$, there exists $x \in L$ such that $0 \neq [x, \alpha(q)] \in L$.
\end{defn}
\begin{re}
Every algebra of quotients of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ is a weak algebra of quotients.
\end{re}
\begin{prop}\label{prop:3.6}
Let $(Q, [\cdot, \cdot], \alpha)$ be a weak algebra of quotients of the Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$.
\begin{enumerate}[(1)]
\item If $I$ is a nonzero Hom-ideal of $Q$, then $I \cap L$ is a nonzero Hom-ideal of $L$.
\item If $L$ is semiprime(prime), so is $Q$.
\end{enumerate}
\end{prop}
\begin{proof}
(1). Since $I$ is a Hom-ideal of $Q$, $\alpha(I) \subseteq I$. Hence $\alpha(I \cap L) \subseteq \alpha(I) \cap \alpha(L) \subseteq I \cap L$.

Note that $I \neq \{0\}$, we can take $0 \neq x_{0} \in I$. There exists $x \in L$ such that $0 \neq [x, \alpha(x_{0})] \in L$. $[x, \alpha(x_{0})] \in I$ since $I$ is a Hom-ideal of $Q$. So we have $0 \neq [x, \alpha(x_{0})] \in I \cap L$.

Moreover, $[I \cap L, L] \subseteq [I, Q] \subseteq I$ and $[I \cap L, L] \subseteq [L, L] \subseteq L$. Hence, $[I \cap L, L] \subseteq I \cap L$. Therefore, $I \cap L$ is a nonzero Hom-ideal of $L$.

(2). Suppose that $L$ is semiprime. If $Q$ isn't semiprime, there exists a nonzero Hom-ideal $I$ of $Q$ such that $[I, \alpha(I)] = \{0\}$. By (1), $I \cap L$ is a nonzero Hom-ideal of $L$ and $[I \cap L, \alpha(I \cap L)] \subseteq [I, \alpha(I)] = \{0\}$. Contradict with $L$ being semiprime.

Similarly, we have $Q$ is prime if $L$ is prime.
\end{proof}
\begin{defn}
Let $(L, [\cdot, \cdot], \alpha)$ be a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Then $Q$ is called ideally absorbed into $L$ if, for every $0 \neq q \in Q$, there exists a Hom-ideal $I$ of $L$ with $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$ such that $\{0\} \neq [I, \alpha(q)] \in L$.
\end{defn}
\begin{prop}\label{prop:3.8}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Take $q \in Q$.

(1) If $Q$ is an algebra of quotients of $L$, then $(L : q)$ is an essential Hom-ideal of $L$. Moreover, $\rm{Ann}\it{_{L}((L : q))} = \{\rm{0}\}$.

(2) If $Q$ is ideally absorbed into $L$, then $(L : q)$ is an essential Hom-ideal of $L$. Moreover, $\rm{Ann}\it{_{L}((L : q))} = \{\rm{0}\}$.

\end{prop}
\begin{proof}
(1). Let $I$ be a nonzero Hom-ideal of $L$. Take $0 \neq y \in I$. Apply that $Q$ is an algebra of quotients of $L$ to find $x \in L$ such that $0 \neq [x, \alpha(y)]$ and $x \in (L : q)$. Since $(L : q)$ is a Hom-ideal of $L$, $[x, \alpha(y)] \in I \cap (L : q)$. This implies that $I \cap (L : q) \neq \{0\}$. Therefore $(L : q)$ is an essential Hom-ideal of $L$.

Suppose that $\rm{Ann}\it{_{L}((L : q))} \neq \{\rm{0}\}$. Then $\rm{Ann}\it{_{L}((L : q))} \cap (L : q) \neq \{\rm{0}\}$ since $(L : q)$ is essential. Take $0 \neq u \in \rm{Ann}\it{_{L}((L : q))} \cap (L : q)$. Apply that $Q$ is an algebra of quotients of $L$ to find $x \in L$ such that $0 \neq [x, \alpha(u)]$ and $x \in (L : q)$. However, $[x, \alpha(u)] = -[u, \alpha(x)] = 0$ since $u \in \rm{Ann}\it{_{L}((L : q))}$. Contradiction.

(2). Since $Q$ is ideally absorbed into $L$, there exists a nonzero Hom-ideal $I$ of $L$ with $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$ such that $\{0\} \neq [I, \alpha(q)] \subseteq L$. According to the proof of Proposition \ref{prop:3.1}, $[I, \alpha(_{L}(q))] \subseteq L$. So $I \subseteq (L : q)$. Hence, $\rm{Ann}\it{_{L}((L : q))} \subseteq \rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$. According to Proposition \ref{prop:2.10} (1), $(L : q)$ is essential.
\end{proof}
\begin{lem}\label{le:3.9}
Let $(Q, [\cdot, \cdot], \alpha)$ be a weak algebra of quotients of Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$. Let $I$ be a Hom-ideal of $L$ with $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$. Then there is no nonzero element $x$ in $Q$ such that $[\alpha(x), I] = \{0\}$.
\end{lem}
\begin{proof}
Suppose that there exists $0 \neq x \in Q$ such that $[\alpha(x), I] = \{0\}$. Apply that $Q$ is a weak algebra of quotients of $L$ to find $y \in L$ such that $0 \neq [y, \alpha(x)] \in L$. Since $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$, $[[y, \alpha(x)], \alpha(I)] \neq \{0\}$. However,
\begin{align*}
&[[y, \alpha(x)], \alpha(I)] = \alpha([[y, x], \alpha(I)]) = \alpha([\alpha(x), [I, y]] + [\alpha(y), [x, I]])\\
&= \alpha([\alpha(x), [I, y]] + [y, [\alpha(x), I]]) = \{0\}.
\end{align*}
This is a contradiction.
\end{proof}
\begin{thm}\label{thm:3.10}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Then $Q$ is an algebra of quotients of $L$ if and only if $Q$ is ideally absorbed into $L$.
\end{thm}
\begin{proof}
Suppose that $Q$ is an algebra of quotients of $L$. For any $0 \neq q \in Q$, $(L : q)$ is a Hom-ideal of $L$ with $\rm{Ann}\it{_{L}((L : q))} = \{\rm{0}\}$ according to Proposition \ref{prop:3.8} (1). By Lemma \ref{le:3.9}, $[\alpha(q), (L : q)] \neq \{0\}$. According to the definition of $(L : q)$, $[(L : q), \alpha(q)] \subseteq L$. Hence, $Q$ is ideally absorbed into $L$.

Now suppose that $Q$ is ideally absorbed into $L$. Then $Q$ is a weak algebra of quotients of $L$. For any $p, q \in Q$ with $p \neq 0$, we have $\rm{Ann}\it{_{L}((L : q))} = \{\rm{0}\}$. Hence, $[\alpha(p), (L : q)] = [p, \alpha((L : q))] \neq \{0\}$. Therefore, there exists $x \in (L : q)$ such that $[\alpha(p), x] \neq 0$. Hence, $Q$ is an algebra of quotients of $L$.
\end{proof}

In the proof of Theorem \ref{thm:3.10}, we can conclude that $Q$ is an algebra of quotients of $L$ if and only if $Q$ is a weak algebra of quotients of $L$ satisfying $\rm{Ann}\it{_{L}((L : q))} = \{\rm{0}\}$ for every $q \in Q$.
\begin{prop}\label{prop:3.11}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $Q$ a weak algebra of quotients of $L$. Then for every Hom-ideal $I$ of $L$, $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$ implies $\rm{Ann}\it{_{Q}(I)} = \{\rm{0}\}$.
\end{prop}
\begin{proof}
Suppose that $0 \neq q \in \rm{Ann}\it{_{Q}(I)}$. By hypothesis, there exists an element $x$ in $L$ such that $0 \neq [x, \alpha(q)] \in L$. Since $\rm{Ann}\it{_{L}(I)} = \{\rm{0}\}$, there exists $y \in I$ such that $[[x, \alpha(q)], \alpha(y)] \neq 0$. However,
\begin{align*}
&[[x, \alpha(q)], \alpha(y)] = \alpha([[x, q], \alpha(y)]) = \alpha([[y, q], \alpha(x)] + [[x, y], \alpha(q)])\\
&= \alpha([[\alpha(y), q], x] + [\alpha([x, y]), q]) \subseteq \alpha([[\alpha(y), q], x] + [\alpha(I), q]) = 0.
\end{align*}
This is a contradiction.
\end{proof}
\begin{no}
Denote by $\mathscr{J}_{e}(L)$ the set of all essential Hom-ideals of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$.
\end{no}

Clearly, given $I, J \in \mathscr{J}_{e}(L)$, we have $I \cap J \in \mathscr{J}_{e}(L)$. If $L$ is semiprime, then $I^{2} \in \mathscr{J}_{e}(L)$, where $I^{2} = [I, \alpha(I)]$. If $L$ is prime, then $[I, \alpha (J)] \in \mathscr{J}_{e}(L)$ for any $I, J \in \mathscr{J}_{e}(L)$.
\begin{prop}\label{prop3.13}
Let $(L, [\cdot, \cdot], \alpha)$ be a semiprime Hom-Lie algebra and $(Q, [\cdot, \cdot], \alpha)$ an algebra of quotients of $L$. Then for every essential Hom-ideal $I$ of $L$, we have that $Q$ is an algebra of quotients of $I$.
\end{prop}
\begin{proof}
For any $p, q \in Q$ with $p \neq 0$, we have $(L : q) \in \mathscr{J}_{e}(L)$. So that $(L : q) \cap I \in \mathscr{J}_{e}(L)$ and consequently, $((L : q) \cap I)^{2} \in \mathscr{J}_{e}(L)$. Then we have $\rm{Ann}\it{_{Q}(((L : q) \cap I)^{2})} = \{\rm{0}\}$. So there exist $y, z \in (L : q) \cap I$ such that $[\alpha([y, \alpha(z)]), p] \neq 0$, i.e., $[[y, \alpha(z)], \alpha(p)] \neq 0$. Moreover,
\begin{align*}
&[[y,\alpha(z)], \alpha(_{I}(q))] = \alpha([[_{I}(q), z], \alpha(y)] + [[y, _{I}(q)], \alpha(z)]) \subseteq \alpha([[_{L}(q), z], \alpha(y)] + [[y, _{L}(q)], \alpha(z)])\\
&= \alpha([[\alpha(_{L}(q)), z], y] + [[y, \alpha(_{L}(q))], z]) \subseteq \alpha([L, I]) \subseteq \alpha(I) \subseteq I,
\end{align*}
which implies that $[y, \alpha(z)] \in (I : q)$. Therefore, $Q$ is an algebra of quotients of $I$.
\end{proof}
\section{The maximal algebra of quotients of a semiprime Hom-Lie algebra}\label{se:4}
In this section, we will construct a maximal algebra of quotients for every semiprime Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ with $\alpha(L)$ is an essential ideal. Maximal in the sense that every algebra of quotients of $L$ can be considered inside this maximal algebra of quotients via an injective map which restricted in $L$ is the identity.
\begin{defn}
Given a Hom-ideal $I$ of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$, we say that a linear map $\delta : I \rightarrow L$ is an $\alpha^{k}$-partial derivation for $k \in \mathbb{N}$ if for any $x, y \in I$ it is satisfied:
\begin{enumerate}[(1)]
\item $\delta \circ \alpha = \alpha \circ \delta$;
\item $\delta([x, y]) = [\delta(x), \alpha^{k}(y)] + [\alpha^{k}(x), \delta(y)]$.
\end{enumerate}
\end{defn}
Denote by $\rm{PDer}_{\it{k}}\it{(I, L)}$ the set of all $\alpha^{k}$-partial derivations of $I$ in $L$.
\begin{lem}\label{le:4.2}
Let $(L, [\cdot, \cdot], \alpha)$ be a semiprime Hom-Lie algebra and consider the set
\[\mathscr{D}_{k} := \{(\delta, I)\;|\;I \in \mathscr{J}_{e}(L), \delta \in \rm{PDer}_{\it{k}}\it{(I, L)}\}.\]
Define on $\mathscr{D}_{k}$ the following relation:

$(\delta, I) \equiv (\mu, J)$ if and only if there exists $K \in \mathscr{J}_{e}(L)$, $K \subseteq I \cap J$ such that
\[\delta|_{K} = \mu|_{K}.\]
Then $\equiv$ is an equivalence relation.
\end{lem}
\begin{proof}
(1). For any $(\delta, I) \in \mathscr{D}_{k}$, $\delta|_{I} = \delta|_{I}$, so $(\delta, I) \equiv (\delta, I)$.

(2). For any $(\delta, I), (\mu, J) \in \mathscr{D}_{k}$ and $(\delta, I) \equiv (\mu, J)$, there exists $K \in \mathscr{J}_{e}(L)$, $K \subseteq I \cap J$ such that $\delta|_{K} = \mu|_{K}$. It's obvious that $\mu|_{K} = \delta|_{K}$. Hence $(\mu, J) \equiv (\delta, I)$.

(3). For any $(\delta, I), (\xi, J), (\beta, K) \in \mathscr{D}_{k}$ and $(\delta, I) \equiv (\xi, J)$ and $(\xi, J) \equiv (\beta, K)$, there exist $P, Q \in \mathscr{J}_{e}(L)$, $P \subseteq I \cap J$, $Q \subseteq J \cap K$ such that $\delta|_{P} = \xi|_{P}$ and $\xi|_{Q} = \beta|_{Q}$. Since $P, Q \in \mathscr{J}_{e}(L)$, $P \cap Q \neq \emptyset$. Take $T \subseteq P \cap Q$. We also get $T \in \mathscr{J}_{e}(L)$. Moreover, $\delta|_{T} = \xi|_{T} = \beta|_{T}$. Therefore $(\delta, I) \equiv (\beta, K)$.

The proof is completed.
\end{proof}
\begin{no}\label{no:4.3}
Denote by $Q_{k}(L)$ the quotient set $\mathscr{D}_{k}/\equiv$. Set $Q(L) = Q_{0}(L)$. Let $\delta_{I}$ denote the equivalence class of $(\delta, I)$ in $Q(L)$.
\end{no}
\begin{thm}\label{thm:4.4}
Let $(L, [\cdot, \cdot], \alpha)$ be a semiprime Hom-Lie algebra over $\mathbb{F}$, and let $Q := Q(L)$ be as in Notation \ref{no:4.3}. Define the following maps:
\[\cdot : \mathbb{F} \times Q \rightarrow Q,\;(p, \delta_{I}) \mapsto (p\delta)_{I}\;\rm{where}\; \it{p}\delta : I \rightarrow L,\; y \mapsto \delta(py),\]
\[+ : Q \times Q \rightarrow Q,\;(\delta_{I}, \mu_{J}) \mapsto (\delta + \mu)_{I \cap J}\;\rm{where}\;\delta + \mu : \it{I} \cap J \rightarrow L,\; x \mapsto \delta(x) + \mu(x),\]
\[[\cdot, \cdot] : Q \times Q \rightarrow Q,\; (\delta_{I}, \mu_{J}) \mapsto [\delta, \mu]_{[I \cap J, I \cap J]}\;\rm{where}\;[\delta, \mu] : \it{[I \cap J, I \cap J]} \rightarrow L,\; x \mapsto \delta\mu(x) - \mu\delta(x),\]
\[\tilde{\alpha} : Q \rightarrow Q,\; \delta_{I} \mapsto (\alpha \circ \delta)_{I}\;\rm{where}\; \alpha \circ \delta : \it{I} \rightarrow L,\; x \mapsto \alpha\delta(x).\]
Then $Q$, with these operations, is a Hom-Lie algebra satisfying
$$\tilde{\alpha}([\delta_{I}, \mu_{J}]) = [\tilde{\alpha}(\delta_{I}), \mu_{J}] = [\delta_{I}, \tilde{\alpha}(\mu_{J})]$$
for any $\delta_{I}, \mu_{J} \in Q$ and containing $L$ as a subalgebra, via the injection
\[\varphi : L \rightarrow Q,\quad x \mapsto (\rm{ad}_{\it{x}})_{\it{L}}.\]
\end{thm}
\begin{proof}
(1). For any $\delta_{I}, \mu_{J} \in Q$ and $x, y \in I \cap J$,
\begin{align*}
&[\delta, \mu]([x, y]) = \delta\mu([x, y]) - \mu\delta([x, y]) = \delta([\mu(x), y] + [x, \mu(y)]) - \mu([\delta(x), y] + [x, \delta(y)])\\
&= [\delta\mu(x), y] + [\mu(x), \delta(y)] + [\delta(x), \mu(y)] + [x, \delta\mu(y)] - [\mu\delta(x), y] - [\delta(x), \mu(y)]\\
&- [\mu(x), \delta(y)] - [x, \mu\delta(y)]\\
&= [[\delta, \mu](x), y] + [x, [\delta, \mu](y)],
\end{align*}
which implies that $[\delta, \mu] \in \rm{PDer}_{0}\it{([I \cap J, I \cap J], L)}$.

For any $\delta_{I} \in Q$ and $x, y \in I$,
\begin{align*}
&\alpha \circ \delta([x, y]) = \alpha([\delta(x), y] + [x, \delta(y)]) = [\alpha \circ \delta(x), y] + [x, \alpha \circ \delta(y)],
\end{align*}
which implies that $\alpha \circ \delta \in \rm{PDer}_{0}\it{(I, L)}$.

Therefore, $[\cdot, \cdot]$, $\tilde{\alpha}$ are well-defined.

(2). It's obvious that $(Q, \cdot, +)$ is a vector space over $\mathbb{F}$ and $[\cdot, \cdot]$ is skew-symmetric. Next we'll show that $Q$ satisfies the Jacobi identity. For any $\delta_{I}, \mu_{J}, \beta_{K} \in Q$, we have
\[[[\delta_{I}, \mu_{J}], \tilde{\alpha}(\beta_{K})] + [[\mu_{J}, \beta_{K}], \tilde{\alpha}(\delta_{I})] + [[\beta_{K}, \delta_{I}], \tilde{\alpha}(\mu_{J})] = 0\]
since $\alpha$ commutes with $\delta, \mu, \beta$. Hence, $Q$ is a Hom-Lie algebra.

(3). For any $\delta_{I}, \mu_{J} \in Q$,
\begin{align*}
&\tilde{\alpha}([\delta_{I}, \mu_{J}]) = \tilde{\alpha}([\delta, \mu]_{[I \cap J, I \cap J]}) = (\alpha \circ [\delta, \mu])_{[I \cap J, I \cap J]} = (\alpha\delta\mu - \alpha\mu\delta)_{[I \cap J, I \cap J]}\\
&= (\alpha\delta\mu - \mu\alpha\delta)_{[I \cap J, I \cap J]} = [\alpha\delta, \mu]_{[I \cap J, I \cap J]} = [(\alpha\delta)_{I}, \mu_{J}] = [\tilde{\alpha}(\delta_{I}), \mu_{J}].
\end{align*}
Similarly, we get $\tilde{\alpha}([\delta_{I}, \mu_{J}]) = [\delta_{I}, \tilde{\alpha}(\mu_{J})]$. Hence, we have $\tilde{\alpha}([\delta_{I}, \mu_{J}]) = [\tilde{\alpha}(\delta_{I}), \mu_{J}] = [\delta_{I}, \tilde{\alpha}(\mu_{J})]$ for any $\delta_{I}, \mu_{J} \in Q$.

(4). According to Remark \ref{re:2.4}, we get for any $x \in L$, $\rm{ad}_{\it{x}} \in \rm{PDer}_{0}\it{(L, L)}$. So $\varphi$ is well-defined. If $\varphi(x) = 0$, we have $(\rm{ad}_{\it{x}})_{\it{L}} = \rm{0}$. Then we have $[x, \alpha(L)] = [\alpha(x), L]= \{0\}$, which implies that $x \in \rm{Ann}\it{(L)}$. Note that $L$ is semiprime, we have $\rm{Ann}\it{(L)} = \{\rm{0}\}$ by Proposition \ref{prop:2.10} (2). Hence $x = 0$, i,e, $\varphi$ is injective.
\end{proof}

For any $X \subseteq L$, write $X^{\varphi}$ to denote the image of $X$ inside $Q(L)$ via the map defined in Theorem \ref{thm:4.4}.
\begin{lem}\label{le:4.5}
For every $\delta_{I} \in Q(L)$, and $(\rm{ad}_{\it{x}})_{\it{L}} \in \it{I}^{\varphi}(x \in I)$ we have $[\delta_{I}, (\rm{ad}_{\it{x}})_{\it{L}}] = (\rm{ad}_{\it{\delta(x)}})_{\it{L}} \in \it{L}^{\varphi}$.
\end{lem}
\begin{proof}
For any $y \in I$,
\begin{align*}
&[\delta, \rm{ad}_{\it{x}}](\it{y}) = \delta\rm{ad}_{\it{x}}(\it{y}) - \rm{ad}_{\it{x}}\delta(\it{y}) = \delta([\alpha(x), y]) - [\alpha(x), \delta(y)]\\
&= [\alpha(x), \delta(y)] + [\delta\alpha(x), y] - [\alpha(x), \delta(y)] = [\delta\alpha(x), y] = [\alpha\delta(x), y] = \rm{ad}_{\it{\delta(x)}}(\it{y}),
\end{align*}
and so $[\delta, \rm{ad}_{\it{x}}]_{\it{I}} = (\rm{ad}_{\it{\delta(x)}})_{\it{I}} = (\rm{ad}_{\it{\delta(x)}})_{\it{L}} \in \it{L}^{\varphi}$.
\end{proof}
\begin{prop}\label{prop:4.6}
Let $(L, [\cdot, \cdot], \alpha)$ be a semiprime Hom-Lie algebra with $\alpha(L)$ is essential. Then $Q(L)$ is semiprime and an algebra of quotients of $L$. Moreover, $Q(L)$ is maximal among the algebras of quotients of $L$, in the sense that if $S$ is an algebra of quotients of $L$, then there exists an injection $\psi : S \rightarrow Q(L)$ which is the identity in $L$. In particular, the map
\[\psi : S \rightarrow Q(L),\quad s \mapsto (\rm{ad}_{\it{s}})_{\it{(L : s)}}\]
is an injective map which is the identity when restricted to $L$.
\end{prop}
\begin{proof}
Take $\delta_{I}, \mu_{I} \in Q(L)$ with $\delta_{I} \neq 0$(we can consider the same $I$ for $\delta$ and $\mu$ because if $J, K \in \mathscr{J}_{e}(L)$ are such that $\delta : J \rightarrow L$, $\mu : K \rightarrow L$, then $I := J \cap K \in \mathscr{J}_{e}(L)$ and so $\delta_{J} = \delta_{I}$ and $\mu_{K} = \mu_{I}$). Choose $a \in I$ such that $\delta(a) \neq 0$. Then $(\rm{ad}_{\it{a}})_{\it{L}} \in \it{L}^{\varphi}$ satisfies $[\tilde{\alpha}(\delta_{\it{I}}), (\rm{ad}_{\it{a}})_{\it{L}}] = (\rm{ad}_{\it{\alpha\delta(a)}})_{\it{L}} \neq \rm{0}$. Otherwise, $[\delta(a), \alpha^{2}(L)] = [\alpha^{2}\delta(a), L] = \{0\}$, which implies that $\delta(a) \in \rm{Ann}\it{_{L}(\alpha(L))} = \{\rm{0}\}$. And for every $\lambda_{I} \in _{L}^{\varphi}(\mu_{I})$, $[\tilde{\alpha}(\lambda_{I}), (\rm{ad}_{\it{a}})_{\it{L}}] = (\rm{ad}_{\it{\alpha\lambda(a)}})_{\it{L}} \in \it{L}^{\varphi}$. This implies that $(\rm{ad}_{\it{a}})_{\it{L}} \in \it{(L^{\varphi} : \mu_{I})}$. Hence, $Q(L)$ is an algebra of quotients of $L^{\varphi}$. The semiprimeness of $Q(L)$ follows by Proposition \ref{prop:3.6} (2).

Now suppose that $S$ is an algebra of quotients of $L$ and consider the map
\[\psi : S \rightarrow Q(L),\quad s \mapsto (\rm{ad}_{\it{s}})_{\it{(L : s)}}.\]
According to Proposition \ref{prop:3.8} (1), $\psi$ is well-defined. To prove the injectivity, suppose that $s \in S$ such that $\psi(s) = 0$, that is $[\alpha(s), K] = \{0\}$ for some Hom-ideal $K \in \mathscr{J}_{e}(L)$, $K \subseteq (L : s)$. This implies that $s \in \rm{Ann}\it{_{L}(K)}$. Note that $K$ is essential, $\rm{Ann}\it{_{L}(K)} = \{\rm{0}\}$, so $s = 0$.
\end{proof}
\begin{defn}
For a semiprime Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ with $\alpha(L)$ is essential, the Hom-Lie algebra constructed in Theorem \ref{thm:4.4} will be called the maximal algebra of quotients of $L$. Denote it by $Q_{m}(L)$.
\end{defn}

The axiomatic characterization of the Martindale ring of quotients given by D. Passman in \cite{P1} has inspired us to give the following description of the maximal algebra of quotients of a semiprime Hom-Lie algebra.
\begin{prop}\label{prop:4.8}
Let $(L, [\cdot, \cdot], \alpha)$ be a semiprime Hom-Lie algebra with $\alpha(L)$ is essential and consider an overalgebra $S$ of $L$. Then there exists an injection between $S$ and $Q_{m}(L)$ which is the identity on $L$, if and only if $S$ satisfies the following properties:
\begin{enumerate}[(1)]
\item For any $s \in S$, there exists $I \in \mathscr{J}_{e}(L)$ such that $[\alpha(I), s] \subseteq L$.
\item For $s \in S$ and $I \in \mathscr{J}_{e}(L)$, $[\alpha(I), s] = \{0\}$ implies that $s = 0$.
\end{enumerate}
\end{prop}
\begin{proof}
Define $\psi : S \rightarrow Q_{m}(L)$, $s \mapsto (\rm{ad}_{\it{s}})_{\it{I}}$ where $I$ is a nonzero essential Hom-ideal of $L$ satisfying $[\alpha(I), s] \subseteq L$; this exists by (1), and so the map is well-defined.

If $\psi(s) = 0$, i.e., $(\rm{ad}_{\it{s}})_{\it{I}} = \rm{0}$, then $[\alpha(s), I] = \{0\}$. By (2), we get $s = 0$. This implies that $\psi$ is an injective map.

Finally, $\psi$ is the identity on $L$, by identifying $L$ with $L^{\varphi}$, where $\varphi$ is the map defined in Theorem \ref{thm:4.4}.

Conversely, we'll prove that $Q(L)$ satisfies the two conditions.

(1). Consider $q \in Q(L)$. According to Propositions \ref{prop:4.6} and \ref{prop:3.8} (1), $(L : q) \in \mathscr{J}_{e}(L)$, and by the definition, $[\alpha((L : q)), q] = [(L : q), \alpha(q)] \subseteq L$.

(2). Take $q \in Q(L)$ and $I \in \mathscr{J}_{e}(L)$ such that $[\alpha(I), q] = \{0\}$. We have $q \in \rm{Ann}\it{_{Q(L)}(I)} = \rm{0}$ according to Propositions \ref{prop:4.6} and \ref{prop:3.11}.
\end{proof}
\section{Algebras of quotients of associative algebras generated by inner derivations of Hom-Lie algebras}\label{se:5}
In \cite{PS}, authors examined how the notion of algebras of quotients for Lie algebras tied up with the corresponding well-known concept in the associative case. Inspired by the method in \cite{PS}, we mainly study the relationship between Hom-Lie algebras and the associative algebras generated by inner derivations of the corresponding Hom-Lie algebras of quotients. First of all, we will give some definitions and basic notations.

We shall denote by $A(L)$ the associative subalgebra (possibly without identity) of $\rm{End}\it{(L)}$ generated by the elements $\rm{ad}\it{_{x}}$ for $x$ in $(L, [\cdot, \cdot], \alpha)$.

By an extension of Hom-Lie algebras $L \subseteq Q$ we will mean that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of the Hom-Lie algebra $(Q, [\cdot, \cdot], \alpha)$.

Let $L \subseteq Q$ be an extension of Hom-Lie algebras and let $A_{Q}(L)$ be the associative subalgebra of $A(Q)$ generated by $\{\rm{ad}\it{_{x}} : x \in L\}$.

Recall that, given an associative algebra $A$ and a subset $X$ of $A$, we define the right annihilator of $X$ in $A$ as
\[rann_{A}(X) = \{a \in A | Xa = 0\},\]
which is always a right ideal of $A$ (and two-sided if $X$ is a right ideal). One similarly defines the left annihilator, which shall be denoted by $lann_{A}(X)$.
\begin{lem}\label{le:5.1}
Let $I$ be a Hom-ideal of a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ with $\rm{Ann}\it{(L)} = \{\rm{0}\}$. Then $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$ if and only if $rann_{A(L)}(A_{L}(I)) = \{0\}$.
\end{lem}
\begin{proof}
Suppose that $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$. For any $\mu$ in $rann_{A(L)}(A_{L}(I))$, we have $\rm{ad}\it{_{y}}\mu = \rm{0}$ for any $y \in I$. In particular if $x \in L$, we get $0 = \rm{ad}\it{_{y}}\mu(x) = [\alpha(y), \mu(x)]$, and this implies that $\mu(x) \in \rm{Ann}_{\it{L}}(\it{I}) = \rm{0}$. Hence, $\mu = 0$.

Conversely, suppose that $rann_{A(L)}(A_{L}(I)) = \{0\}$. For any $x$ in $\rm{Ann}_{\it{L}}(\it{I})$, $y$ in $I$ and $z$ in $L$, we have
\[\rm{ad}\it{_{y}}\rm{ad}\it{_{x}}(z) = [\alpha(y), [\alpha(x), z]] = \alpha([\alpha(y), [x, z]]) = \alpha([[z, y], \alpha(x)] + [[y, x], \alpha(z)]) = \rm{0},\]
which implies that $\rm{ad}\it{_{x}} \in rann_{A(L)}(A_{L}(I)) = \rm{0}$. Since by assumption $\rm{Ann}\it{(L)} = \{\rm{0}\}$, we obtain $x = 0$.
\end{proof}
For a subset $X$ of an associative algebra $A$, denote by $\langle X \rangle_{A}^{l}$, $\langle X \rangle_{A}^{r}$ and $\langle X \rangle_{A}$ the left, right and two sided ideal of $A$, respectively, generated by $X$.
\begin{lem}\label{le:5.2}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $I$ a Hom-ideal of $(L, [\cdot, \cdot], \alpha)$. Then
\[\langle A_{Q}(I) \rangle_{A_{Q}(L)}^{l} = \langle A_{Q}(I) \rangle_{A_{Q}(L)}^{r} = \langle A_{Q}(I) \rangle_{A_{Q}(L)}.\]
\end{lem}
\begin{proof}
Obviously, we have $\langle A_{Q}(I) \rangle_{A_{Q}(L)}^{l} = A_{Q}(L)A_{Q}(I) + A_{Q}(I)$ and $\langle A_{Q}(I) \rangle_{A_{Q}(L)}^{r} = A_{Q}(I)A_{Q}(L) + A_{Q}(I)$. Notice that given $x \in L$ and $y \in I$, we have
\[\rm{ad}\it{_{x}}\rm{ad}\it{_{y}} = \rm{ad}\it{_{\alpha([x, y])}} + \rm{ad}\it{_{y}}\rm{ad}\it{_{x}}.\]
Then we have $\langle A_{Q}(I) \rangle_{A_{Q}(L)}^{l} = \langle A_{Q}(I) \rangle_{A_{Q}(L)}^{r}$. Hence we come to the conclusion.
\end{proof}
\begin{lem}\label{le:5.3}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $I$ a Hom-ideal of $(L, [\cdot, \cdot], \alpha)$. Write $\tilde{I}$ to denote the ideal of $A_{Q}(L)$ generated by $A_{Q}(I)$. Then
\begin{enumerate}[(1)]
\item $rann_{A_{Q}(L)}(\tilde{I}) = rann_{A_{Q}(L)}(A_{Q}(I))$.
\item $lann_{A_{Q}(L)}(\tilde{I}) = lann_{A_{Q}(L)}(A_{Q}(I))$.
\end{enumerate}
\end{lem}
\begin{proof}
(1) Since $A_{Q}(I) \subseteq \tilde{I}$, we have $rann_{A_{Q}(L)}(\tilde{I}) \subseteq rann_{A_{Q}(L)}(A_{Q}(I))$. So it's enough to show that $rann_{A_{Q}(L)}(A_{Q}(I)) \subseteq rann_{A_{Q}(L)}(\tilde{I})$. Let $\lambda \in rann_{A_{Q}(L)}(A_{Q}(I))$. By Lemma \ref{le:5.2} we know that, if $\mu \in \tilde{I}$ there exist a natural number $n$, elements $x_{1, i}, \cdots, x_{r_{i}, i} \in L$ and $y_{1, i}, \cdots, y_{s_{i}, i} \in I$ with $0 \leq r_{i} \in \mathbb{N}$ for all $i$ and $\emptyset \neq \{s_{1}, \cdots, s_{n}\} \subseteq \mathbb{N}$, such that
\[\mu = \sum_{i}^{n}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{r_{i}, i}}}\rm{ad}\it{_{y_{1, i}}} \cdots \rm{ad}\it{_{y_{s_{i}, i}}}.\]
Since $\rm{ad}\it{_{y_{s_{i}, i}}}\lambda = \rm{0}$, we see that $\mu\lambda = 0$.

(2) The proof is similar to (1).
\end{proof}
\begin{lem}\label{le:5.4}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ such that $Q$ is a weak algebra of quotients of $L$. Let $I$ be a Hom-ideal of $(L, [\cdot, \cdot], \alpha)$. If $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$, then $rann_{A(Q)}(A_{Q}(I)) = \{0\}$.
\end{lem}
\begin{proof}
According to Proposition \ref{prop:3.11}, we have $\rm{Ann}_{\it{Q}}(\it{I}) = \{\rm{0}\}$. For any $\mu \in rann_{A(Q)}(A_{Q}(I))$, we have $\rm{ad}\it{_{y}}\mu = \rm{0}$ for every $y \in I$. If $q \in Q$, we then have $0 = \rm{ad}\it{_{y}}\mu(q) = [\alpha(y), \mu(q)]$. This implies that $\mu(q) \in \rm{Ann}_{\it{Q}}(\it{I}) = \rm{0}$, and so $\mu = 0$.
\end{proof}
\begin{lem}\label{le:5.5}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and let $x_{1}, \cdots, x_{n}$, $y$ in $L$. Then we have, in $A(Q)$:
\[\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{y}} = \rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{n}}} + \sum_{i = 1}^{n}\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{\alpha([x_{i}, y])}}\cdots\rm{ad}\it{_{x_{n}}}.\]
In particular, if $I$ is a Hom-ideal of $L$ and $x_{1}, \cdots, x_{n} \in I$, then
\[\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{y}} = \rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{n}}} + \delta\]
where $\delta \in span\{\rm{ad}\it{_{z_{1}}}\cdots\rm{ad}\it{_{z_{n}}}\;|\;z_{i} \in I\}$.
\end{lem}
\begin{proof}
We'll prove it by induction on $n$.

When $n = 1$, we get
\[\rm{ad}\it{_{x}}\rm{ad}\it{_{y}} = \rm{ad}\it{_{\alpha([x, y])}} + \rm{ad}\it{_{y}}\rm{ad}\it{_{x}}\]
for any $x, y \in L$.

Suppose that $\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}}\rm{ad}\it{_{y}} = \rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}} + \sum_{i = 1}^{k}\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{\alpha([x_{i}, y])}}\cdots\rm{ad}\it{_{x_{k}}}$ for any $x_{1}, \cdots, x_{k}, y \in L$. Then when $n = k + 1$, for any $x_{1}, \cdots, x_{k}, x_{k + 1}, y \in L$, we get
\begin{align*}
&\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}}\rm{ad}\it{_{x_{k + 1}}}\rm{ad}\it{_{y}} = \rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}}(\rm{ad}\it{_{y}}\rm{ad}\it{_{x_{k + 1}}} + \rm{ad}\it{_{\alpha([x_{k + 1}, y])}})\\
&= (\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}}\rm{ad}\it{_{y}})\rm{ad}\it{_{x_{k + 1}}} + \rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}}\rm{ad}\it{_{\alpha([x_{k + 1}, y])}}\\
&= (\rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}} + \sum_{i = 1}^{k}\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{\alpha([x_{i}, y])}}\cdots\rm{ad}\it{_{x_{k}}})\rm{ad}\it{_{x_{k + 1}}} + \rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}}\rm{ad}\it{_{\alpha([x_{k + 1}, y])}}\\
&= \rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{k}}}\rm{ad}\it{_{x_{k + 1}}} + \sum_{i = 1}^{k + 1}\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{\alpha([x_{i}, y])}}\cdots\rm{ad}\it{_{x_{k + 1}}}.
\end{align*}
The proof is completed.
\end{proof}
Let $(L, [\cdot, \cdot], \alpha)$ be a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Denote by $A_{0}$ the associative subalgebra of $A(Q)$ whose elements are those $\mu$ in $A(Q)$ such that $\mu(L) \subseteq L$. We obviously have the containments:
\[A_{Q}(L) \subseteq A_{0} \subseteq A(Q).\]
\begin{lem}\label{le:5.6}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $I$ a Hom-ideal of $(L, [\cdot, \cdot], \alpha)$. Let $q_{1}, \cdots, q_{n}$ in $Q$ such that $[\alpha(q_{i}), I] \subseteq L$ for every $i = 1, \cdots, n$. Then for $\mu = \rm{ad}\it{_{q_{1}}} \cdots \rm{ad}\it{_{q_{n}}}$ in $A(Q)$, we have that $\mu(\tilde{I})^{n} + (\tilde{I})^{n}\mu \subseteq A_{0}$ (where $(\tilde{I})^{n}$ denotes the $n$-th power of $\tilde{I}$ in the associative algebra $A_{Q}(L)$).
\end{lem}
\begin{proof}
According to Lemma \ref{le:5.2}, we have
\[(\tilde{I})^{n} = A_{Q}(L)(A_{Q}(I))^{n} + (A_{Q}(I))^{n} = (A_{Q}(I))^{n}A_{Q}(L) + (A_{Q}(I))^{n}.\]
Thus it's enough to prove that, for any $y = \rm{ad}\it{_{x_{1}}} \cdots \rm{ad}\it{_{x_{n}}}$ where $x_{i} \in I$, both $\mu y$ and $y \mu$ belong to $A_{0}$.

We'll use induction on $n$. For $n = 1$, we have $\rm{ad}\it{_{x}}\rm{ad}\it{_{q}} = \rm{ad}\it{_{\alpha([x, q])}} + \rm{ad}\it{_{q}}\rm{ad}\it{_{x}}$ and since $\alpha([x, q]) = [x, \alpha(q)] \in L$ we see that $\rm{ad}\it{_{\alpha([x, q])}} \in A_{\rm{0}}$. On the other hand, $\rm{ad}\it{_{q}}\rm{ad}\it{_{x}}(L) = [\alpha(q), [\alpha(x), L]] \subseteq [\alpha(q), I] \subseteq L$, and so $\rm{ad}\it{_{q}}\rm{ad}\it{_{x}} \in A_{\rm{0}}$.

Assume that the result is true for $n - 1$. Now, by Lemma \ref{le:5.5} we have
\begin{align*}\label{eq:*}
&\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{1}}}\cdots\rm{ad}\it{_{q_{n}}} = (\rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{x_{1}}})\rm{ad}\it{_{x_{2}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{2}}}\cdots\rm{ad}\it{_{q_{n}}}\\
&+ \sum_{i = 1}^{n}\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{\alpha([x_{i}, q_{1}])}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{2}}}\cdots\rm{ad}\it{_{q_{n}}}.\tag{*}
\end{align*}
The first summand on the right side belongs to $A_{0}$ because, as proved before, $\rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{x_{1}}} \in A_{\rm{0}}$ and $\rm{ad}\it{_{x_{2}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{2}}}\cdots\rm{ad}\it{_{q_{n}}} \in A_{\rm{0}}$ by induction hypothesis. On the other hand, for each of the terms $\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{\alpha([x_{i}, q_{1}])}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{2}}}\cdots\rm{ad}\it{_{q_{n}}}$ we have that $x_{i} \in I$ and $\alpha([x_{i}, q_{1}]) = [x_{i}, \alpha(q_{1})] \in L$. Using Lemma \ref{le:5.5}, we may write this as:
\[\rm{ad}\it{_{\alpha([x_{i}, q_{1}])}}\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{x_{i - 1}}}\rm{ad}\it{_{x_{i + 1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{2}}}\cdots\rm{ad}\it{_{q_{n}}} + \delta \cdot \rm{ad}\it{_{x_{i + 1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{2}}}\cdots\rm{ad}\it{_{q_{n}}},\]
where $\delta \in span\{\rm{ad}\it{_{z_{1}}}\cdots\rm{ad}\it{_{z_{i - 1}}}\;|\;z_{j} \in I\}$. The induction hypothesis applies again to show that this belongs to $A_{0}$. Hence, $y\mu \in A_{0}$.

If we continue to develop in the expression (\ref{eq:*}), we get, for some $\delta_{0} \in A_{0}$,
\[\rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{q_{2}}}\rm{ad}\it{_{x_{1}}}\rm{ad}\it{_{x_{2}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{3}}}\cdots\rm{ad}\it{_{q_{n}}} + \sum_{i = 1}^{n}\rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{\alpha([x_{i}, q_{2}])}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{3}}}\cdots\rm{ad}\it{_{q_{n}}} + \delta_{\rm{0}}.\]
Using Lemma \ref{le:5.5} we can write each term of the form
\[\rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{x_{1}}} \cdots\rm{ad}\it{_{\alpha([x_{i}, q_{2}])}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{3}}}\cdots\rm{ad}\it{_{q_{n}}}\]
as:
\[\rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{\alpha([x_{i}, q_{2}])}}\rm{ad}\it{_{x_{1}}} \cdots\it{_{x_{i - 1}}}\rm{ad}\it{_{x_{i + 1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{3}}}\cdots\rm{ad}\it{_{q_{n}}} + \rm{ad}\it{_{q_{1}}}\delta \cdot \rm{ad}\it{_{x_{i + 1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{3}}}\cdots\rm{ad}\it{_{q_{n}}},\]
where $\delta \in span\{\rm{ad}\it{_{z_{1}}}\cdots\rm{ad}\it{_{z_{i - 1}}}\;|\;z_{j} \in I\}$. Notice that
\[\rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{\alpha([x_{i}, q_{2}])}}\rm{ad}\it{_{x_{1}}} = \rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{\alpha([\alpha([x_{i}, q_{2}]), x_{1}])}} + \rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{x_{1}}}\rm{ad}\it{_{\alpha([x_{i}, q_{2}])}}.\]
Hence, using $\alpha([x_{i}, q_{2}]) \in L$ and $x_{i} \in I$, we see that the first summand above belongs to $A_{0}$. For the second summand, assuming that $\delta = \rm{ad}\it{_{z_{1}}}\cdots\rm{ad}\it{_{z_{i - 1}}}$ with $z_{j} \in I$, we have $(\rm{ad}\it{_{q_{1}}}\rm{ad}\it{_{z_{1}}})\rm{ad}\it{_{z_{2}}}\cdots\rm{ad}\it{_{z_{i - 1}}}\rm{ad}\it{_{x_{i + 1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{3}}}\cdots\rm{ad}\it{_{q_{n}}}$, which is also an element of $A_{0}$. Continuing in this way, we find that
\[\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{n}}}\rm{ad}\it{_{q_{1}}}\cdots\rm{ad}\it{_{q_{n}}} - \rm{ad}\it{_{q_{1}}}\cdots\rm{ad}\it{_{q_{n}}}\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{n}}} \in A_{\rm{0}}\]
and by what we have just proved, we see that $\rm{ad}\it{_{q_{1}}}\cdots\rm{ad}\it{_{q_{n}}}\rm{ad}\it{_{x_{1}}}\cdots\rm{ad}\it{_{x_{n}}} \in A_{\rm{0}}$, as was to be shown.
\end{proof}
\begin{cor}\label{cor:5.7}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $I$ a Hom-ideal of $(L, [\cdot, \cdot], \alpha)$. Let $q_{1}, \cdots, q_{n}$ in $Q$ such that $[\alpha(q_{i}), I] \subseteq L$ for every $i = 1, \cdots, n$. Then for $\mu = \rm{ad}\it{_{q_{1}}} \cdots \rm{ad}\it{_{q_{n}}}$ in $A(Q)$, we have that $\mu\tilde{I^{n}} \in A_{0}, \tilde{I^{n}}\mu \subseteq A_{0}$ (where $I^{1} = I$ and $I^{k} = [I^{k - 1}, \alpha(I)]$ for $k \geq 2$).
\end{cor}
\begin{proof}
It's straightforward to show that $I^{n}$ is a Hom-ideal of $L$ for each $n \geq 1$.

Claim that $\widetilde{I^{n}} \subseteq (\tilde{I})^{n}$. We'll show it by induction on $n$. When $n = 1$, it's obvious. Suppose that $\widetilde{I^{k}} \subseteq (\tilde{I})^{k}$. When $n = k + 1$, for any $y \in I^{k}, z \in I$,
\[\rm{ad}\it{_{[y, \alpha(z)]}} = \rm{ad}\it{_{\alpha([y, z])}} = \rm{ad}\it{_{y}}\rm{ad}\it{_{z}} - \rm{ad}\it{_{z}}\rm{ad}\it{_{y}} \in \tilde{I^{k}}\tilde{I} \subseteq (\tilde{I})^{k}\tilde{I} = (\tilde{I})^{k + 1}.\]
According to Lemma \ref{le:5.6}, we come to the conclusion.
\end{proof}
\begin{lem}\label{le:5.8}
Let $(L, [\cdot, \cdot], \alpha)$ be a semiprime Hom-Lie algebra. If $I$ is a Hom-ideal of $(L, [\cdot, \cdot], \alpha)$ with $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$, then $\rm{Ann}_{\it{L}}(\it{I^{s}}) = \{\rm{0}\}$for any $s \geq 1$. Any finite intersection of ideals with zero $\alpha$-annihilator will also have zero $\alpha$-annihilator.
\end{lem}
\begin{proof}
We'll show it by induction on $s$. When $s = 1$, it's obvious. Suppose that $\rm{Ann}_{\it{L}}(\it{I^{k}}) = \{\rm{0}\}$, i.e., $I^{k}$ is essential. When $s = k + 1$, for any nonzero Hom-ideal $J$ of $L$, we have $I^{k} \cap J \neq \{0\}$ since $I^{k}$ is essential. So
\[\{0\} \neq [I^{k} \cap J, \alpha(I^{k} \cap J)] \subseteq [I^{k}, \alpha(I)] \cap J = I^{k + 1} \cap J,\]
which implies that $I^{k + 1}$ is essential and so $\rm{Ann}_{\it{L}}(\it{I^{k + 1}}) = \{\rm{0}\}$.

Similarly, we can show that any finite intersection of ideals with zero $\alpha$-annihilator will also have zero $\alpha$-annihilator by induction.
\end{proof}
\begin{prop}\label{prop:5.9}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a semiprime Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Then the following conditions are equivalent:

(1) $Q$ is an algebra of quotients of $L$;

(2) $\rm{Ann}(\it{Q}) = \{\rm{0}\}$ and, if $\mu \in A(Q)\setminus\{0\}$, there exists a Hom-ideal $I$ of $L$ with $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$ such that $\mu\tilde{I} \subseteq A_{0}$ and $\{0\} \neq \tilde{I}\mu \subseteq A_{0}$. If $\mu = \rm{ad}\it{_{q}}$, then we also have $\mu\tilde{I}(L) \neq \{0\}$.
\end{prop}
\begin{proof}
(2) $\Rightarrow$ (1) Let $q \in Q\setminus\{0\}$. Then $\mu = \rm{ad}\it{_{q}} \neq \rm{0}$ since $\rm{Ann}(\it{Q}) = \{\rm{0}\}$. Let $\tilde{I}$ be as in (2), so it satisfies $\mu\tilde{I}(L) \neq \{0\}$ and $\mu\tilde{I} \subseteq A_{0}$. Set
\[I_{0} := span\{\delta(x)\;|\;x \in L\; and\; \delta \in \tilde{I}\}.\]
Then $I_{0}$ is a Hom-ideal of $L$ such that $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$. Indeed, for any $\delta \in \tilde{I}$, there exist a natural number $n$, elements $x_{1, i}, \cdots, x_{r_{i}, i} \in L$ and $y_{1, i}, \cdots, y_{s_{i}, i} \in I$ with $0 \leq r_{i} \in \mathbb{N}$ for all $i$ and $\emptyset \neq \{s_{1}, \cdots, s_{n}\} \subseteq \mathbb{N}$, such that
\[\delta = \sum_{i}^{n}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{r_{i}, i}}}\rm{ad}\it{_{y_{1, i}}} \cdots \rm{ad}\it{_{y_{s_{i}, i}}}.\]
For any $x, y \in L$, if $r_{i} = 0$ and $s_{i} = 1$, we get
\begin{align*}
&[y, \rm{ad}\it{_{y_{1, i}}}(x)] = [y, [\alpha(y_{1, i}), x]] = [\alpha(y_{1, i}), [y, x]] + [\alpha([y, y_{1, i}]), x]\\
&= \rm{ad}\it{_{y_{1, i}}}([y, x]) + \rm{ad}\it{_{[y, y_{1, i}]}}(x) \in I_{\rm{0}},
\end{align*}
if $r_{i} > 0$ and $s_{i} = 1$, we'll prove
\[[y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{r_{i}, i}}}\rm{ad}\it{_{y_{1, i}}}(x)] \in I_{\rm{0}}\]
by induction on $r_{i}$.
When $r_{i} = 1$, we get
\begin{align*}
&[y, \rm{ad}\it{_{x_{1, i}}}\rm{ad}\it{_{y_{1, i}}}(x)] = [y, \rm{ad}\it{_{y_{1, i}}}\rm{ad}\it{_{x_{1, i}}}(x) + \rm{ad}\it{_{\alpha([x_{1, i}, y_{1, i}])}}(x)]\\
&= [y, \rm{ad}\it{_{y_{1, i}}}\rm{ad}\it{_{x_{1, i}}}(x)] + [y, \rm{ad}\it{_{\alpha([x_{1, i}, y_{1, i}])}}(x)]\\
&= [\alpha(y), \rm{ad}\it{_{y_{1, i}}}([x_{1, i}, x])] + [y, \rm{ad}\it{_{\alpha([x_{1, i}, y_{1, i}])}}(x)] = \rm{ad}\it{_{y}}\rm{ad}\it{_{y_{1, i}}}([x_{1, i}, x]) + [y, \rm{ad}\it{_{\alpha([x_{1, i}, y_{1, i}])}}(x)] \in I_{\rm{0}}.
\end{align*}
Suppose that when $r_{i} = k$, $[y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}\rm{ad}\it{_{y_{1, i}}}(x)] \in I_{\rm{0}}$ for any $x_{1, i}, \cdots, x_{k, i} \in L$ and $y_{1, i} \in I$. When $r_{i} = k + 1$, we get
\begin{align*}
&[y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k + 1, i}}}\rm{ad}\it{_{y_{1, i}}}(x)]\\
&= \left[y, \left(\rm{ad}\it{_{y_{1, i}}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k + 1, i}}} + \sum_{j = 1}^{k + 1}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{\alpha([x_{j, i}, y_{1, i}])}}\cdots\rm{ad}\it{_{x_{k + 1, i}}}\right)(x)\right]\\
&= [y, \rm{ad}\it{_{y_{1, i}}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k + 1, i}}}(x)] + \left[y, \sum_{j = 1}^{k + 1}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{\alpha([x_{j, i}, y_{1, i}])}}\cdots\rm{ad}\it{_{x_{k + 1, i}}}(x)\right]\\
&= [\alpha(y), \rm{ad}\it{_{y_{1, i}}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}([x_{k + 1, i}, x])] + \sum_{j = 1}^{k}[\alpha(y), \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{\alpha([x_{j, i}, y_{1, i}])}}\cdots\rm{ad}\it{_{x_{k, i}}}([x_{k + 1, i}, x])]\\
&+ [y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}\rm{ad}\it{_{\alpha([x_{k + 1, i}, y_{1, i}])}}(x)]\\
&= \rm{ad}\it{_{y}}\rm{ad}\it{_{y_{1, i}}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}([x_{k + 1, i}, x]) + \sum_{j = 1}^{k}\rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{\alpha([x_{j, i}, y_{1, i}])}}\cdots\rm{ad}\it{_{x_{k, i}}}([x_{k + 1, i}, x])\\
&+ [y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}\rm{ad}\it{_{\alpha([x_{k + 1, i}, y_{1, i}])}}(x)]\\
&= \rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}\rm{ad}\it{_{y_{1, i}}}([x_{k + 1, i}, x]) - \sum_{j = 1}^{k}\rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{\alpha([x_{j, i}, y_{1, i}])}}\cdots\rm{ad}\it{_{x_{k, i}}}([x_{k + 1, i}, x])\\
&+ \sum_{j = 1}^{k}\rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{\alpha([x_{j, i}, y_{1, i}])}}\cdots\rm{ad}\it{_{x_{k, i}}}([x_{k + 1, i}, x])  + [y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}\rm{ad}\it{_{\alpha([x_{k + 1, i}, y_{1, i}])}}(x)]\\
&= \rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}\rm{ad}\it{_{y_{1, i}}}([x_{k + 1, i}, x]) + [y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{k, i}}}\rm{ad}\it{_{\alpha([x_{k + 1, i}, y_{1, i}])}}(x)] \in I_{\rm{0}},
\end{align*}
if $s_{i} > 1$, we have
\begin{align*}
&[y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{r_{i}, i}}}\rm{ad}\it{_{y_{1, i}}} \cdots \rm{ad}\it{_{y_{s_{i}, i}}}(x)] = [y, \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{r_{i}, i}}}\rm{ad}\it{_{y_{1, i}}} \cdots \rm{ad}\it{_{y_{s_{i} - 1, i}}}([\alpha(y_{s_{i}, i}), x])]\\
&= [\alpha(y), \rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{r_{i}, i}}}\rm{ad}\it{_{y_{1, i}}} \cdots \rm{ad}\it{_{y_{s_{i} - 1, i}}}([y_{s_{i}, i}, x])]\\
&= \rm{ad}\it{_{y}}\rm{ad}\it{_{x_{1, i}}} \cdots \rm{ad}\it{_{x_{r_{i}, i}}}\rm{ad}\it{_{y_{1, i}}} \cdots \rm{ad}\it{_{y_{s_{i} - 1, i}}}([y_{s_{i}, i}, x]) \in I_{\rm{0}}.
\end{align*}
Therefore we get $[y, \delta(x)] \in I_{0}$, i.e., $[L, I_{0}] \subseteq I_{0}$. It's straightforward to show that $\alpha(I_{0}) \subseteq I_{0}$.

If now $[x, \alpha(I_{0})] = \{0\}$ for $x \in L$, then $\rm{ad}\it{_{x}}\tilde{I}(L) = \{\rm{0}\}$. In particular, for $y, z \in I$, we have $\rm{ad}\it{_{x}}\rm{ad}\it{_{y}}(z) = \rm{0}$ and so $x \in \rm{Ann}_{\it{L}}(\it{I^{2}}) = \{\rm{0}\}$, which is zero by Lemma \ref{le:5.8}.

Finally, $\{0\} \neq [\alpha(q), I_{0}] = \rm{ad}\it{_{q}}\tilde{I}(L) \subseteq A_{\rm{0}}(L) \subseteq L$, which implies that $Q$ is ideally absorbed into $L$. According to Theorem \ref{thm:3.10}, $Q$ is an algebra of quotients of $L$.

(1) $\Rightarrow$ (2) Since $Q$ is an algebra of quotients of $L$, we have $\rm{Ann}(\it{Q}) = \{\rm{0}\}$. Next let $\mu = \sum_{i \geq 1}\rm{ad}\it{_{q_{i, 1}}} \cdots \rm{ad}\it{_{q_{i, r_{i}}}} \in A(Q)\setminus\{\rm{0}\}$. We may of course assume that all $q_{i, j}$ are nonzero elements in $Q$. Set $s = \sum_{i \geq 1}r_{i}$. As $Q$ is an algebra of quotients of $L$, there exists, for every $i$ and $j$, a Hom-ideal $J_{i, j}$ of $L$ such that $\rm{Ann}_{\it{L}}(\it{J_{i, j}}) = \{\rm{0}\}$, and $\{0\} \neq [\alpha(q_{i, j}), J_{i, j}] \subseteq L$. By Lemma \ref{le:5.8}, the Hom-ideal $J = \cap_{i, j}J_{i, j}$ and hence also $I = J^{s}$ will have zero $\alpha$-annihilator in $L$. Moreover, $[\alpha(q_{i, j}), I] \subseteq [\alpha(q_{i, j}), J_{i, j}] \subseteq L$. According to Corollary \ref{cor:5.7} and taking into account that $J^{s} \subseteq J^{r_{i}}$ for every $i$, we have $\mu\tilde{I} \subseteq A_{0}$ and $\tilde{I}\mu \subseteq A_{0}$.

If $\tilde{I}\mu = \{0\}$, then $\mu \in rann_{A(Q)}(A_{Q}(I))$ which is zero by Lemma \ref{le:5.4}.

Finally, suppose that $\mu = \rm{ad}\it{_{q}}$ where $q \in A(Q)\setminus\{0\}$. Note that $Q$ is also an algebra of quotients of $I^{2}$, this implies that there exist elements $y, z \in I$ such that $0 \neq [\alpha(q), [y, \alpha(z)]]$. But $[\alpha(q), [y, \alpha(z)]] = \rm{ad}\it{_{q}}\rm{ad}\it{_{y}}(z) \in \rm{ad}\it{_{q}}\tilde{I}(L)$.
\end{proof}
Recall that an associative algebra $S$ is said to be a left quotient algebra of a subalgebra $A$ if whenever $p$ and $q \in S$, with $p \neq 0$, there exists $x$ in $A$ such that $xp \neq 0$ and $xq \in A$. An associative algebra $A$ has a left quotient algebra if and only if it has no total right zero divisors different from zero. (Here, an element $x$ in $A$ is said to be a total right zero divisor if $Ax = \{0\}$.)
\begin{lem}\label{le:5.10}\cite{S1}
Let $A$ be a subalgebra of an associative algebra $Q$. Then $Q$ is a left quotient algebra of $L$ if and only if for every nonzero element $q \in Q$ there exists a left ideal $I$ of $A$ with $rann_{A}(I) = \{0\}$ such that $\{0\} \neq Iq \subseteq A$.
\end{lem}
\begin{thm}\label{thm:5.11}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a semiprime Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. Moreover, suppose that $Q$ is an algebra of quotients of $L$. Then $A(Q)$ is a left quotient algebra of $A_{0}$.
\end{thm}
\begin{proof}
Let $\mu \in A(Q)\setminus\{0\}$ and let $I$ be a Hom-ideal of $L$ satisfying condition (2) in Proposition \ref{prop:5.9}. Set $J = A_{0}\tilde{I} + \tilde{I}$, a left ideal of $A_{0}$ that satisfies $\{0\} \neq J\mu \subseteq A_{0}$ (because $\{0\} \neq \tilde{I}\mu \subseteq A_{0}$).

Since also $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$, we obtain from Lemma \ref{le:5.4} that $rann_{A(Q)}(A_{Q}(I)) = \{0\}$. This, together with the fact that $\tilde{I}$ contains $A_{Q}(I)$, implies that $rann_{A(Q)}(\tilde{I}) = \{0\}$. Since $rann_{A_{0}}(\tilde{I}) \subseteq rann_{A(Q)}(\tilde{I})$ we get that also $rann_{A_{0}}(\tilde{I}) = \{0\}$. Hence we get $rann_{A_{0}}(J) = \{0\}$ since $\tilde{I} \subseteq J$. This concludes the proof according to Lemma \ref{le:5.10}.
\end{proof}
\begin{re}\label{re:5.12}
As established in the proof of the previous result, if $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and if $(L, [\cdot, \cdot], \alpha)$ is semiprime and $I$ is a Hom-ideal of $(L, [\cdot, \cdot], \alpha)$ with $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$, then $A_{0}\tilde{I} + \tilde{I}$ is a left ideal of $A_{0}$ with zero right annihilator.
\end{re}
\section{Algebras of quotients of Hom-Lie algebras with dense extensions}\label{se:6}
In this section, we mainly study algebras of quotients of Hom-Lie algebras via their dense extensions which is introduced by Cabrera in \cite{C1}. We show that dense extension can be lifted from a Hom-Lie algebra to its Hom-ideals if the extension is also an algebra of quotients. For any Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$, let $M(L)$ be the associative algebra generated by the identity map together with inner derivations of $L$. Moreover, $M(L)$ is nothing but the unitization of $A(L)$.

Following \cite{C1}, given an extension of Hom-Lie algebras $L \subseteq Q$, the annihilator of $L$ in $M(Q)$ is defined by
\[L^{ann} := \{\mu \in M(Q)\; |\; \mu(x) = 0,\; x \in L\}.\]
If $L^{ann} = \{0\}$, then $L$ is said to be a dense Hom-subalgebra of $Q$, and we will say that $L \subseteq Q$ is a dense extension of Hom-Lie algebras.
\begin{lem}\label{le:6.1}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ with $\rm{Ann}(\it{Q}) = \{\rm{0}\}$. Then the following conditions are equivalent:
\begin{enumerate}[(1)]
\item $L$ is a dense Hom-subalgebra of $Q$.
\item If $\mu(L) = \{0\}$ for some $\mu$ in $A(Q)$, then $\mu = 0$.
\end{enumerate}
\end{lem}
\begin{proof}
Clearly, (1) implies (2). Conversely, suppose that $\mu \in M(Q)$ satisfies $\mu(L) = \{0\}$. If $\mu(p) \neq 0$ for some $p$ in $Q$, then use $\rm{Ann}(\it{Q}) = \{\rm{0}\}$ to find a nonzero elements $q$ in $Q$ satisfying $\rm{ad}\it{_{q}}\mu(p) \neq \rm{0}$. But then $\rm{ad}\it{_{q}}\mu(L) = \{\rm{0}\}$ and since $A(Q)$ is a two-sided ideal of $M(Q)$, we have that $\rm{ad}\it{_{q}}\mu \in A(Q)$. Hence, condition (2) yields $\rm{ad}\it{_{q}}\mu = \rm{0}$, a contradiction.
\end{proof}
Following Cabrera and Mohammed in \cite{CM}, we say that a Hom-Lie algebra $(L, [\cdot, \cdot], \alpha)$ is multiplicatively semiprime whenever $L$ and its multiplication algebra $M(L)$ are semiprime. Observe that in this situation, and if $L$ is a Hom-Lie algebra, then $A(L)$, being an ideal of $M(L)$, will also be a semiprime algebra.
\begin{lem}\label{le:6.2}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a dense Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $I$ a Hom-ideal of $L$. If $\mu$ is an element of $M(Q)$ such that $\mu(I) = \{0\}$, then $\mu(M(Q)(I)) = \{0\}$.
\end{lem}
\begin{proof}
Consider the set
\[\mathcal{H} = \{\varphi \in M(Q)\;|\;\mu\omega\varphi(I) = \{0\}\;\rm{for\;each}\;\omega \in \it{M(Q)}\;\rm{such\;that}\;\mu\omega(\it{I}) = \{\rm{0}\}\}.\]
It's clear that $\mathcal{H}$ is a subalgebra of $M(Q)$. Indeed, for any $\varphi, \psi \in \mathcal{H}$ and each $\omega \in M(Q)$ such that $\mu\omega(I) = \{0\}$, we get that $\mu\omega\varphi\psi(I) = \{0\}$ since $\omega\varphi \in M(Q)$ satisfying $\mu\omega\varphi(I) = \{0\}$ and $\psi \in \mathcal{H}$, which implies that $\varphi\psi \in \mathcal{H}$.

Furthermore, for $\omega \in M(Q)$ such that $\mu\omega(I) = \{0\}$, $x \in I$ and $y \in L$ we see that
\[0 = \mu\omega([\alpha(x), y]) = \mu\omega\rm{ad}\it{_{x}}(y).\]
Therefore $\mu\omega\rm{ad}\it{_{x}}$ lies in $L^{ann}$, and so it is equal to zero. Consequently for each $z \in Q$, it can be seen that
\[0 = \mu\omega\rm{ad}\it{_{x}}(z) = \mu\omega([\alpha(x), z]) = -\mu\omega([\alpha(z), x]) = -\mu\omega\rm{ad}\it{_{z}}(x).\]
From this it follows that $\mu\omega\rm{ad}\it{_{z}}(I) = \{\rm{0}\}$, and hence $\rm{ad}\it{_{z}}$ belongs to $\mathcal{H}$. Thus $\mathcal{H} = M(Q)$. Finally the statement follows directly from the definition of $\mathcal{H}$ and from the fact that, by assumption, $\mu\rm{id}_{\it{Q}}(\it{I}) = \{\rm{0}\}$.
\end{proof}
\begin{cor}\label{cor:6.3}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a dense Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $I, J$ two Hom-ideals of $L$. If $[\alpha(I), J] = \{0\}$, then $[M(Q)(I), \alpha(M(Q)(J))] = \{0\}$.
\end{cor}
\begin{proof}
First, apply Lemma \ref{le:6.2} to the Hom-ideal $J$ taking $\mu = \rm{ad}\it{_{x}}$ for any $x \in I$ to obtain the result $[\alpha(I), M(Q)(J)] = \{0\}$. Now, to conclude the proof, apply once more Lemma \ref{le:6.2} to the Hom-ideal $I$ taking $\mu = \rm{ad}\it{_{z}}$ where $z \in M(Q)(J)$.
\end{proof}
\begin{cor}\label{cor:6.4}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a dense Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. If $(Q, [\cdot, \cdot], \alpha)$ is semiprime, then $(L, [\cdot, \cdot], \alpha)$ is also semiprime.
\end{cor}
\begin{proof}
Otherwise, there exists a nonzero Hom-ideal $I$ such that $[I, \alpha(I)] = \{0\}$. By Corollary \ref{cor:6.3}, $[M(Q)(I), \alpha(M(Q)(I))] = \{0\}$ where $M(Q)(I)$ is a Hom-ideal of $Q$, contradiction.
\end{proof}
\begin{lem}\label{le:6.5}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a dense Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$. If $Q$ is multiplicatively semiprime, then $L$ is also multiplicatively semiprime.
\end{lem}
\begin{proof}
It's sufficient to show that $M(L)$ is semiprime by Corollary \ref{cor:6.4}. Suppose that $\mu$ in $M(L)$ satisfies $\mu M(L)\mu = 0$. Then $\mu(I) = \{0\}$ where $I$ denotes the Hom-ideal of $L$ generated by $\mu(L)$. Now, choose an element of $M(Q)$, say $\hat{\mu}$, satisfying $\hat{\mu}|_{L} = \mu|_{L}$. Hence $\hat{\mu}(I) = \{0\}$. By Lemma \ref{le:6.2}, $\hat{\mu}(M(Q)(I)) = \{0\}$, and so $\hat{\mu}M(Q)\hat{\mu}(L) = \{0\}$. Consequently we have $\hat{\mu}M(Q)\hat{\mu} = 0$ since $L \subseteq Q$ is dense. Note that $M(Q)$ is semiprime, we get $\hat{\mu} = 0$, and so $\mu = 0$.
\end{proof}
\begin{re}
Given a dense extension of Hom-Lie algebras $L \subseteq Q$, where $Q$ is multiplicatively semiprime, we have that $L$ is also multiplicatively semiprime from Lemma \ref{le:6.5}.
\end{re}
\begin{prop}\label{prop:6.7}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a dense Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $Q$ a multiplicative semiprime algebra of quotients of $L$. Then $I \subseteq Q$ is a dense extension for every essential Hom-ideal of $I$ of $L$.
\end{prop}
\begin{proof}
We first observe that $\rm{Ann}(\it{Q}) = \{\rm{0}\}$ since $Q$ is an algebra of quotients of $L$ and hence Lemma \ref{le:6.1} applies. Thus, let $\mu$ be in $A(Q)$ such that $\mu(I) = \{0\}$, and by way of contradiction assume that $\mu \neq 0$. According to Theorem \ref{thm:5.11}, $A(Q)$ is a left quotients algebra of $A_{0}$ and hence there exists $\lambda$ in $A_{0}$ such that $0 \neq \lambda\mu \in A_{0}$. Since the extension $L \subseteq Q$ is dense, $\lambda\mu(L) \neq \{0\}$, and since $\rm{Ann}_{\it{L}}(\it{I}) = \{\rm{0}\}$, there exists a nonzero element $y$ in $I$ such that $\rm{ad}\it{_{y}}\lambda\mu(L) \neq \{\rm{0}\}$. Using now that $A(Q)$ has no total right zero divisor according to Lemma \ref{le:5.1}, we get that $A(Q)\rm{ad}\it{_{y}}\lambda\mu \neq \rm{0}$, and this, coupled with the semiprimeness of $A(Q)$, implying that $A(Q)\rm{ad}\it{_{y}}\lambda\mu A(Q)\rm{ad}\it{_{y}}\lambda\mu \neq \rm{0}$. A second application of the fact that $L \subseteq Q$ is a dense extension yields $A(Q)\rm{ad}\it{_{y}}\lambda\mu A(Q)\rm{ad}\it{_{y}}\lambda\mu(L) \neq \{\rm{0}\}$. However, $\mu(I) = \{0\}$ by assumption and thus implies that $\mu M(Q)(I) = \{0\}$. But this is a contradiction, because of the containments
\[\mu A(Q)\rm{ad}\it{_{y}}\lambda\mu(L) \subseteq \mu A(Q)([\alpha(I), L]) \subseteq \mu A(Q)(I) \subseteq \mu M (Q)(I) = \{\rm{0}\}.\]
This completes the proof.
\end{proof}
\begin{cor}\label{cor:6.8}
Suppose that $(L, [\cdot, \cdot], \alpha)$ is a dense Hom-subalgebra of $(Q, [\cdot, \cdot], \alpha)$ and $Q$ a multiplicative semiprime algebra of quotients of $L$. Then for every essential Hom-ideal $I$ of $L$, $lann_{A(Q)}(\tilde{I}) = \{0\}$.
\end{cor}
\begin{proof}
Let $\mu \in lann_{A(Q)}(\tilde{I})$. Then, if $y \in I$, we have $\mu\rm{ad}\it{_{y}}(I) = \{\rm{0}\}$. This implies that $\mu(I^{2}) = \{0\}$. By Proposition \ref{prop:6.7} applied to the essential Hom-ideal $I^{2}$ of $L$, the extension $I^{2} \subseteq Q$ is dense, so $\mu = 0$.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Q: Migrating ANT sqlplus execution from MacOSX to Windows, getting TNS error I have a working ANT build.xml that executes sqlplus to compile PL/SQL.  I develop on MacOSX.  My coworker is going to start contributing to the same application.  He develops on WinXP.  The sqlplus exec from ANT returns a TNS ORA-12560 error on WinXP.  The identical build.xml works on MacOSX.  The run string that ANT should be constructing on WinXP works in the DOS shell.  We have set up the TNS_ADMIN environment variable in WinXP.  What could I be missing?

A: I'm not sure from your description whether you're using the ANT SQL Task (http://ant.apache.org/manual/Tasks/sql.html) or if your using the exec task to call Sqlplus.  Can you please clarify?
If you're calling sqlplus as an external command, you will be at the mercy of two main things:
a) environment variables
b) tnsnames.ora file
Windows and unix are different in how they handle specifying where you want to connect.  In OSX you'll have the ORACLE_HOME and ORACLE_SID environment variables set.  In Windows this would be in the registry.  You may also have TNS_ADMIN set which determines where your tnsnames.ora file will be found.
First thing to check would be if you are both using the same tnsnames.ora file.
If you are, have both of you try to run tnsping to verify that the alias is configured properly.
Get back to us with the results of those tests.

redpajama_set_name: RedPajamaStackExchange


text:
Press release – Vatican convenes ecological leadership summit – CIDSE
WP_Post Object ( [ID] => 4449 [post_author] => 26 [post_date] => 2018-07-04 08:05:49 [post_date_gmt] => 2018-07-04 08:05:49 [post_content] => Political leaders and grassroots activists from around the world unite to signal urgent need for increased ambition on climate change and sustainability. On July 5 and 6, the Vatican will host a major international conference on protection of the environment. The occasion of the conference is the third anniversary of Laudato Si', Pope Francis' landmark encyclical on care of the Earth, our common home. The conference will discuss the moral imperative for protecting the environment and events this fall that may deliver progress in achieving it. Participants will develop orientations on how to protect our planet including at forthcoming international events, such as the UN climate change conference COP24, and non-state events, such as the Global Climate Action Summit. Contributions by regions with heavily Catholic populations, such as Latin America, Australia, Poland, the Philippines, and the U.S. states of Florida, Ohio, and Pennsylvania will be crucial to the success of these events. This conference will provide a venue for Church leaders, political leaders, and grassroots organisers to more clearly formulate how to achieve that progress. Outcomes are expected to include a statement from Pope Francis and statements of support for increased global ambition on emissions, clean energy, and climate justice from unusually diverse perspectives. A broad range of allies – including politicians, indigenous leaders, youth activists, and economists – are uniting under the Catholic Church's vision of action to protect the Earth. From the Church itself, cardinals and other representatives from communities around the world, including educators, leaders of spiritual movements, and men and women religious will be present. Together, these allies share hope of justice for all who share our common home. The conference programme, including the full list of speakers, is available here. Speakers include:
His Holiness Pope Francis
Leaders of international negotiations (Dr. Patricia Espinosa, Exec.. Sec., UNFCCC; Dr. Michał Kurtyka, Pres., COP24)
Politicians with a conservative perspective (Hon. John Gummer, Lord Deben, Parliament of UK; Minister Jadwiga Emilewicz, Polish Ministry of Entrepreneurship and Technology)
Economic thinkers (Prof. Nicholas Stern, London School of Economics; Mr. Pierre Larrouturou, Pacte Finance Climat pour l'Europe)
Youth organisers from the continents of Africa, South America, Europe, Oceania, and Asia
Representatives of grassroots movements (Mr. Bill McKibben, 350.org; Mr. Naderev Saño, Greenpeace Southeast Asia)
Catholic authorities from around the world (Cardinal Pietro Parolin, Vatican Sec. State; Cardinal Peter Turkson, Prefect of the Dicastery for Promoting Integral Human Development; Cardinal Pedro Barreto, Archbishop of Huancayo, Perú; Cardinal Charles Bo, Archbishop of Yangon, Myanmar; Fr. Robert Athickal, India; Fr. Rigobert Minani, Democratic Republic of the Congo)
Recordings and presentations will be available here. STATEMENTS A statement about the conference from the Vatican is available here. Michel Roy, Secretary General of Caritas Internationalis, said "As Pope Francis says in his encyclical, the climate crisis is not only an environmental issue, but also a human and ethical one. Environmental degradation is accompanied by poverty, loss of livelihoods and the perspective of a better future. We have a responsibility towards the next generations. It is urgent to change our over consumption and commit to the care of our common home." Josianne Gauthier, Secretary General of CIDSE, said "CIDSE is honored to be a partner of this conference on such an important and timely issue. For three years, Laudato Si' has been an incredible source of strength and inspiration in our work. What we want to see now is real political commitment to climate justice and an active response to Pope Francis' call to respect our common home. CIDSE is working with people around the world, to fight for a safe climate in which they can live and thrive, and where global warming is limited to 1.5 °C. Guided by their courage, we are asking world leaders to make serious commitments this year. Another, and a better world is possible, but only if we start putting people and the planet first!" Tomás Insua, Executive Director of Global Catholic Climate Movement, said "Around the globe, people look to the Church for leadership. The Vatican, and actually Catholics from all around the world, are saying clearly, climate change is a faith issue. If we work together, we can protect the most vulnerable from climate change and provide a just transition to clean energy for hard-working miners and oil workers. There is not a moment to lose. The climate crisis is with us, and we must take up the solutions within our grasp–right now." Additional statements from partners attending the conference: "Laudato Si' was so important for us because we are Christians at our core, for instance where I am from in Samoa, our motto is 'Samoa is founded on God'. Having Laudato Si' come out was a validation for us that our Church has recognised our struggle, that Pope Francis and the Vatican know what's happening to us, and they know that it's wrong and it's not God's will for us to suffer. It's really through us that churches can be part of the mobilisation of this climate movement, and that there is a lot of power that comes from faith". Brianna Fruean, Pacific climate warrior. "To commit to Laudato Si' on the American continent means, as a priority, to take a clear position on extractivism. In their recent pastoral letter, CELAM bishops declared that extractivism is a tendency without control of the economic system, aiming at making profit from nature." Fr. Dario Bossi, Church & Mining network Media contact: CIDSE, Valentina Pavarotti, pavarotti@cidse.org CIDSE Press Release Vatican conference on LSI 3rd anniversary [post_title] => Press release – Vatican convenes ecological leadership summit [post_excerpt] => [post_status] => publish [comment_status] => closed [ping_status] => open [post_password] => [post_name] => press-release-vatican-convenes-ecological-leadership-summit [to_ping] => [pinged] => [post_modified] => 2019-09-04 13:55:49 [post_modified_gmt] => 2019-09-04 13:55:49 [post_content_filtered] => [post_parent] => 0 [guid] => https://zvitez.com/cidse/press-release-vatican-convenes-ecological-leadership-summit/ [menu_order] => 184 [post_type] => post [post_mime_type] => [comment_count] => 0 [filter] => raw )
Climate Negotiation
Press release – Vatican convenes ecological leadership summit
Political leaders and grassroots activists from around the world unite to signal urgent need for increased ambition on climate change and sustainability. On July 5 and 6, the Vatican will host a major international conference on protection of the environment. The occasion of the conference is the third anniversary of Laudato Si', Pope Francis' landmark encyclical on care of the Earth, our common home.
The conference will discuss the moral imperative for protecting the environment and events this fall that may deliver progress in achieving it. Participants will develop orientations on how to protect our planet including at forthcoming international events, such as the UN climate change conference COP24, and non-state events, such as the Global Climate Action Summit. Contributions by regions with heavily Catholic populations, such as Latin America, Australia, Poland, the Philippines, and the U.S. states of Florida, Ohio, and Pennsylvania will be crucial to the success of these events.
This conference will provide a venue for Church leaders, political leaders, and grassroots organisers to more clearly formulate how to achieve that progress. Outcomes are expected to include a statement from Pope Francis and statements of support for increased global ambition on emissions, clean energy, and climate justice from unusually diverse perspectives.
A broad range of allies – including politicians, indigenous leaders, youth activists, and economists – are uniting under the Catholic Church's vision of action to protect the Earth. From the Church itself, cardinals and other representatives from communities around the world, including educators, leaders of spiritual movements, and men and women religious will be present. Together, these allies share hope of justice for all who share our common home.
The conference programme, including the full list of speakers, is available here. Speakers include:
Economic thinkers (Prof. Nicholas Stern, London School of Economics; Mr. Pierre Larrouturou, Pacte Finance Climat pour l'Europe)
Recordings and presentations will be available here.
A statement about the conference from the Vatican is available here.
Michel Roy, Secretary General of Caritas Internationalis, said "As Pope Francis says in his encyclical, the climate crisis is not only an environmental issue, but also a human and ethical one. Environmental degradation is accompanied by poverty, loss of livelihoods and the perspective of a better future. We have a responsibility towards the next generations. It is urgent to change our over consumption and commit to the care of our common home."
Josianne Gauthier, Secretary General of CIDSE, said "CIDSE is honored to be a partner of this conference on such an important and timely issue. For three years, Laudato Si' has been an incredible source of strength and inspiration in our work. What we want to see now is real political commitment to climate justice and an active response to Pope Francis' call to respect our common home. CIDSE is working with people around the world, to fight for a safe climate in which they can live and thrive, and where global warming is limited to 1.5 °C. Guided by their courage, we are asking world leaders to make serious commitments this year. Another, and a better world is possible, but only if we start putting people and the planet first!"
Tomás Insua, Executive Director of Global Catholic Climate Movement, said "Around the globe, people look to the Church for leadership. The Vatican, and actually Catholics from all around the world, are saying clearly, climate change is a faith issue. If we work together, we can protect the most vulnerable from climate change and provide a just transition to clean energy for hard-working miners and oil workers. There is not a moment to lose. The climate crisis is with us, and we must take up the solutions within our grasp–right now."
Additional statements from partners attending the conference:
"Laudato Si' was so important for us because we are Christians at our core, for instance where I am from in Samoa, our motto is 'Samoa is founded on God'. Having Laudato Si' come out was a validation for us that our Church has recognised our struggle, that Pope Francis and the Vatican know what's happening to us, and they know that it's wrong and it's not God's will for us to suffer. It's really through us that churches can be part of the mobilisation of this climate movement, and that there is a lot of power that comes from faith". Brianna Fruean, Pacific climate warrior.
"To commit to Laudato Si' on the American continent means, as a priority, to take a clear position on extractivism. In their recent pastoral letter, CELAM bishops declared that extractivism is a tendency without control of the economic system, aiming at making profit from nature." Fr. Dario Bossi, Church & Mining network
Media contact: CIDSE, Valentina Pavarotti, pavarotti@cidse.org
CIDSE Press Release Vatican conference on LSI 3rd anniversary
Pope Francis: in Paris every effort should be made to stop climate change December 6, 2015
Sharing energy to change December 5, 2019
Sarah Fayolle: garder la foi en la justice climatique December 7, 2012


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