text:
\section{Introduction}

Our main motivation for this work is the
celebrated Skolem-Noether theorem. We will state its version as given, for example, in \cite{Her}.
But first, a word on conventions. All our algebras are assumed to be unital algebras over a  fixed field $F$, subalgebras  are assumed to contain the same unity, and all homomorphisms send $1$ to $1$.

\begin{theorem} {\bf (Skolem-Noether)} Let $A$ be simple artinian algebra with center $F$. If $R$ is a finite-dimensional simple $F$-subalgebra of $A$ and $\varphi$ is an $F$-algebra homomorphism from $R$ into $A$, then there exists an invertible element $c\in A$ such that $\varphi(x)=cxc^{-1}$ for all $x\in R$. (In other words, $\varphi$ can be extended to an inner automorphism of $A$.)
\end{theorem}

Recall that
an algebra is said to be {\bf central} if its center consists of
scalar multiples of unity.
As usual, we will use the term {\bf central simple algebra} for an algebra that is central, simple, and also finite-dimensional.

\begin{definition}\label{def}
An algebra $S$ is a {\bf Skolem-Noether algebra} ({\bf SN algebra} for short) if for every central simple algebra $R$ and every homomorphism $\varphi:R\to R\otimes S$ there exists an invertible  element $c\in R\otimes S$ such that $\varphi(x)= cxc^{-1}$  for every $x\in R$. (Here, $R$ is identified with $R\otimes 1$).
\end{definition}

The  Skolem-Noether theorem, together
with the well-known fact that the class  of central simple algebras is closed under tensor products, implies that every central simple algebra $S$ is an  SN algebra.  A partial converse is also true: the assertion that  central simple algebras are SN algebras implies
an important special case of the Skolem-Noether theorem where $A$ is a central simple algebra and $R$ is its central simple subalgebra. This is because, under these assumptions,
$A$  is isomorphic to $R\otimes S$ where $S$ is also a central simple subalgebra of $A$ \cite[Corollary 4.49]{INCA}.

SN algebras naturally arise from the problem of understanding automorphism groups of tensor products of algebras.
Unlike the case of derivations on tensor products \cite{Bre}, the general solution to this problem seems far out of reach.
For instance, while automorphisms of univariate and bivariate polynomial algebras
are well understood \cite{Jun}, already the trivariate case is wild \cite{SU}.
In another direction, functional analysts consider the question when the flip automorphism $A\otimes A\to A\otimes A$ is (approximately) inner for operator algebras $A$, see \cite{S,ER,Izu}. In this paper we settle the following special case of the above problem. If $S$ is an SN algebra and $R$ is a central simple algebra, then automorphisms of $R\otimes S$ are just compositions of inner automorphisms and automorphisms of $S$; see Proposition \ref{prop:aut}. While
the class of SN algebras looks restrictive, our main results show that various classical and important families of algebras satisfy the SN property, for example semilocal (in particular artinian and finite-dimensional) algebras, unique factorization domains, free algebras, etc.

Some of the readers might be interested only in the case where $R=M_n(F)$,  the algebra of $n\times n$ matrices with entries in $F$.  Let us therefore mention that since $M_n(F)\otimes S$ can be identified with $M_n(S)$, the condition that $S$ is an SN algebra implies that every homomorphism from $M_n(F)$ into $M_n(S)$ can be extended to an inner automorphism of $M_n(S)$. Moreover, we show in Proposition \ref{p:mat} that the latter condition implies the SN property.
However, this does not lead to any simplifications of our proofs, so  we persist with  central simple algebras as in Definition \ref{def}.

\subsection*{Main results and guide to the paper}
The short Section \ref{s:prelim} on preliminaries
includes Proposition \ref{p:mat}:  $S$ is an SN algebra if and only if all homomorphisms $M_n(F)\to M_n(S)$ extend to inner automorphisms. Section \ref{s:auto}
positions SN algebras  into a wider context of automorphisms of tensor products. For instance, Proposition \ref{prop:aut} proves that given an
SN algebra $S$ and a central simple algebra
$R$, every automorphism of $R\otimes S$ is the composition of an inner automorphism and an automorphism of $S$. In particular, this applies to matrix algebras over SN algebras.

We then identify classes of algebras which satisfy the SN property. In Section \ref{s:basic} we derive Lemma \ref{l}, which is the main technical tool for proving subsequent results. Section \ref{s:semiloc} culminates in Theorem \ref{thm:semiLoc} showing that semilocal algebras are SN. Hence all artinian algebras and thus all finite-dimensional algebras are SN. Section \ref{s:findim}
refines the latter result. Namely,
every homomorphism from a central simple subalgebra $R$ of a finite-dimensional algebra $A$
into $A$ extends to an  inner
automorphism of $A$ (see Theorem \ref{tfd}). In Section \ref{s:dom} we give examples of
domains which are SN algebras, such as
unique factorization domains (UFDs) and free algebras, see Corollary \ref{ufd}
and Corollary \ref{cor:free}.
Section~\ref{s:matrixpoly} uses the Quillen-Suslin theorem to prove that matrix algebras over polynomial algebras are SN.
The paper concludes with Section \ref{s:poly},
where we show that an algebra $S$ is SN if and only if the formal power series algebra $S[[\xi]]$ is SN.

\section{Preliminaries}\label{s:prelim}

The purpose of this   section is to introduce the notation and terminology, and
prove a proposition that yields a characterization of SN algebras.

Let $R$ be a central simple algebra. Given  $w, z \in R$, we define
the left and right multiplication operators
 $L_w, R_z:R\to R$
by
 $$L_w(x)=wx\quad\mbox{and}\quad R_z(x)=xz.$$
As is well-known, every linear map from $R$ into $R$ can be written as a sum of maps of the form $L_w R_z$, $w,z\in R$  \cite[Lemma 1.25]{INCA}. Accordingly, given a basis $\{r_1,\dots,r_d\}$ of $R$, there exists $w_j,z_j\in R$ such that
$h=\sum_j L_{w_j}R_{z_j}$ satisfies
$h(r_1)=1$ and $h(r_k)=0$, $k\ne 1$. That is,
\begin{equation}\label{oh}
\sum_j w_jr_1z_j = 1\quad \mbox{and}\quad \sum_j w_jr_kz_j = 0\,\,\,\mbox{if $k> 1$.}
\end{equation}

We will be mostly concerned with  tensor product algebras $R\otimes S$.
Here $R,S$ are algebras over a field $F$ and the tensor product is taken over $F$.
As usual,
 we identify  $R$ by $R\otimes 1$, and, accordingly, often write $r\otimes 1\in R\otimes 1$ simply as $r$.
 Let us  point out an elementary fact that will be used freely  without further reference. If the $r_i$'s are linearly independent elements
in $R$, then for all  $p_j\in R$ and  $s_j, t_i\in S$,
\begin{equation}\label{nov}
\sum_i r_i\otimes t_i = \sum_j  p_j\otimes s_j
\end{equation}
implies that each $t_i$ lies in the linear span of the $s_j$'s \cite[Lemma 4.9]{INCA}. Similarly,  assuming  that the $t_i$'s are linearly independent, it follows from \eqref{nov}  that each $r_i$ lies in the linear span of the $p_j$'s.

By rad$(S)$ we denote the {\bf Jacobson radical} of the algebra $S$. Recall that $S$ is called a {\bf semilocal algebra}
if $S/{\rm rad}(S)$ is a semisimple algebra, i.e., isomorphic to a finite direct product of simple artinian algebras.
In the special case where  $S/{\rm rad}(S)$ is a division algebra, $S$ is called a {\bf local algebra}. Finally, we say that  $S$ is a {\bf stably finite algebra} if for all $n\ge 1$ and all $x,y\in M_n(S)$, $xy=1$ implies $yx=1$.

To conclude the section we  give an alternative characterization of the SN property.
In order to show that $S$ is an SN algebra it suffices to
verify
the condition of Definition \ref{def} for $R=M_n(F)$, i.e.,
all $F$-algebra homomorphisms $M_n(F)\to M_n(S)$
are given by conjugation.

\begin{proposition}\label{p:mat}
Let $S$ be an algebra and suppose that for every $n\in\mathbb{N}$ and a homomorphism $\varphi:M_n(F)\to M_n(S)$ there exists $c\in M_n(S)$ such that $\varphi(x)=cxc^{-1}$ for every $x\in R$. Then $S$ is an SN algebra.
\end{proposition}

\begin{proof}
Let $R$ be a central simple algebra and $\varphi:R\to R\otimes S$ a homomorphism. Let $e\in\mathbb{N}$ be the exponent of $R$, i.e., the order of $R$ as an element of the Brauer group of $F$ \cite[Definition 4.5.12]{GS}. Let
$$\tilde{\varphi}={\rm id}^{e-1}\otimes \varphi: R^{\otimes e}\to R^{\otimes e}\otimes S.$$
Since
$$R^{\otimes e}\cong M_{(\deg R)^e}(F),$$
by assumption there exists $c\in R^{\otimes e}\otimes S$ such that
$\tilde{\varphi}(x)=cxc^{-1}$ for every $x\in R^{\otimes e}$.

If $e>1$, we can write $c$ as
$$c=\sum_{i_2,\dots,i_e,j} a_{i_2,\dots,i_e,j}\otimes r_{i_2}\otimes\cdots \otimes r_{i_e}\otimes s_j$$
for some $a_{i_2,\dots,i_e,j},r_i\in R$ and $s_j\in S$ where $\{r_i\}_i\subset R$ and $\{s_j\}_j\subset S$ are linearly independent sets. If $x=x_1\otimes 1\otimes\cdots\otimes 1$ for $x_1\in R$, then $\tilde{\varphi}(x)c-cx=0$ becomes
$$ \sum_{i_2,\dots,i_e,j}
(xa_{i_2,\dots,i_e,j}-a_{i_2,\dots,i_e,j}x)\otimes r_{i_2}\otimes\cdots r_{i_e}\otimes s_j=0.$$
Since the elements $r_{i_2}\otimes\cdots r_{i_e}\otimes s_j$ form a linearly independent set in $R^{\otimes (e-1)}\otimes S$, we conclude that $xa_{i_2,\dots,i_e,j}=a_{i_2,\dots,i_e,j}x$ for all $a_{i_2,\dots,i_e,j}\in R$ and $x\in R$. As $R$ is central we have $a_{i_2,\dots,i_e,j}\in F$ and therefore $c\in 1\otimes R^{\otimes (e-1)}\otimes S\cong R^{\otimes (e-1)}\otimes S$. Consequently the homomorphism
$$\hat{\varphi}={\rm id}^{e-2}\otimes \varphi: R^{\otimes (e-1)}\to R^{\otimes (e-1)}\otimes S$$
satisfies $\hat{\varphi}(x)=cxc^{-1}$ for every $x\in R^{\otimes (e-1)}$. Continuing by induction we conclude that $c\in R\otimes S$ and $\varphi(x)=cxc^{-1}$ for all $x\in R$, so $S$ is an SN algebra.
\end{proof}

While Proposition \ref{p:mat} seemingly facilitates demonstrating that $S$ is an SN algebra, it does not  simplify our proofs in the sequel.

\section{SN algebras and automorphisms}\label{s:auto}

In this section we give a few motivating
results and prove that every automorphism of
a matrix algebra over an SN algebra $S$
is an inner automorphism composed
with an automorphism of $S$, see Corollary \ref{cor:autMn}.

We begin with a  proposition which justifies the requirement in  Definition \ref{def} that the algebra  $R$
is central simple.

\begin{proposition}\label{rjes}
Let $R$ be a subalgebra of an algebra $S$. If the homomorphism $x\otimes 1\mapsto 1\otimes x$ from $R=R\otimes 1$
into $R\otimes S$ can be extended to an inner automorphism of $R\otimes S$, then
 $R$ is a central simple algebra.
\end{proposition}

\begin{proof}
 By assumption, there exists an invertible element $a\in R\otimes S$ such that
$$1\otimes x =a(x\otimes 1)a^{-1}$$ for all $x\in R$.
 Let us write $a=\sum_{i=1}^m u_i\otimes v_i$ and $a^{-1}= \sum_{j=1}^n w_j\otimes z_j$.
Accordingly,
\begin{align}\label{k}
1\otimes x   =& \Big(\sum_{i=1}^m u_i\otimes v_i\Big)(x\otimes 1)\Big( \sum_{j=1}^n w_j\otimes z_j\Big) \\
=&\sum_{i=1}^m \sum_{j=1}^n u_ixw_j\otimes v_iz_j.\nonumber
\end{align}
This implies that every $x\in R$ lies in the linear span of all $v_iz_j$, $i=1,\dots,m$, $j=1,\dots,n$. Thus, $R$ is finite-dimensional.
On the other hand,  \eqref{k} implies that for every nonzero  $x\in R$, $1$ lies in $RxR$. This means that $R$ is simple.
Finally, if $z$ lies in the center of $R$, then $1\otimes z = a(z\otimes 1)a^{-1} = z\otimes 1$, which readily implies that $z$ is a scalar multiple of $1$,  as desired.
\end{proof}

The question of when the automorphism $x\otimes y\mapsto y\otimes x$ of $R\otimes R$ is inner was initiated by Sakai \cite{S}  in the C$^*$-algebra context, and investigated further by Bunce \cite{Bu}. The following corollary is an extension of \cite[Theorem 2]{Bu}.

\begin{corollary}\label{rjes2}
Let $R$ be an arbitrary algebra. The homomorphism $x\otimes 1\mapsto 1\otimes x$ from $R=R\otimes 1$ into $R\otimes R$ can be extended to an inner automorphism of $R\otimes R$ if and only if
 $R$ is a  central simple algebra.
\end{corollary}

\begin{proof}
If $R$ is a central simple algebra, then so is $R\otimes R$ \cite[Corollary 4.44]{INCA}, and so every homomorphism from $R$ into $R\otimes R$ can be extended to an inner automorphism by the Skolem-Noether theorem. The converse follows from Proposition \ref{rjes}.
 \end{proof}

The next proposition yields another motivation for considering SN algebras.

\begin{proposition}\label{prop:aut}
Let $R$ be a central simple algebra and let $S$ be an SN algebra. Then every automorphism $\varphi$ of $R\otimes S$ is the composition of an inner automorphism and an automorphism of the form ${\rm id}_R\otimes \sigma$ where $\sigma$ is an automorphism of $S$.
\end{proposition}

\begin{proof}
By assumption, the restriction of $\varphi$ to $R$ can be extended to an inner automorphism $x\mapsto cxc^{-1}$ of $R\otimes S$. Considering the automorphism $x\mapsto
c^{-1}\varphi(x)c$ we thus see that there is no loss of generality in assuming that $\varphi$ acts as the identity on $R$. Note that the proposition will be proved by showing
that $\varphi$
maps $1\otimes S$ into itself. Pick $s\in S$. We can write $\varphi(1\otimes s)$ as
$\sum_{j} p_j\otimes s_j$
where the $s_j$'s are linearly independent.
Since $1\otimes s$ commutes with $x\otimes 1$ for every $x\in R$ it follows that so does $\varphi(1\otimes s)$. This implies that
$$\sum_j (p_jx-xp_j)\otimes s_j =0.$$
As the $s_j$'s are linearly independent it follows that $p_jx-xp_j=0$ for each $j$ and each $x\in R$.
Hence, since $R$ is central,  each $p_j$ is a scalar multiple of $1$. Consequently, $\varphi(1\otimes s)\in 1\otimes S$.
\end{proof}

If $R=M_n(F)$, then
 $R\otimes S$ can be identified with $M_n(S)$, and the proposition gets the following form.

\begin{corollary}\label{cor:autMn}
If $S$ is an SN algebra, then every automorphism $\varphi$ of $M_n(S)$ is of the form $$\varphi\big((s_{ij})\big) = c\big(\sigma(s_{ij})\big)c^{-1}$$ where $c$ is an invertible element in $M_n(S)$ and $\sigma$ is an automorphism of $S$.
\end{corollary}

This result is known in the case where $S$ is either an artinian algebra \cite[Theorem 3.13]{BO}, a UFD \cite[Corollary 15]{I}, or a commutative local algebra (see, e.g., \cite[p. 163]{K}). As we will see,  all these algebras are SN algebras. On the other hand, \cite{I}
shows that the commutative domain $\mathbb Z[\sqrt{-5}]$ does not satisfy the conclusion of Corollary \ref{cor:autMn}.
We give an algebra with the same property in Example \ref{ex:elliptic}.

\section{Basic lemma}\label{s:basic}

All our main results will be derived from the following technical lemma. Its proof  will use some ideas from the proof of the (special case of)   Skolem-Noether theorem  given in \cite[pp. 13--14]{INCA}.

\begin{lemma}\label{l}
Let $R$ be a  central simple algebra with basis $\{r_1,\dots,r_d\}$ and let $S$ be an arbitrary algebra. Then  $\varphi:R\to R\otimes S$ is a homomorphism
if and only if there exist   $c_1,\dots,c_d\in R\otimes S$ such that
\begin{enumerate}
\item[(a)] $\varphi(x)=\sum_{k=1}^d c_k x r_{k}$ for all $x\in R$,
\item[(b)] $\varphi(x)c_k=c_kx$ for all $x\in R$, and
\item[(c)] $\sum_{k=1}^d c_kr_k = 1.$
\end{enumerate}
Moreover, writing $c_k= \sum_{l=1}^{d} r_{l}\otimes s_{kl}$, we
have that for each $k$ and $l$ there exists $b_{kl}\in R\otimes S$ such that $$b_{kl}c_k = 1\otimes s_{kl}.$$ Accordingly, if $S$ is stably finite and there exist $k$ and $l$ such that  $s_{kl}$ is invertible in $S$, then $c=c_k$ is invertible in $R\otimes S$ and $$\varphi(x)=cxc^{-1}$$ for all $x\in R$.
\end{lemma}

\begin{proof}
Since $R$ is finite-dimensional, there exist finitely many $s_i\in S$ and
linear maps $f_i:R\to R$
 such that
$$\varphi(x)= \sum_i f_i(x)\otimes s_i
$$
for all $x\in S$. By  \cite[Lemma 1.25]{INCA}  there exist $w_{ij}, z_{ij}\in R$ such that
$$f_i = \sum_{j} L_{w_{ij}}R_{z_{ij}}.$$
 Consequently, for every $x\in R$ we have
\begin{align*}
\varphi(x) &=\sum_i \Big( \sum_{j} w_{ij}xz_{ij}\Big)\otimes s_i\\
&=\sum_i \sum_{j} (w_{ij}\otimes s_i)x z_{ij}.
\end{align*}
Writing each $z_{ij}$ as a linear combination of $r_1,\dots,r_d$ we see that $\varphi$ is of the form described in (a).

 We now use the multiplicativity of  $\varphi$, i.e.,
 $\varphi(xy)= \varphi(x)\varphi(y)$ for all $x,y\in R$. In view of (a) we can rewrite this as
 \begin{equation} \label{nm} \sum_{k=1}^d c_kxyr_k = \sum_{k=1}^d  \varphi(x) c_k yr_k. \end{equation}
 Pick  $w_j,z_j\in R$  such that \eqref{oh} holds.
Setting $y= w_j$ in  \eqref{nm},
multiplying  the identity, so obtained, from the right by $z_j$,  and then summing up over all $j$ we get
 $$
 \sum_j  \sum_{k=1}^d c_kx w_j r_k z_j = \sum_j  \sum_{k=1}^d \varphi(x) c_k w_j r_k z_j,
 $$
 that is,
  $$
 \sum_{k=1}^d c_kx \Big(\sum_j w_j r_k z_j\Big) =  \sum_{k=1}^d \varphi(x) c_k \Big(\sum_j w_j r_k z_j\Big).
 $$
 By \eqref{oh} this reduces to $c_1x = \varphi(x)c_1$. Of course,  the same proof applies  to every $c_k$, so (b) holds. Finally, (c) follows
from $\varphi(1)=1$.

A direct verification shows that  (a), (b), and (c) imply that $\varphi$ is a homomorphism.

Let us write $c_1 = \sum_l r_{l}\otimes s_{1l}$, and let $w_j,z_j$ be   as above. Using (b) we obtain
\begin{align*}
\sum_j w_j\varphi(z_j) c_1 &= \sum_j w_jc_1z_j\\
&= \sum_j\sum_l w_j ( r_{l}\otimes s_{1l})z_j\\
&=\sum_l\Big(\sum_j w_jr_{l}z_j\Big)\otimes s_{1l}\\
&= 1\otimes s_{11}.
\end{align*}
Thus, $b_{11}= \sum_j w_j\varphi(z_j)$ satisfies $b_{11}c_1 = 1\otimes s_{11}$. Similarly we find other $b_{kl}$'s.

Finally, assume that $s_{kl}$ is invertible in $S$ for some  $k$ and $l$. Then
$(1\otimes s_{kl}^{-1})b_{kl}$ is a left inverse of $c_k$. If $S$ is stably finite, then the result by Montgomery  \cite[Theorem 1]{Mon} implies that
this element is also a right inverse. Therefore, (b) shows that $c=c_k$ satisfies  $\varphi(x)=cxc^{-1}$ for all $x\in R$.
\end{proof}

We continue with a simple application
of Lemma \ref{l}, showing that local algebras
are SN. This result will be generalized
to semilocal algebras (with a considerably more involved proof)
in the next section, see Theorem \ref{thm:semiLoc}.

\begin{corollary}\label{t1}
Every local algebra  is an SN algebra.
\end{corollary}

\begin{proof}
Let $r_k,s_{kl}$ be elements from Lemma \ref{l}. From (c) it follows that  $$\sum_{k,l} r_{l}r_k\otimes s_{kl} = 1\otimes 1.$$
This implies that
$1$ lies in the linear span of $s_{kl}$. Consequently, at least one $s_{kl}$  does not lie in rad$(S)$. Since $S$ is local it follows that $s_{kl}$  is invertible in $S$.
As  $S$ is stably finite  \cite[Theorem 20.13]{Lam},  the last assertion of Lemma \ref{l} shows that there exists $c\in R\otimes S$ such that
$\varphi(x)=cxc^{-1}$ for all $x\in R$.
\end{proof}

\section{Semilocal algebras}\label{s:semiloc}

The main result of this section is Theorem
\ref{thm:semiLoc} showing that
semilocal algebras are SN. We begin with a simple lemma.

\begin{lemma}\label{prod}
If  $S_1$ and $S_2$ are SN algebras, then so is their direct product $S_1\times S_2$.
\end{lemma}

\begin{proof}
Recall that $R\otimes (S_1\times S_2)$ can be identified with $(R\otimes S_1)\times (R\otimes S_2)$. Take a homomorphism $$\varphi:R\to (R\otimes S_1)\times (R\otimes S_2).$$
Writing $$\varphi(x)=(\varphi_1(x),\varphi_2(x))$$ it is immediate that $\varphi_i$ is a homomorphism from $R$ into $R\otimes S_i$, $i=1,2$. By assumption,  there exist $c_i\in R\otimes S_i$  such that
$\varphi_i(x)= c_ixc_i^{-1}$ for all $x\in R$, $i=1,2$. Hence, $$c=(c_1,c_2)\in (R\otimes S_1)\times (R\otimes S_2)$$ satisfies $\varphi(x)=cxc^{-1}$ for all $x\in R$.
\end{proof}

As mentioned in the introduction, the Skolem-Noether theorem implies that every central simple algebra is an SN algebra.
With a little extra effort we can extend this to  semisimple algebras.

\begin{lemma}\label{lsemi}
Every semisimple algebra is an SN algebra.
\end{lemma}

\begin{proof}
In view of Lemma \ref{prod} it suffices to consider the case where $S$ is  simple artinian.
Let $R$ be a central simple algebra. The algebra $R\otimes S$
is then simple  \cite[Theorem 4.42]{INCA}.
We claim that it is  also artinian. Indeed, considering $R\otimes S$ as a left $S$-module in the natural way  we see that
it is isomorphic to the  left $S$-module $S^d$ where $d$ is the dimension of $R$, and that
 a descending chain of left ideals of $R\otimes S$
is also a descending chain of left $S$-submodules. The desired conclusion thus follows from the fact that   $S^d$ is  artinian.

Let $K$ be the center of $S$.
The center of $R\otimes S$  is equal to $1\otimes K$ \cite[Corollary 4.32]{INCA}, which we     identify with $K$. Consider $R\otimes K$ as an algebra over $K$ in the usual way. Clearly,
it is finite-dimensional and, again by \cite[Theorem 4.42]{INCA}, simple.

Now, given a homomorphism $\varphi:R\to R\otimes S$, we define $$\Phi:R\otimes K\to R\otimes S$$ by $$\Phi(x\otimes k)= \varphi(x)(1\otimes k).$$ Note that $\Phi$ is a $K$-algebra homomorphism. The Skolem-Noether theorem thus tells us that there exists $c\in R\otimes S$ such that
$\Phi(x\otimes k)= c(x\otimes k)c^{-1}$ for all $x\in R$, $k\in K$. Setting $k=1$ we get the desired conclusion.
\end{proof}

Our goal is to show that semilocal algebras are SN algebras by reducing the general case to the semisimple case. We will actually prove a general reduction theorem whose possible applications are not limited to  semilocal algebras. To this end, we need the following lemma. From its nature one would expect that it is known,  but we were unable to find a good reference.
We include a short proof for the sake of completeness.

\begin{lemma}\label{lemrad}
If $R$ is a central simple algebra, then ${\rm rad}(R\otimes S) = R\otimes {\rm rad}(S)$ for every algebra $S$.
\end{lemma}

\begin{proof}
	As an ideal of $R\otimes S$, ${\rm rad}(R\otimes S)$ is necessarily of the form
	$R\otimes I$ for some ideal $I$ of $S$ \cite[Theorem 4.42]{INCA}.
	We will show that $I\subseteq {\rm rad}(S)$, by making use of the following  characterization of rad$(A)$:
	 $v\in {\rm rad}(A)$ if and only if
	$1- vx$ is invertible for every $x\in A$.
Take  $u\in I$. Since $1\otimes u \in {\rm rad}(R\otimes S)$ it follows that

	$$1\otimes (1-ux)=1\otimes 1 - 1\otimes ux = 1\otimes 1 - (1\otimes u)(1\otimes x)$$
	is invertible in $R\otimes S$ for every $x\in S$. However, this is possible only if $1-ux$ is invertible, implying that $u\in {\rm rad}(S)$. Thus, $I\subseteq {\rm rad}(S)$,
	and so $ {\rm rad}(R\otimes S)\subseteq R\otimes {\rm rad}(S)$.

	As the lemma is well-known if $R=M_n(F)$ (see, e.g., \cite[pp. 57-58]{Lam}),
we will establish the converse inclusion  by reducing the general case to this one.
Take a splitting field $K$ for $R$ which is a finite separable extension of $F$ (see, e.g., \cite[Proposition 4.5.4]{GS}). Then  $K\otimes R$ may be identified with  $M_n(K)$ for some $n\ge 1$, and, therefore, $K\otimes R\otimes S$ may be identified with $M_n(K\otimes S)$. Thus, by what we pointed out at the beginning of the paragraph, we have
$${\rm rad}(K\otimes R\otimes S) = M_n({\rm rad}(K\otimes S)).$$
By \cite[Theorem 5.17]{Lam}, ${\rm rad}(K\otimes S) = K\otimes {\rm rad}(S)$, so that
\begin{equation}\label{rs}{\rm rad}(K\otimes R\otimes S) = M_n(K)\otimes {\rm rad}(S).\end{equation}
According to  \cite[Theorem 5.14]{Lam},
$${\rm rad}(R\otimes S) = (R\otimes S)\cap {\rm rad}(K\otimes R\otimes S),$$
and hence, by \eqref{rs},
$${\rm rad}(R\otimes S) = (R\otimes S)\cap  (M_n(K)\otimes {\rm rad}(S)).$$
Since  both $R\otimes S$ and $M_n(K)\otimes {\rm rad}(S)$ readily contain $R\otimes {\rm rad}(S)$,
 it follows that $R\otimes {\rm rad}(S)\subseteq {\rm rad}(R\otimes S)$.
\end{proof}

We can now prove the announced reduction theorem.

\begin{theorem}\label{trad} If an algebra $S$ is stably finite and  $S/{\rm rad}(S)$ is  an  SN algebra, then
$S$  is  an  SN algebra.
\end{theorem}

\begin{proof}
Let $R$ be central simple and
 write   $$J=R\otimes {\rm rad}(S).$$
  Take a homomorphism $\varphi:R\to R\otimes S$. We define
	$$\Phi:R\to (R\otimes S)/J$$ by
	$$\Phi(x)=\varphi(x)+J.$$
	Since $(R\otimes S)/J$ is canonically isomorphic to $R\otimes (S/{\rm rad}(S))$,  and $S/{\rm rad}(S)$ is  assumed to be an SN algebra, it follows that
	there exists an invertible element $a\in (R\otimes S)/J$ such that \begin{equation*}\label{bg}\Phi(x)= a(x+J)a^{-1}\,\,\mbox{ for all $x\in R$.}\end{equation*}
As $J$ is, by Lemma \ref{lemrad}, the Jacobson radical of $R\otimes S$,  it follows that there exists an invertible element $b\in R\otimes S$ such that $a = b +J$. Obviously, we have
	$$\varphi(x)-bxb^{-1}\in J\,\,\mbox{ for all $x\in R$,}$$
that is,
	$$b^{-1}\varphi(x)b -x\in J\,\,\mbox{ for all $x\in R$.}$$
Replacing the role of $\varphi$ by the homomorphism $x\mapsto b^{-1}\varphi(x)b$	we see that without loss of generality we may assume that $b=1$. Thus,
\begin{equation}\label{kk}\varphi(x) -x\in J\,\,\mbox{ for all $x\in R$.}\end{equation}
Now apply Lemma \ref{l}. Picking a basis $\{r_1,\dots,r_d\}$ of $R$, we can thus find  $s_{kl}\in S$, $k,l=1,\dots,p$, such that
\begin{equation}\label{ll}\varphi(x)=\sum_{k=1}^d \sum_{l=1}^d r_l x r_k \otimes s_{kl} \,\,\mbox{ for all $x\in R$,}\end{equation}
and our goal is to show that at least one $s_{kl}$ is invertible in $S$.

Let $\lambda_k\in F$ be such that  $1=\sum_{k=1}^d {\lambda_k}r_k$. Then
$$x=1\cdot x\cdot 1 = \sum_{k=1}^d \sum_{l=1}^d (\lambda_k\lambda_l) r_l x r_k \,\,\mbox{ for all $x\in R$}.$$
Using \eqref{kk} and \eqref{ll} we thus obtain
\begin{equation}\label{ro}
\sum_{k=1}^d \sum_{l=1}^d r_l x r_k \otimes (s_{kl} - \lambda_k\lambda_l\cdot 1)\in J \,\,\mbox{ for all $x\in R$}.
\end{equation}
We may assume that $\lambda_1\ne 0$. Choose  $w_j,z_j\in R$ that satisfy \eqref{oh}.
Denoting the expression in \eqref{ro} by $\rho(x)$, we have
\begin{align*}
\sum_j w_j\rho(z_j) &= \sum_j \sum_{k=1}^d \sum_{l=1}^d w_jr_l z_j r_k \otimes (s_{kl} - \lambda_k\lambda_l \cdot1)\\
&= \sum_{k=1}^d \sum_{l=1}^d \Big(\sum_j w_jr_l z_j \Big) r_k \otimes (s_{kl} - \lambda_k\lambda_l\cdot 1)\\
&= \sum_{k=1}^d   r_k \otimes (s_{k1} - \lambda_k\lambda_1 \cdot1).
\end{align*}
As $\rho$ maps into $J$ it follows that
$$\sum_{k=1}^d   r_k \otimes (s_{k1} - \lambda_k\lambda_1 \cdot 1)\in J =R\otimes {\rm rad}(S).$$
Since the $r_k$'s are linearly independent, we must have  $s_{k1} - \lambda_k\lambda_1 \cdot 1 \in {\rm rad}(S)$
for each $k$. In particular, $s_{11} = \lambda_1^2\cdot 1 + u$ for some $u\in  {\rm rad}(S)$. Since $\lambda_1\ne 0$ it follows that $s_{11}$ is invertible, as desired.
\end{proof}

 We are now in a position to give our main result.
\begin{theorem}\label{thm:semiLoc}
Every semilocal algebra $S$ is an SN algebra.
\end{theorem}

\begin{proof}Since $S$  is stably finite \cite[Theorem 2.13]{Lam} and the algebra $S/{\rm rad}(S)$ is semisimple,
the theorem follows from Lemma \ref{lsemi} and  Theorem \ref{trad}.
\end{proof}

\begin{corollary} \label{art}
Every artinian algebra is an SN algebra.
\end{corollary}

\section{Finite-dimensional algebras}\label{s:findim}

Corollary \ref{art}  shows that every finite-dimensional algebra  is an   SN algebra.
The next result gives a strengthening of this property.

\begin{theorem}\label{tfd}
Let $A$ be a finite-dimensional algebra and let $R$ be its central simple subalgebra. Then every homomorphism from $R$ into $A$ can be extended to an inner automorphism of $A$.
\end{theorem}

\begin{proof}
Assume first that $R=M_n(F)$. Then $A$ contains a set of $n\times n$ matrix units and is therefore isomorphic to $M_n(S)\cong R\otimes S$ for some subalgebra $S$ of $A$ \cite[Lemma 2.52]{INCA}. Since $S$ is also finite-dimensional, the desired conclusion follows from Corollary \ref{art}.

Now let $R$ be an arbitrary central simple algebra. We may assume that the field $F$ is infinite, for otherwise $R\cong M_n(F)$ by   Wedderburn's theorem on finite division rings.
Let
$\varphi$ be a homomorphism from $R$ into $A$.
Take  a splitting field $K$ for $R$. Identifying $K\otimes R$  with  $M_n(K)$, $n \ge 1$,  it follows from the preceding paragraph that there exists
$b\in K\otimes A$ such that \begin{equation*} \label{y} ({\rm id}_K\otimes \varphi)(y) = byb^{-1}\end{equation*} for all $y\in K\otimes R$.
In particular, $$(1\otimes \varphi(x))b = b(1\otimes x)$$ for all $x\in R$.
 Writing $b=\sum_{i=1}^m k_i\otimes a_i$ with the $k_i$'s  linearly independent, it follows that
$$\sum_{i=1}^m k_i \otimes (\varphi(x)a_i - a_ix)=0,$$
and so
$\varphi(x)a_i = a_ix$
for all $x\in R$ and every $i$. Hence we see that it suffices to show that span$_F\{a_1,\dots,a_m\}$ contains an  element which is invertible in $A$.

As a finite-dimensional algebra, $A$ can be considered as a subalgebra of $M_N(F)$ for some $N\ge 1$. Take the polynomial
$$f(\xi_1,\dots,\xi_m)=\det\Big(\sum_{i=1}^m \xi_i a_i\Big) \in F[\xi_1,\dots,\xi_m].$$
Note that $K\otimes A$ can be viewed as a subalgebra of $M_N(K)$.
Since $b$ is invertible in $K\otimes A$, we know that span$_K\{a_1,\dots,a_m\}$ contains an invertible element in $K\otimes A$. This clearly implies that $f$ is a nonzero polynomial.
As $F$ is infinite,  there exist $\lambda_i\in F$ such that $f(\lambda_1,\dots,\lambda_m)\ne 0$. That is, span$_F\{a_1,\dots,a_m\}$ contains an element $c$ which is invertible in $M_N(F)$. However,
since we are in finite dimensions, $c^{-1}$ is a polynomial in $c$. Thus, $c$ is invertible in  $A$.
\end{proof}

Using the standard homomorphism construction we will now see that Theorem \ref{tfd} can be used for  showing that all derivations from $R$
into any $R$-bimodule $M$ are inner (in accordance with the conventions mentioned at the very beginning of the paper, we assume that our bimodules are unital).
This is, of course,  a  well-known result. Another way of stating it is that central simple algebras are separable.

\begin{corollary}\label{cfd}
Every derivation from a central simple algebra $R$ into an arbitrary $R$-bimodule $M$ is inner.
\end{corollary}

\begin{proof} Let $d:R\to M$ be  a derivation.
As  a finite-dimensional subspace of $M$, $d(R)$ generates a finite-dimensional subbimodule of $M$. Therefore, there is no loss of generality in assuming that $M$ itself is finite-dimensional.

Let $\widetilde{A}$ be the set of all matrices of the form $\left[
    \begin{smallmatrix}
      x & u\\
      0&x
    \end{smallmatrix}\right]$, where $x\in R$ and $u\in M$. Note that  $\widetilde{A}$ is a (finite-dimensional!) algebra under the standard matrix operations.
Let $\widetilde{R}$ be its subalgebra consisting of all matrices of the form $\left[
    \begin{smallmatrix}
      x & 0\\
      0&x
    \end{smallmatrix}\right]$,  $x\in R$. Of course,  $\widetilde{R}\cong R$.

Define $\varphi:\widetilde{R}\to \widetilde{A}$ by
$$\varphi\left(\left[
    \begin{matrix}
      x & 0\\
      0&x
    \end{matrix}\right]\right)=\left[
    \begin{matrix}
      x & d(x)\\
      0&x
    \end{matrix}\right].$$
		One immediately checks  that $\varphi$ is a homomorphism. By Theorem \ref{tfd} there exists an invertible element $c= \left[
    \begin{smallmatrix}
      t & v\\
      0&t
    \end{smallmatrix}\right] \in \widetilde{A}$ such that $\varphi(\tilde{x})= c\tilde{x}c^{-1}$ for all $\tilde{x}\in\widetilde{R}$. Consequently,  $\varphi(\tilde{x})c= c\tilde{x}$, that is,
		$$\left[
    \begin{matrix}
      x & d(x)\\
      0&x
    \end{matrix}\right]\cdot \left[
    \begin{matrix}
      t & v\\
      0&t
    \end{matrix}\right] = \left[
    \begin{matrix}
      t & v\\
      0&t
    \end{matrix}\right]\cdot \left[
    \begin{matrix}
      x & 0\\
      0&x
    \end{matrix}\right]$$
		for all $x\in R$. This yields $$xt = tx\,\,\,\mbox{ and}\,\,\, xv + d(x)t = vx$$   for all $x\in R$. Since $R$ is central, the first identity shows that $t\in F$. Moreover, $t\ne 0$ for $c$ is invertible.  Hence $w=t^{-1}v$ satisfies  $d(x) = wx - xw$ by the second identity.
\end{proof}

\section{Domains}\label{s:dom}

In this section we give classes of domains which are SN algebras. For instance,
UFDs and free algebras are SN algebras (Corollaries \ref{ufd} and \ref{cor:free}).

As the coordinate ring of an elliptic curve demonstrates (see Example \ref{ex:elliptic}),
not every commutative domain is an SN algebra.
However, the following proposition shows that
every domain $S$ embedded into a division algebra satisfies
a certain weaker condition.

\begin{proposition}\label{p}
Let $R$ be a  central simple algebra, and let an algebra $S$ be a domain which can be embedded into a division algebra $D$. If $\varphi$ is a homomorphism from $R$ into
 $R\otimes S$, then there exists  $c\in R\otimes S$ which is  invertible in $R\otimes D$, in fact $$c^{-1} = (1\otimes s^{-1})b\in R\otimes D$$ for some nonzero $s\in S$ and $b\in R\otimes S$,  such that
$$\varphi(x)=cxc^{-1}$$ for all $x\in R$. Moreover, if $\{r_1,\dots,r_d\}$ is a basis of $R$, $b=\sum_{k=1}^d
r_k\otimes s_k$ for some $s_k\in S$, and
$c=\sum_{l=1}^d r_l\otimes t_l$  for some $t_l\in S$, then   $$t_ls^{-1}s_k\in S$$ for all $k$ and $l$.
\end{proposition}

\begin{proof}
Not every $c_k$ from Lemma \ref{l} can be $0$ (in view of (c)), and so $s_{kl}\ne 0$ for some $k$ and $l$.
Set $c=c_k$ and $s=s_{kl}$. By the lemma we have $\varphi(x)c=cx$ for all $x\in R$ and $bc=1\otimes s$ for some
$b\in R\otimes S$. Of course,  $s$ is invertible in $D$. Therefore $(1\otimes s^{-1})b$ is a left inverse of $c$ in $R\otimes D$. By \cite[Theorem 1]{Mon}, a left inverse in $R\otimes D$ is also a right inverse, so $c^{-1}=(1\otimes s^{-1})b$.

Now take a basis  $\{r_1,\dots,r_d\}$ of $R$, and let us write
$b=\sum_{k=1}^d r_k\otimes s_k$  and
$c=\sum_{l=1}^d r_l\otimes t_l$. Then
$$\varphi(x)= cxc^{-1}=\sum_{k=1}^d\sum_{l=1}^d r_lxr_k\otimes t_ls^{-1}s_k.$$
for all $x\in R$. Pick $w_j,z_j\in R$ satisfying \eqref{oh}.
 We have
\begin{align*}
\sum_j w_j\varphi(z_j) &= \sum_j \sum_{k=1}^d \sum_{l=1}^d w_jr_l z_j r_k \otimes t_ls^{-1}s_k\\
&=  \sum_{k=1}^d \sum_{l=1}^d \Big( \sum_j w_jr_l z_j \Big)r_k \otimes t_ls^{-1}s_k\\
&=  \sum_{k=1}^d r_k \otimes t_1s^{-1}s_k.
\end{align*}
Since the left hand side, i.e.  $\sum_j w_j\varphi(z_j)$, lies in $R\otimes S$, so does the right hand side. This readily yields that
$t_1s^{-1}s_k\in S$.
\end{proof}

\begin{corollary}\label{ufd}
Every UFD is an SN algebra.
\end{corollary}

\begin{proof}
Let $S$ be a UFD and let $R$, $\varphi$, $b$, $c$, $s$, $s_k$, $t_l$
be as in Proposition \ref{p}. Since $S$ is a UFD, $t_1,\dots,t_d$ have a greatest common divisor $e$ and $c$ can be replaced with $(1\otimes e^{-1})c$, we can without loss of generality assume that $t_1,\dots,t_d$ are coprime. Then it suffices to prove that $s^{-1}s_k\in S$ for every $k$. Since $t_ls^{-1}s_k\in S$ for every $k,l$, we see that $s$ divides $t_ls_k$ for every $l,k$. Let $p$ be a prime such that $p^n$ divides $s$. Suppose that $p^n$ does not divide $s_{k_0}$ for some $k_0$. Since $p^n$ divides $t_ls_{k_0}$ for every $l$, we conclude that $p$ divides $t_l$ for every $l$, which contradicts the assumption about $t_l$ being coprime. Hence $s$ divides $s_k$ for every $k$.
\end{proof}

We now move to the noncommutative setting. Since embeddings of noncommutative domains into division rings can be ill-behaved or nonexistent, one needs stronger assumptions than in Corollary \ref{ufd}. Let $S$ be an arbitrary ring. The {\bf inner rank} of $A\in S^{m\times n}$ is the least $r$ such that $A=BC$ for some $B\in S^{m\times r}$ and $C\in S^{r\times n}$. We write $\rho A=r$. For example, if $S$ is a division ring, then $\rho A$ is just the rank of $A$. We say that $S$ is a {\bf Sylvester domain} \cite[Section 5.5]{Coh} if for any $P\in S^{\ell\times m}$ and $Q\in S^{m\times n}$ such that $PQ=0$, it follows that $\rho P+\rho Q\le m$.

We say that an element $s\in S$ {\bf right divides} $a\in S$ if $a=a's$ for some $a'\in S$. If $S$ is a domain and $a,b\in S\setminus\{0\}$, then $s$ is a {\bf highest common right factor} (HCRF) of $a$ and $b$ if $s$ right divides $a,b$ and every $s'\in S$ that right divides $a,b$ also right divides $s$. We say that $S$ is an {\bf HCRF domain} if every pair of nonzero elements in $S$ admits a HCRF. Special examples of HCRF domains are filtered rings satisfying the 2-term weak algorithm \cite[Section 2.8]{Coh} or more generally, 2-firs with right ACC$_1$  (ascending chain condition on principal right ideals) \cite[Exercise 3.2.1]{Coh}.

\begin{theorem}\label{t:sylv}
If $S$ is an HCRF domain and a Sylvester domain, then $S$ is an SN algebra.
\end{theorem}

\begin{proof}
Since $S$ is a Sylvester domain, it admits a universal skew field of fractions $D$ and this embedding preserves the inner rank by \cite[Theorem 7.5.13]{Coh}. Let $R$ be a central simple algebra and $\varphi:R\to R\otimes S$ a homomorphism. By Proposition \ref{p} there exists $c\in R\otimes S$ invertible in $R\otimes D$ such that $\varphi(x)=cxc^{-1}$ for all $x\in R$. Furthermore, if $\{r_1,\dots,r_d\}$ is a basis of $R$ and
$$c=\sum_l r_l\otimes t_l,\qquad c^{-1} =\sum_k r_k\otimes u_k$$
for $t_l\in S$ and $u_k\in D$, then $t_lu_k=s_{lk}\in S$ for all $1\le l,k\le d$ (here $u_k=s^{-1}s_k$ from Proposition \ref{p}). Since $S$ is an HCRF domain, we can assume that $t_1,\dots,t_d$ have no non-trivial common right factors (otherwise they have a nontrivial HCRF $e$ and we can replace $c$ with $c(1\otimes e^{-1})$). Fix $k$ such that $u_k\neq 0$. Then $(u_k,-1)^{\rm t}\in D^2$ belongs to the right kernel of the matrix
$$\begin{pmatrix}
t_1 & s_{1k} \\
\vdots & \vdots \\
t_d & s_{dk}
\end{pmatrix}\in S^{d\times 2}$$
which is therefore of (inner) rank $1$ over $D$. Since the embedding $S\subseteq D$ is inner rank preserving, this matrix is also of inner rank 1 over $S$, so
$$\begin{pmatrix}
t_1 & s_{1k} \\
\vdots & \vdots \\
t_d & s_{dk}
\end{pmatrix}=
\begin{pmatrix}
v_1 \\ \vdots \\ v_d
\end{pmatrix}
\begin{pmatrix}
w_1 & w_2
\end{pmatrix}$$
for some $v_i,w_j\in S$. Since $w_1$ right divides $t_l$ for every $l$ and $t_1,\dots,t_d$ are right coprime by assumption, we conclude that $w_1$ is invertible in $S$. By taking some $t_l\neq0$ we get
$$u_k=t_l^{-1}s_{lk}=w_1^{-1}v_l^{-1}v_lw_2=w_1^{-1}w_2\in S.$$
Consequently $c^{-1}\in R\otimes S$.
\end{proof}

\begin{corollary}\label{cor:free}
Every free algebra $F\langle X\rangle$ is an SN algebra.
\end{corollary}
\begin{proof} A free algebra is a filtered ring with a weak algorithm \cite[Theorem 2.5.3]{Coh}, so it is a HCRF domain and a fir (free ideal ring) by \cite[Theorem 2.4.6]{Coh} and hence a Sylvester domain by \cite[Proposition 5.5.1]{Coh}.
\end{proof}

Theorem \ref{t:sylv} has the following form for commutative rings.
\begin{corollary}\label{bez}
Every B\'ezout domain  is an SN algebra.
\end{corollary}
\begin{proof}
Every B\'ezout domain is a GCD domain, which is just a commutative HCRF domain. Moreover, by \cite[Proposition 2.3.17]{Coh} it is also a semifir and hence a Sylvester domain by \cite[Proposition 5.5.1]{Coh}. Therefore Theorem \ref{t:sylv} applies.
\end{proof}

In the next example we present a domain that is not an SN algebra;
cf.~\cite[Theorem 15]{RZ}.

\begin{example}\label{ex:elliptic}
Let $S=F[x,y]/(y^2-x^3-x)$. Then $S$ is a domain,
$$a=\begin{pmatrix}y & x \\ x^2 & y\end{pmatrix}\in M_2(S)$$
is invertible as a matrix over the field of fractions of $S$ and
$$a^{-1}=\begin{pmatrix}\frac{y}{x} & -1 \\ -x & \frac{y}{x}\end{pmatrix}.$$
Since every product of an entry in $a$ and an entry in $a^{-1}$ lies in $S$, it follows that
$$\varphi\colon M_2(F)\to M_2(F)\otimes S,\qquad u\mapsto aua^{-1}$$
is a well-defined homomorphism. Suppose there exists an invertible $c\in M_2(S)$ such that $\varphi(u)=cuc^{-1}$ for all $u\in M_2(F)$. Then $\gamma=\det(c)$ is invertible in $S$ and it is easy to see that this implies $\gamma\in F\setminus\{0\}$. Since $c^{-1}a$ commutes with every $u\in M_2(F)$ by the definition of $\varphi$, we have $c^{-1}a=f I_2$ for some $f\in S$. But then
$$\gamma f^2=\gamma \det(c^{-1}a)=\det(a)= x$$
contradicts the irreducibility of $x$ in $S$.
\end{example}

\section{Polynomial matrix algebras}\label{s:matrixpoly}

In this section we prove that matrix algebras over polynomial algebras are SN.

\begin{theorem}\label{t:matrixpoly}
	$M_n(F[\xi_1,\dots,\xi_s])$ is an SN algebra.
\end{theorem}
Besides the Quillen-Suslin theorem, saying that over $F[\xi_1,\dots,\xi_s]$ every finitely generated projective module is free, see \cite{Q,Sus}, the proof of this theorem is mostly based on the following simple factorization lemma.
\begin{lemma}
	\label{l:denominators}
	Let $A$ be a commutative algebra, which is a domain with field of fractions $K$, $R$ a central simple algebra and $a\in R\otimes M_n(A)$.
	Suppose that $a$ is invertible in $R\otimes M_n(K)$ and that $a(x\otimes 1)a^{-1}\in R\otimes M_n(A)$ for all $x\in R$. Then for $c\in M_n(A)$ the following are equivalent:
	\begin{enumerate}
		\item[(i)] There exists a factorization $a=u(1\otimes c)$ for some invertible $u\in R\otimes M_n(A)$;
		\item[(ii)] The left ideal
			\[
				I(a):=\{\, m\in M_n(A) \mid (1\otimes m)a^{-1}\in R\otimes M_n(A)\,\}\subseteq M_n(A)
			\]
			is generated by $c$;
		\item[(iii)] The rows of $c$ form a basis of the $A$-module
			\[
				M(a):=\{\, r\in A^{1\times n} \mid (1\otimes r)a^{-1}\in R\otimes A^{1\times n}\,\}.
			\]
	\end{enumerate}

\end{lemma}
\begin{proof}
	(i)$\Rightarrow$(ii): Let $u\in R\otimes M_n(A)$ be invertible such that $a=u(1\otimes c)$. Then for any $m\in M_n(A)$ we have that $(1\otimes m)a^{-1}=(1\otimes mc^{-1})u^{-1}$ lies in $R\otimes M_n(A)$ if and only if $mc^{-1}\in M_n(A)$, i.e., $m\in M_n(A)c$.

	(ii)$\Rightarrow$(i):
	Let $\{r_1,\dots,r_d\}$ be a basis of $R$ and for fixed $i$ take $w_k,z_k\in R$ such that $\sum_k w_k r_j z_k=\delta_{ij}$, see \eqref{oh} at the beginning of Section \ref{s:prelim}. If $a=\sum_jr_j\otimes a_j$, then
	\[
		(1\otimes a_i)a^{-1}=\sum_k (w_k\otimes 1)a(z_k\otimes 1)a^{-1}\in R\otimes M_n(A).
	\]
	Since $i$ was arbitrary, this shows that all coefficients $a_i$ of $a$ lie in $I(a)$. Assuming that $I(a)=M_n(A)c$, we can thus factor $a=u(1\otimes c)$ for some $u\in R\otimes M_n(A)$. But then $u^{-1}=(1\otimes c)a^{-1}\in M_n(A)$, i.e., $u$ is invertible.

	(ii)$\Rightarrow$(iii): Suppose $I(a)=M_n(A)c$. Since $a$ is invertible over $K$, there exists a nonzero $e\in A$ such that $e \cdot 1\in I(a)$. Hence the rows of $c$ are clearly linearly independent. Given any $r\in M(a)$, we can extend $r$ by zero to form a matrix $m\in I(a)$ which has $r$ as one of its rows. By assumption $m\in M_n(A)c$. In particular, $r$ is a linear combination of the rows of $c$, which shows that they form a basis of $M(a)$.

	(iii)$\Rightarrow$(ii): Conversely, suppose that the rows of $c$ form a basis of $M(a)$. In particular, $c\in I(a)$. Moreover, for any $m\in I(a)$ the rows of $m$ lie in $M(a)$ and are, therefore, linear combinations of the rows of $c$, which implies that $m\in M_n(A)c$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{t:matrixpoly}]
	Let $A:= F[\xi_1,\dots,\xi_s]$, $K$ its field of fractions, $R$ a central simple algebra and $\varphi\colon R\to R\otimes M_n(A)$ a homomorphism. Since $M_n(K)$ is SN, there exists (after clearing denominators) $a\in R\otimes M_n(A)$, invertible in $R\otimes M_n(K)$, such that $\varphi(x)=a(x\otimes 1)a^{-1}$ for all $x\in R$.

	Fix any prime ideal $P$ of $A$. Then $M_n(A_P)/{\rm rad}(M_n(A_P))$ is canonically isomorphic to the simple algebra $M_n(F)\otimes A_P/PA_P$, see Lemma~\ref{lemrad}. Therefore, $M_n(A_P)$ is semilocal and by Theorem~\ref{thm:semiLoc} it is also an SN algebra.

	It follows that there exists an invertible $u_P\in R\otimes M_n(A_P)$ such that $\varphi(x)=u_P(x\otimes 1)u_P^{-1}$. Then $u_P^{-1}a$ commutes with all elements of $R\otimes 1$ and thus lies in $1\otimes M_n(A)$. This means, $a$ can be factored as $a=u_P(1\otimes c_P)$ for some $c_P\in M_n(A)$. By Lemma~\ref{l:denominators}, this implies that the $A_P$-module $A_PM(a)$ is free of rank $n$. Since the prime ideal $P$ was arbitrary, this shows that $M(a)$ is locally free of rank $n$. As being projective is a local property, this implies $M(a)$ is projective. By the Quillen-Suslin theorem $M(a)$ is free of rank $n$. We choose $c\in M_n(A)$ such that its rows form a basis of $M(a)$. Then again from Lemma~\ref{l:denominators}, we get a factorization $a=u(1\otimes c)$ where $u\in R\otimes M_n(A)$ is invertible. Now $\varphi(x)=u(x\otimes 1)u^{-1}$ for all $x\in R$.
\end{proof}

\section{Formal power series}\label{s:poly}

The aim of this section is to show that the property of being an SN algebra transfers from  $S$ to the formal power series algebra $S[[\xi]]$.

\begin{theorem}\label{pow}
$S$ is an SN algebra if and only if $S[[\xi]]$ is an SN algebra.
\end{theorem}

\begin{proof}
$(\Rightarrow)$ Let $R$ be a central simple algebra and let $\varphi:R\to R\otimes S[[\xi]]$ be a homomorphism. Since $R$ is finite-dimensional, we can identify $R\otimes S[[\xi]]$ with $(R\otimes S)[[\xi]]$ and write
$$\varphi(x)= \varphi_0(x) + \varphi_1(x)\xi+ \varphi_2(x)\xi^2 + \dots$$
where $\varphi_i:R\to R\otimes S$. Note that $\varphi_0$ is an algebra homomorphism. By assumption, there exists an invertible element $a\in R\otimes S$ such that
$\varphi_0(x)=axa^{-1}$ for all $x\in R$. Considering the map $x\mapsto a^{-1}\varphi(x)a$
we see that without loss of generality we may assume that $\varphi_0(x)=x$ for all $x\in R$, so that
\begin{equation}\label{ena1}\varphi(x)= x + \varphi_1(x)\xi+ \varphi_2(x)\xi^2 + \dots\end{equation}
Now apply Lemma \ref{l}. Thus, let $\{r_1,\dots,r_d\}$ be  a basis of $R$ and let $c_1,\dots,c_d\in R\otimes S[[\xi]]$ be such that \begin{equation}\label{ena2}
\sum_{k=1}^d c_kr_k =1\,\,\,\mbox{  and
}\,\,\,\varphi(x)c_k = c_kx\end{equation} for all $x\in R$  and all  $k$. Writing $$c_k = \sum_{j=0}^\infty c_{kj}\xi^j,$$
where $c_{kj}\in R\otimes S$, it follows from \eqref{ena1} and  \eqref{ena2}  that
\begin{equation}\label{ena3}xc_{k0} = c_{k0}x\end{equation}
 for all $x\in R$. Let us write $c_{k0} = \sum_j p_{kj}\otimes s_{kj}$
with the $s_{kj}$'s linearly independent. From \eqref{ena3} we infer that $$\sum_j (xp_{kj} - p_{kj}x)\otimes s_{kj} =0$$
for all $x\in R$, yielding $xp_{kj} - p_{kj}x=0$. Since $R$ is central this means that  each $p_{kj}$ is a scalar multiple of $1$. Accordingly, each $c_{k0}$ is of the form
$1\otimes t_k$ for some $t_k\in S$. From the first identity in $\eqref{ena2}$ one easily deduces that $\sum_{k=1}^d r_k \otimes t_k = 1\otimes 1$. Writing $1=\sum_{k=1}^d\lambda_k r_k$, where $\lambda_k\in F$, it follows that $t_k=\lambda_k 1$. We may assume that $\lambda_1\ne 0$. Accordingly, $c_{10}$ is a nonzero scalar multiple of unity of $R\otimes S$, implying that $c_1$ is invertible in $(R\otimes S)[[\xi]]$. Applying $\eqref{ena2}$ we arrive at $\varphi(x)= c_1xc_1^{-1}$ for all $x\in R$.

$(\Leftarrow)$ Straightforward; more generally, the SN property is clearly preserved by retractions. Here an algebra $S'$ is a retract of $S$ if $S'\subset S$ and there exists a homomorphism $\pi:S\to S'$ that restricts to the identity map on $S'$.
\end{proof}


text:
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News for Anime Festival Orlando 2008:
Anime Festival Orlando 9 announces YouTube contest
Orlando, FL (May 15, 2008) - Anime Festival Orlando is giving fans of the event a chance to win two weekend passes to its convention being held at the Wyndham Orlando Resort from August 14 - 17, 2008. Each year, attendees of AFO come from all over the world, so AFO wants to know: "How Do You Plan To Get To AFO?"
To participate in the contest, fans will make a fun and entertaining video about how they plan to get to AFO9. Their vignette can be up to 3 minutes long and as Anime Festival Orlando is always a family event, all entries must be PG. Videos must be uploaded to YouTube™ by the deadline of 11:59pm on June 30, 2008. AFO staff will choose the winner, but entries will be screened for viewers at Anime Festival Orlando.
The winner of the AFO9 YouTube™ Contest will receive two 3-day membership passes to AFO9 with the "Passport to Orlandia" upgrade, two vintage AFO t-shirts as well as a $50.00 Visa Gift card: over $200 worth of prizes!
For a complete listing of contest details, rules and a complete description of the prize package visit www.AnimeFestivalOrlando.com
About Anime Festival Orlando
AFO is a multi-day celebration of all things related to Japanese animation and pop culture held at the Wyndham Orlando Resort August 15, 16 and 17, 2008. Ticket prices are only $45.00 for an adult 3-day membership badge, $20.00 for a Friday or Sunday Only wristband and $25.00 for a Saturday Only wristband. Growing in attendance nearly 20% each year since its inception in 2000, Anime Festival Orlando has earned the title of "Florida's Favorite Anime Convention" as well as being voted one of the "Top 10 Things To Do In Orlando" by DK Publishing's Travel Guide.
This year, Anime Festival Orlando's staff has crafted an original story/game called "Swords of Orlandia" and will feature an interactive RPG story for our attendees to voluntarily participate in; in addition to the traditional fare offered at anime conventions.
For more information about Anime Festival Orlando, please visit http://www.animefestivalorlando.com/
The above content is a press release which was submitted to FanCons.com and appears as submitted without modification. Statements within the press release should not be taken as any sort of endorsement by FanCons.com. This site claims no responsibility as to verifying the accuracy of the information of this press release.
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redpajama_set_name: RedPajamaCommonCrawl


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Oxaprozin is a tablet that is taken orally, once or twice a day with water and with or without food. The dosing regimen will be determined by your healthcare provider depending on the condition being treated and your medical profile. Do not take more or less than the prescribed amount.
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Oxaprozin is the generic version of the brand name medication Daypro. At HelpRx, we have oxaprozin price-reducing coupons available for those who have been prescribed this medication. When used at local pharmacies, these coupons can lower oxaprozin costs by up to 43% when filling a prescription. Download or print yours and present it to take advantage of the savings.


What is the difference between the gaming gears of ASUS?
Speaking about the gaming gears, STRIX of course will be the number one choice for gaming usage. I would use this kind of comparison, such as for when you are in a game you want to be the king, you want to win the game so STRIX will be the best choice. But when we talk about Republic of Gamers, at the beginning it is actually starting from the overclocking so it actually represents the fuller performance, so it is the king of kings. Echelon is actually at the same level as the STRIX series, but it is oriented towards the shooting games, designed for the army lovers and for gamers who love especially this kind of games.
When can we expect Republic of Gamers keyboard to hit the market?
For the Republic of Gamers keyboard we estimated that we will launch it on the Bulgarian market in 2016. The reason it takes longer is that we actually receive a lot of user feedback and questions like: "when will we have Republic of Gamers keyboard?", but to make sure that we will be the best of the best, it takes a little bit longer to hit the market, but we are pretty sure that we will be the best performers for the product.
When will we see Republic of Gamers Spatha mouse on the Bulgarian market?
Right now we are in the final stage for Spatha. We expect it to hit the market at the same time as ROG keyboard.
Not too long ago you introduced STRIX sound cards. Why do we need discreet sound card, when we have one embedded in the motherboard?
The embedded solution, especially on the ROG motherboard is absolutely fantastic, but why we need discreet sound card is easy to tell. For example with it we can have the signal isolated and operating and we can have better power control.
Tell us a little bit more about the STRIX 7.1 headphones.
I know a lot of people will compare out product with those of our competitors. It is really difficult to make a real multi channel sound for a headset, but we chose a different approach, compared with other brands. For example we use a USB direct connection which makes the sound truly multi channel, compared to other brands using the analog connection. This is the main difference. Also for the 7.1 you actually listen more carefully and distinguish the different sounds and the direction they are coming from. This is a major benefit for the gamers.
What else is in your gaming gears portfolio?
Besides the previously mentioned STRIX, Echelon and ROG we also have new series which is named Cerberus. It is the most basic gear, but it is still a high end, for the high cost and performance ratio. For example for the Cerberus in terms of keyboards we think that the most important thing is back to basic. In this range of the product we introduced the rubber foot and also there is a metal plate in the back and anti-splashing function that can help. We aim to build loyalty to our users. We want the mouse to last longer for gaming usage, because we think that for long time usage comfort is the most important thing. So in this kind of range we focus on the basic functions. In terms of headphones, the Cerberus aims to be used not only for games or at home, but also the user can listen while walking, because when we are talking about multy function you must be able to use them for gaming or just connect it to your phone.
Are all of those products available at the Bulgarian market?
Cerberus is available since the beginning of December. We really look forward to get feedback from the local users. STRIX is also available and when we launch the ROG, we will make sure that Bulgaria is our first option.


text:
Reasons why the sky isn't falling (yet) for FC Dallas
Oscar Pareja has done a great job at advancing things forward in 2018.
By Drew Epperley@wvhooligan Jun 26, 2018, 12:30pm CDT
Share All sharing options for: Reasons why the sky isn't falling (yet) for FC Dallas
Two straight weeks of losses for FC Dallas (one in the US Open Cup, the other being a blow out in New York), the sky seems to be getting darker for fans as we recall what happened this time last year.
While it is easy to believe the sky is going to fall on this season based on last year, 2018 has felt different from day one with this group. Here are some things that stand out to me as I look at the summer and how FC Dallas should be able to handle things better moving forward.
Depth is actually there
I know the recent sell of Anton Nedyalkov is a tough one to wrap your head but aside from any potential money that comes in from that transfer, FC Dallas is actually in a decent position here with Ryan Hollingshead and Maynor Figueroa as capable stand-ins until either a full-time replacement is brought in this summer or someone steps up and really takes this starting role for their own.
The depth is strong with this roster, most so than what we've seen in the past. Just look at what Oscar Pareja has been able to do in the central midfield this season with the rotation of Jacori Hayes, Kellyn Acosta, Carlos Gruezo and Victor Ulloa. The same can be said for how the wings have been rotated before all the injuries popped up with Santiago Mosquera.
I'd also throw in the fact that we still haven't seen some of the new Homegrown signings yet in league play, but unlike last year, they're getting a solid run in the USL. Their time will come soon enough.
Formation rotation
The one thing that stuck out to me last year was Pareja's lack of willingness to move away from certain formations and looks. Half of the season (when the team was successful) was in a 4-4-2. The rest of the year was mostly in a 4-2-3-1 when Mauro Diaz came back.
This season has been massively different with a new look almost every two or three games. The rotation of the roster has kept players fresh (maybe too fresh in some cases), and has given some extra motivation in training to players to step up and battle for their spot.
Not all looks have worked out, as we saw last weekend in New York with the 3-5-2. But the 4-3-3 with Diaz as a false wing has been something that other teams haven't sorted out just yet.
Speaking of Diaz...
I know it is easy to get frustrated by the constant resting and holding him out of games, especially road games but the reasoning behind it makes sense. You want him for the long haul, most importantly, for the playoffs.
Do you really need him for a random game in the middle of the summer on the road in New York when he may not be 100%? Sure, but wouldn't it be better to have him at 100% for four or five games in a row when it matters? Yes.
FC Dallas is taking a more careful approach to his health this season, which is understandable when you consider the types of injuries he has had since he arrived here in Dallas. The four games on, two games off approach isn't the worst if he is able to return each time and contribute the way he has. Each time he has not made the field for a game this season, the very next time out he has either brought in a goal or an assist.
Mauro Diaz stats by each game in 2018
Like some folks are saying, the club has lost two games this season. Just two! It is easy to get sucked into the dark thoughts on this club based on what we've seen in the past but Pareja and company have done a good enough job this season and making sure history doesn't get repeated for a second straight summer.


Usually where danger is involved we simply say something like "please know what you're doing" or "If you need to ask how this is done, this is not for you", but for this particular piece, I thought a bigger warning should be in place. And I darn well mean it.
WARNING: This article involves dealing with High Voltage, up to 300v, which is stored in big capacitors. THIS VOLTAGE CAN KILL YOU. Never touch any of the components and always discharge the capacitor before working on it. Always keep in a well insulated project enclosure.
BY FOLLOWING THESE INSTRUCTIONS YOU AGREE TO BE THE ONLY RESPONSIBLE FOR ANY DAMAGE THAT MIGHT BE CAUSED BY DOING SO. WORK CAREFULLY AND FOLLOW WARNINGS.
This article will help you understand the basics of how a flash circuit is working, and the second part will show you how to build your own optical triggered portable flash rated at 50 watt/seconds!
How Does A Flash Work In A Nutshell?
There really is no explanation of electronics without a scheme, so we placed a scheme below. While it does not go in depth about each of the parts it explains in a nut shell what's its name and what it does. The magic of it, of course, is that you can create quite an effective burst of light, from a few 1.5V AA batteries.
The power from a low voltage battery (1) is fed into the inverter transformer (3) through a transistor (2), that is switching the current on-and-off, in a few Kilo-Hertz frequency (thousands of times per second).
The transformer (3) outputs high voltage of about 300 volts to a big capacitor (4) which will charge during a few seconds and serves as an energy storing device.
Meanwhile (as they say in comic books), the small trigger capacitor (5) is being charged. As soon as a burst of light is detected (from firing another flash) the photo-transistor (6) is translating the light-burst to a small voltage, that will trigger the Trigger Coil (7). The coil then produces a short pulse of very high voltage (6000 volts), which will ignite ionize the Xenon gas inside the Flash tube (8). Once ionized, it will draw all the energy stored in the big capacitor (4) and will output this energy in the form of light (the final flash).
What Are The Steps To Building A Strobe?
The first step is to go back and read the warning at the top of the post.
Then, start with deciding how much power (or how much light) you want your flash to output. Since the power to the flash tube is derived from the big capacitor (4 in the previous drawing), you control the power by controlling the size of that capacitor. Here is how to calculate the capacitor size.
As all things energy, flash output works kinda like exposure on P mode. You start with a base value for flash power, say 100w/s flash (or 100 Joules). To get this kind of flash you need to feed the bulb from a 2000uF capacitor at 330v.
Cut the capacitor in half, and you get half the power: a 1000uF capacitor will give you about 50w/s.
Actually, as a thumb rule, every 1000uF capacitor added will give you about 50w/s of energy.
Of course, once you have the power calculated you need to decide on a bulb capable on carrying that amount of power without exploding (yes Exploding).
Unlike incandescent lamps, flash tubes are rated for their MAXIMUM energy.
That means that a flash tube that is rated 200w/s MAX, can be fed with energy anywhere from 0 to 200 w/s. This exact lamp with a 1000uF capacitor will output 50w/s, and with FOUR(4) 1000uF capacitors will output 200w/s – its maximum allowed energy per flash.
So to recap, the capacitors are defining the flash energy, and the lamp has to have a compatible rating.
The rest are fixed components that have little dependency on the flash output which you wouldn't have to change, we are listing them here as a reference.
We like to use the XFT-1610 as the inverter transformer, and TC-36 as trigger coil. They are very compact and always work great.
The following circuit diagram shows an exact basic circuit. It works reliably and gives good results, accepting 4xAA batteries as a portable power source, or a 5v transformer for external power, and delivering 50 w/s of light.
This basic circuit is great, because you can easily modify it – add more capacitors for higher flash power, change the flash tube to fit in reflectors, or even connect several flash tubes in parallel to place them in a circular reflector and create a Ring-flash.
Just a small reminder, the most important of course is, staying alive. Please take great care when trying to assemble this high voltage circuit.
After powering it, never touch any of the soldering or the components, always use an insulated plastic project box, and always use a volt meter to verify that the big capacitor is EMPTY before you are working on the circuit.
Avner Richard is a fashion photographer and an electronics tinkerer. You can find his photography here, and his tinkering here.
Fantastick article, is the prin available?
thanks great article! This is a very interesting post for me. I have one question, how can I extend the circuit to have a variable flash energy? So may 100%, 50% and 25% of the flash energy. I think I have to add other capacitors but don't know how.
I want to use a micro controller to enable different flash energy 100%, 50% and 25%.
So how can I extend this circuit to change the flash energy by micro controller?
How can I activate different capacitors by my micro controller?
its me again. Is there any reflector available that I can use this flash inside my DIY camera?
any suggestions on how to tweak the circuit to work on a large power source?
To convert to 220V You don't need any additional cirquitry, just get rid of first 3 parts in cirquit diagram. All these parts do is converting DC to 220V AC anyway.
You could die if you try this. You'd need galvanic separation and current limiting and a fuse at least. Symmetric power usage would be nice, too.
That said, I'm not an electrician. A specialist might see even more problems.
But how do you connect this to a camera???
Hi, do you think it's possible to create a remote, portable flash unit that straps to your wrist and has the head in the palm of your hand? Flash street photographers could use it to stay a bit more stealthy.
if I want to power this from a 12 volt battery, would I need to do anything besides change to an XFT-1934 transformer?


text:
Sam Burns signs on loan from Blackburn Rovers
Scunthorpe United is delighted to have completed the loan signing of forward Sam Burns from Championship side Blackburn Rovers for the rest of the season.
The 19-year-old, who has been with Rovers since Under-14s level, has netted 11 goals for their Under-23s in Premier League 2 and the Premier League Cup so far this campaign and is tipped to have a bright future in the game.
Talking about his move to United, he told iFollow Iron: "The move came about quite quickly, over the duration of a week really, but it's great to be here and get going.
"It's been good training with the players already. It's been top drawer and different to what I'm used to at Under-23s level, but I'm buzzing to get in and work hard, give what the manager wants and help the team.
"He's told me to work hard and work my way into the side, and to show my quality. I have pace and I'm quick at running in behind, and can hopefully bring some goals to the team.
"I'm looking forward to playing first team football and getting some goals while I'm here. I can play anywhere across the front but I see myself as a striker."
Burns enjoyed a loan spell with FC United of Manchester in the Northern Premier League Premier Division earlier this term, scoring three times during his month there, and getting used to men's football.
He added: "The physicality at FC United of Manchester earlier this season was different to what I'm used to, but I enjoyed it and it was really beneficial for me, gearing me up for league football. This is now the next step for me and I can't wait to get started."
The transfer is subject to EFL and FA ratification.
See more from Sam Burns on iFollow Iron.
Sam Burns
Loan signing
📹 iFollow: Sam Burns speaks after signing on loan for the Iron
Striker Sam Burns speaks after joining United on loan from Blackburn Rovers for the rest of the season.


Touchstone is a charity that specialises in delivering drama projects all over the UK to learners with Autistic Spectrum Condition (ASC).
At Touchstone we believe that every learner should be allowed to express, play, create and explore their imagination.
Our highly skilled practitioners have developed exercises and workshop structures to allow the ASC Learner a safe and happy environment to flourish.
We deliver workshops all over the UK and beyond.
Please get in touch with us and let us know what sort of workshop you would like us to deliver for you, and when you might like us to visit.


text:
enum Verbosity {
    V_NONE =0,
    V_FATAL,
    V_CRITICAL,
    V_ERROR,
    V_WARN,
    V_INFO,
    V_DEFAULT = V_INFO,
    V_DEBUG,
    V_TRACE,
    V_MAX = V_TRACE
};

enum LogOption {
    LOG_ECHO    =0x01,          // Also log to:
    LOG_STDERR  =0x02,          //  stderr (stdout is default)
    LOG_SYNC    =0x04,          // Force sync after each write
    LOG_UTC     =0x08,          // Log time in UTC
    LOG_FILE    =0x10,          // Log to file
    LOG_NOECHO  =0x20,          // Don't echo log to stdout/stderr
};

// Simplified log4j style interface without the overhead of hierarchical loggers,
// configurable handlers and formatters etc. etc.

extern void log_init(int flags, enum Verbosity verbosity, char const *logpath);
extern void log_rotate();               // close current log (if any), start a new one
extern char const *log_name();          // current log name (NULL if none)

// Logging functions
int log_log(enum Verbosity verbosity, char const *fmt, va_list args);
extern void log_fatal(int ec, char const *fmt, ...) __attribute__((format (printf, 2, 3))) __attribute__((noreturn));

extern int log_critical(char const *fmt, ...) __attribute__((format (printf, 1, 2)));
extern int log_error(char const *fmt, ...) __attribute__((format (printf, 1, 2)));
extern int log_warning(char const *fmt, ...) __attribute__((format (printf, 1, 2)));
extern int log_info(char const *fmt, ...) __attribute__((format (printf, 1, 2)));
extern int log_debug(char const *fmt, ...) __attribute__((format (printf, 1, 2)));
extern int log_trace(char const *fmt, ...) __attribute__((format (printf, 1, 2)));
extern int log_message(enum Verbosity verbose, char const *fmt, ...) __attribute__((format (printf, 2, 3)));

#endif //GENERIC_LOGGING_H

redpajama_set_name: RedPajamaGithub


Worried that owning up to using the drug could jeopardize their VA benefits — even if they're participating in a medical marijuana program approved by California or 28 other states and the District of Columbia — veterans have often kept mum. That may be changing under a new directive from the Veterans Health Administration urging vets and their physicians to open up on the subject. California also this month began allowing the sale of recreational marijuana.
Moves by states to legalize marijuana for medical or recreational use have created a confusing landscape for patients to navigate. Attorney General Jeff Sessions announced last week he would rescind an Obama-era policy that discouraged federal prosecution for marijuana use in states where it is legal. That action has further clouded the issue.


The new PE tops/jumpers are only a requirement for current Year 7 students and is available through Uniform Direct.
Current years 8, 9, 10 & 11 can continue to wear the old PE top until it needs replacing.
Religious headwear (headscarves, topis, turbans) must be plain black or white with no decoration on them. These must be tucked into the collar of the shirt so that the full school tie is visible at all times.
The college is a professional environment. Make-up must be discreet and kept to a minimum. Hair colour should be within the natural range of colours. Students must not have extreme hairstyles or colours which would be deemed unacceptable in a business environment. The academy reserves the right to deem what is considered acceptable.
The college allows students to wear a discreet, single nose stud, nose rings are not allowed. No other visible facial piercings are allowed, this includes clear retainers. We suggest that any new piercings are done over the summer holidays so that they can heal and can be removed for school. Earrings must be discreet and no bigger than a two pence piece. Pupils are permitted to wear a necklace for religious purposes but this must be kept inside the shirt at all times. All jewellery must be removed for PE. Only school badges should be worn on blazers.
Students may choose to wear an outdoor coat, fitting to the weather conditions. This should be black or navy, with no obvious logos and must be removed once indoors. Hoodies are not allowed to be worn.


All > Accessories > Premiumpatibles Inc.
Shop our selection of Premiumpatibles Inc. accessories at Unbeatablesale.com and get the biggest markdowns today! Unbeatablesale.com has over 700,000 products in stock with 182 of them being Premiumpatibles Inc. accessories and related items.
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Memberships at the Sponsor level and above include a Friends book bag; $5 is included for shipping.
Membership is valid for one year, and your contribution is tax-deductible to the full extent allowed by law.
If you also want to make a contribution to the Friends' Book Fund, click here.
A card confirming your gift will be sent to the recipient. Please note that donations at the Sponsor level and above include a Friends of the Chula Vista Library gift bag. Rate includes $5 for shipping.
Pressing the "Submit" button will take you to our payment page and secure website.


text:
package com.github.epiicthundercat.tameablemobs.mobs.itementities;

import com.github.epiicthundercat.tameablemobs.mobs.TameableBat;

import net.minecraft.entity.EntityLiving;
import net.minecraft.entity.EntityLivingBase;
import net.minecraft.entity.player.EntityPlayer;
import net.minecraft.entity.projectile.EntityThrowable;
import net.minecraft.potion.Potion;
import net.minecraft.potion.PotionEffect;
import net.minecraft.util.DamageSource;
import net.minecraft.util.EnumParticleTypes;
import net.minecraft.util.math.RayTraceResult;
import net.minecraft.world.World;

public class EntityBatPee extends EntityThrowable {

	public static boolean entity;
	public static boolean player;
	public static boolean block;
	private EntityLivingBase shootingEntity;

	public EntityBatPee(World worldIn) {
		super(worldIn);
	}

	public EntityBatPee(World worldIn, EntityLivingBase shooter) {
		this(worldIn, shooter.posX, shooter.posY + (double) shooter.getEyeHeight() - 0.10000000149011612D,
				shooter.posZ);
		this.shootingEntity = shooter;

	}

	public EntityBatPee(World worldIn, double x, double y, double z) {
		this(worldIn);
		this.setPosition(x, y, z);
	}

	protected float getGravityVelocity() {
		return -0.01F;

	}

	@Override
	public void onUpdate() {
		super.onUpdate();
		World world = this.worldObj;
		int x = this.ticksExisted;
		if ((this.ticksExisted % 2) == 0) {
			world.spawnParticle(EnumParticleTypes.REDSTONE, this.posX, this.posY, this.posZ, 1.0, 1.0, 0.3);
		}

	}

	@Override
	protected void onImpact(RayTraceResult result) {
		if (result.entityHit != null && result.entityHit != this.shootingEntity) {
			result.entityHit.attackEntityFrom(DamageSource.causeMobDamage((TameableBat) this.shootingEntity),
					10.0F);
			}
		World world = this.worldObj;
		if (result.typeOfHit.equals(result.typeOfHit.ENTITY)) {
			if (result.entityHit instanceof EntityPlayer && player) {
				EntityLivingBase entity = (EntityLivingBase) result.entityHit;
				if (!world.isRemote) {
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(1), 700, 5, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(5), 700, 5, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(6), 700, 5, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(10), 700, 5, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(21), 700, 5, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(22), 700, 4, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(22), 700, 4, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(23), 700, 4, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(26), 700, 4, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(11), 700, 5, false, false));
				}
			}
			if (result.entityHit instanceof EntityLiving && entity) {
				EntityLivingBase entity = (EntityLivingBase) result.entityHit;
				if (!world.isRemote) {
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(2), 400, 3, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(20), 400, 4, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(24), 300, 5, false, false));
					entity.addPotionEffect(new PotionEffect(Potion.getPotionById(25), 500, 5, false, false));

				}
			}
			if (result.typeOfHit.equals(result.typeOfHit.BLOCK) && block) {
				if (!world.isRemote) {
					this.setDead();
				}
			}

		}
	}
}


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PEOPLE OF THE PHILIPPINES, APPELLEE, VS. NARCISO RAMOS Y MATIAS, RAMON SAN ROQUE Y DELA CRUZ, EULALIA SAN ROQUE DE FRANCISCO Y DELA CRUZ ALIAS LALING, WILLIAM RAMOS ALIAS WILFREDO RAMOS, (PROVISIONALLY DISMISSED), AND THREE (3) OTHER JOHN DOES, ACCUSED.
EULALIA SAN ROQUE DE FRANCISCO y DELA CRUZ alias "LALING", appellant.
Appeal by Eulalia San Roque de Francisco y dela Cruz from the Decision dated April 24, 1998 of the Regional Trial Court, Branch 122, Caloocan City, in Criminal Case No. C-46010, declaring her guilty beyond reasonable doubt of the crime of murder and sentencing her to suffer the penalty of reclusion perpetua. She was also adjudged to pay the heirs of the victim, P50,000.00 as civil indemnity.
Upon arraignment on December 15, 1994, appellant, assisted by counsel, pleaded not guilty to the crime charged.
The case against Wilfredo Ramos was provisionally dismissed. The other accused, Narciso Ramos and Ramon San Roque, have remained at large.
On February 11, 1993 at around 7:00 o'clock in the evening, Bernie Ambal was standing outside his store at 168 De Paro St., Caloocan City. Narciso Ramos, Ramon San Roque and three (3) others passed by. They proceeded to the house of William Lomida and appellant, who were then live-in partners. Narciso and Ramon stood by the door, while one of their companions, holding an armalite, positioned himself behind Narciso. Their other companion, armed with a pistol, stayed on the street, and the third one, also armed with a pistol, went to the backyard. At a distance of ten (10) meters away, Ambal saw Narciso pulling out his .45 caliber pistol and knocking at the door. Appellant then opened the door and Ramon went inside.
Soon thereafter, William and appellant, accompanied by Narciso, Ramon, and three (3) others left the house. As they were passing by the store, Ramon stopped and borrowed Ambal's jacket. At that instance, William suddenly held his arm and whispered, "Samahan mo naman ako, baka kung ano ang gawin sa akin ng mga ito, tutal barkada mo naman si Ramon." But Ambal was scared and hesitant. William then requested him to look for Saturnino Rivera.
The group headed to Narciso's house, about ½ kilometer away from Ambal's store. Unknown to them, Ambal trailed behind. Hiding himself behind a tree fifteen meters away, Ambal saw one of their companions poking his armalite at William. Then, they tied William to a santol tree. He was pleading to appellant, but she simply turned her back. Ramon stabbed William twice at the stomach with a 29" bladed knife. Then Narciso shot William five to seven times with his .45 caliber pistol. When William was already dead, Ramon and Wilfredo Ramos untied his body and brought it to a dumpsite (of used tires) twenty five meters away. There they placed William's body atop a pile of rubber tires. Ramon poured gasoline on his body and set it on fire. Appellant and the others were closely watching. After thirty minutes, appellant and the men left. Ambal immediately reported the incident to Saturnino Rivera. They proceeded to the dumpsite where they saw the charred body. According to Ambal, Ramon, appellant and Narciso's sister are brother and sisters.
Saturnino Rivera declared on the witness stand that he considered William his best friend. William and appellant frequently quarreled and sometimes, he maltreated her. Saturnino corroborated Ambal's testimony that they went to the scene of the crime and saw the charred body of William; and that they reported the gruesome incident to the NBI.
Mariano Lomida testified that on February 19, 1993, or eight days after the incident, appellant suddenly arrived in Atimonan, Quezon looking for William. She told Mariano that William left their house on February 9, 1993 without her knowledge. She borrowed P3,000.00 from him (Mariano) with a promise to pay on March 27, 1993. But since then, he never saw her again. Mariano further testified that due to the death of his son, he suffered wounded feelings.
Special Investigator Laurence M. Nidera of the NBI Anti-Organized Crime Division conducted the investigation. He took the statements of Bernie Ambal, Saturnino Rivera and Mariano Lomida. Upon the arrest of appellant and Narciso Ramos by the Capital Command (CAPCOM) of the Philippine National Police, they were turned over to the NBI.
After the prosecution rested its case, appellant filed a demurrer to evidence but was denied. Meanwhile, appellant jumped bail. On the basis of the evidence presented by the prosecution, the case was submitted for decision.
"WHEREFORE, judgment is hereby rendered, finding the accused Eulalia San Roque de Francisco y dela Cruz alias 'Laling' GUILTY beyond reasonable doubt of the crime of murder as charged in the Information and hereby sentences her to suffer the penalty of reclusion perpetua with accessory penalties as provided by the law and to indemnify the heirs of the victim in the sum of P50,000.00.
Appellant appeared during the promulgation of the Decision.
"THE TRIAL COURT ERRED IN HOLDING THAT THE ACCUSED EULALIA SAN ROQUE DE FRANCISCO Y DELA CRUZ alias LALING CONSPIRED AND CONFEDERATED WITH HER CO-ACCUSED IN PERPETRATING THE CRIME OF MURDER, WHEN THE FACT OF SUCH CONSPIRACY HAS NOT BEEN SATISFACTORILY PROVEN TO EXIST BEYOND REASONABLE DOUBT DURING THE TRIAL OF THE CASE.
"THE TRIAL COURT ERRED IN DECIDING THE CASE AGAINST THE ACCUSED-APPELLANT DESPITE THE WEAK EVIDENCE OF THE PROSECUTION, CONSIDERING THAT THE PROSECUTION HAS FAILED TO PROVE ALL THE NECESSARY ELEMENTS OF THE CRIME AND THE CORPUS DELICTI.
"THE TRIAL COURT ERRED IN HOLDING THE ACCUSED GUILTY OF THE CRIME OF MURDER BY THE MERE FACT THAT SHE FAILED TO APPEAR AFTER THE CASE WAS REVIVED SUCH FAILURE HAVING BEEN APPRECIATED BY THE TRIAL COURT AS A CLEAR INDICATION OF HER GUILT.
We shall discuss the above assignments of error jointly.
With treachery, taking advantage of superior strength, with the aim of armed men, or employing means to weaken the defense or of means or persons to insure or afford impunity.
And in the evening of that date, could you recall at about 7:00 in the evening of February 11, 1993, if there was any unusual incident that happened?
Will you please tell us what was that unusual incident all about that you observed and noticed?
Narciso Ramos, Ramon San Roque and three men of Narciso Ramos got the victim and Eulalia San Roque was also with the group.
William Lomida was taken by these people.
From where the accused picked up by these people?
Lomida was taken from the house where Eulalia San Roque and William Lomida were living.
William Lomida is the victim in this case. Now, how is he related to Eulalia San Roque?
They are not married. They are live-in partners.
Now, how far from the store where this house of William Lomida and Eulalia San Roque is located?
More or less, ten meters away.
Is it located in front of the store, on the left side or at the back of the store?
The house of Eulalia is alongside of our house.
What part of your store where you were at the time you saw these people picked up Lomida?
I was outside the store standing.
After the accused picked up the victim, what happened or where did they go?
William Lomida, the victim, was brought to the house of Narciso Ramos.
So the persons whom you saw got Lomida were Narciso Ramos, Ramon San Roque, William Ramos alias Wilfredo Ramos and three men of Narciso Ramos and Eulalia San Roque?
Now, you said that the accused brought William Lomida to the house of Narciso Ramos. How far is this house of Narciso Ramos located in relation to the store, your store?
Now, when the victim was brought to the house of Narciso Ramos, what happened next?
The victim was tied to a santol tree.
Where is this santol tree where the victim Lomida was tied? How far is this from the house of Narciso Ramos?
Ramon San Roque, Wilfredo Ramos and the three others unidentified men.
After the accused tied William Lomida on a santol tree, what happened next?
Ramon San Roque stabbed William Lomida.
And with what kind of weapon was used by Ramon San Roque?
I cannot say how long, sir.
How many times did you see Ramon San Roque stabbed the victim with the 29?
And did you see if Lomida was hit with that two stab thrust?
He was stabbed on the stomach.
What happened after Ramon San Roque stabbed the victim?
The victim, William Lomida 'nangisay' and then Narciso Ramos came near the victim and shot him.
About 5 to 7 shots.
And did you see if the victim was hit?
It seems that the shots were directed to his face because he was bloodied all over his face.
With what kind of firearm, if you remember?
When the investigator showed me some guns, I noticed that the one gun that Narciso Ramos used was a .45 caliber.
Was it magazine or revolver?
After Narciso Ramos shot the victim 7 times on the head, what happened next?
When the group made sure that the victim was dead already, they untied him and brought the body where there was a pile of tires, piles of pieces of rubber tires.
By the way, at the time when the victim was shot by Narciso Ramos and stabbed by Ramon San Roque, where was Willy Ramos, Narciso Ramos and Eulalia San Roque?
They were just there standing and afterwards, turned their back around.
How far was Eulalia San Roque from Lomida when Lomida was shot by Narciso Ramos and stabbed by Ramon San Roque?
This place up to that wall. Distance of about 8 meters.
And what was Eulalia doing at the time when the victim was stabbed and shot?
She turned her back and face the kitchen of the house of Narciso Ramos.
Did she not try to intervene, this Eulalia, being the live-in partner of William Lomida or did not exert effort to pacify?
Just turning her back, what else that she did, being live-in partner of William Lomida?
She did not do anything anymore.
After the victim was stabbed and shot, he was untied from the santol tree. Who untied him?
It was Ramon San Roque, Willy Ramos and the three others.
Willy refers to Wilfredo Ramos?
Where was Eulalia at the time?
She was still there standing.
Now, you said after the victim was untied from the santol tree, he was brought to where pieces of rubber tires were piled, recycled, how far is this from the santol tree where the victim was tied?
Now, after the accused brought the victim to the piles of recycled tires, what happened next?
Ramos San Roque got a can and poured something on the body of the victim and then, lighted it up.
What happened to the body of the victim, William Lomida?
At the time when this Ramon San Roque poured something on the body of the victim and then lighted the same, where were Wilfredo Ramos, Eulalia San Roque and Narciso Ramos and the other unidentified persons?
There were – They were there in front of the burning body.
It took a long time, about more than one hour.
After one hour, what happened next?
They left the place. The group left the place.
What happened to the tires?
And where was the body of Lomida in relation to the tires?
We are not persuaded by appellant's contention that the prosecution failed to adduce sufficient evidence to establish the existence of conspiracy among the accused. She vigorously contends that she did not participate in the killing of the victim.
In determining the existence of conspiracy, it is not necessary to show that all the conspirators actually hit and killed the victim. The presence of conspiracy among the accused can be proven by their conduct before, during or after the commission of the crime showing that they acted in unison with each other, evincing a common purpose or design. There must be a showing that appellant cooperated in the commission of the offense, either morally, through advice, encouragement or agreement or materially through external acts indicating a manifest intent of supplying aid in the perpetration of the crime in an efficacious way. In such case, the act of one becomes the act of all, and each of the accused will thereby be deemed equally guilty of the crime committed.
The series of events in this case convincingly show that appellant and her co-accused acted in unison and cooperated with each other in killing William Lomida. Appellant was the one who opened the door and allowed the other accused to enter the house. She joined them in bringing the victim to the residence of Narciso Ramos, her brother-in-law. While her co-accused dragged the helpless victim, tied him to a santol tree, stabbed him twice by a bladed knife, and shot him 5 to 7 times, appellant merely watched intensely. She even "turned her back" as the lifeless body of the victim was being burned. And after attaining their purpose, she fled with the other accused.
The above circumstances clearly show the common purpose and concerted efforts on the part of appellant and her co-accused. We agree with the trial court in concluding that their acts were indications of a criminal conspiracy to commit the crime of murder.
The only remaining question is whether the crime was attended by aggravating circumstances.
The killing of the victim was attended by treachery. Treachery exists "when the offender commits a crime against persons, employing means, methods or forms in the execution thereof which tend directly and specifically to insure its execution, without risk to himself arising from any defense or retaliatory act which the victim might make." Here, appellant and her co-accused tied William to a santol tree before they stabbed and shot him to death, thus, insuring the execution of the crime without risk to themselves. Obviously, he could not retaliate. This aggravating circumstance qualifies the crime to murder.
However, we cannot sustain the trial court's appreciation of the aggravating circumstance of superior strength as this is absorbed in treachery.
It bears stressing that this crime of murder was committed on February 11, 1993. The law applicable is Article 248 of the Revised Penal Code then penalizing murder with reclusion temporal in its maximum period to death. Under Article 64 (1) of the Revised Penal Code, in cases in which the penalties prescribed by law contain three periods, whether it be a single divisible penalty or composed of three different penalties, and there are neither aggravating nor mitigating circumstances that attended the commission of the crime, the penalty prescribed by law in its medium period shall be imposed.
The range of the imposable penalty, i.e., reclusion temporal in its maximum period to death, is 17 years, 4 months and 1 day to death. Applying the Indeterminate Sentence Law, and there being no aggravating or mitigating circumstance that attended the commission of the crime, the maximum period is the medium of the imposable penalty, which is reclusion perpetua. The minimum period is one degree lower, or prision mayor in its maximum period to reclusion temporal in its medium period, the range of which is 10 years and 1 day to 17 years and 4 months. The minimum period of the penalty imposable is anywhere within this range, or 10 years and 1 day. Hence, appellant should be sentenced to 10 years and 1 day of prision mayor, as minimum, to reclusion perpetua, as maximum.
Regarding damages, the trial court correctly awarded P50,000.00 as civil indemnity to the victim's heirs. When death occurs as a result of a crime, appellant should be ordered to pay the heirs of the victim P50,000.00 as civil indemnity, without need of any evidence or proof of damages.
We likewise award temperate damages, in lieu of actual damages. Here, the prosecution failed to present any proof of the expenses incurred by the victim's heirs. However, as they actually incurred funeral expenses, we award P25,000.00 by way of temperate damages.
Anent moral damages, we award the victim's heirs the amount of P50,000.00. For verily, moral damages are not intended to enrich the victim's heirs; rather they are awarded to allow them to obtain means for diversion that could serve to alleviate their moral and psychological sufferings. Mariano Lomida, victim's father, equivocally described how he suffered untold wounded feelings for the loss of his son.
We also award the victim's heirs P25,000.00 as exemplary damages. This is pursuant to our ruling in People vs. Catubig that if a crime is committed with an aggravating circumstance, either qualifying or generic, an award of P25,000.00 as exemplary damages is justified.
WHEREFORE, the assailed Decision dated April 24, 1998 of the Regional Trial Court, Branch 122, Caloocan City, in Criminal Case No. C-46010, is hereby AFFIRMED with MODIFICATION in the sense that appellant EULALIA SAN ROQUE DE FRANCISCO is sentenced to suffer the penalty of 10 years and 1 day of prision mayor, as minimum, to reclusion perpetua, as maximum. She is ordered to pay the victim's heirs (a) P50,000.00 as civil indemnity; (b) P25,000.00 as temperate damages; (c) P50,000.00 as moral damages and (d) P25,000.00 as exemplary damages.
 Penned by Judge Remigio E. Zari, Rollo at 12-15.
 TSN, February 13, 1996 at 3-17.
 G.R. No. 138609, January 17, 2001, 349 SCRA 385, 399, citing People vs. Benito, 303 SCRA 468 (1999).
 People vs. Amazan, G.R. Nos. 136251, 138606 & 138607, January 16, 2001, 349 SCRA 218, 234, citing People vs. Alib, 322 SCRA 93 (2000).
 People vs. Visaya, G.R. No. 136967, February 26, 2001, 352 SCRA 713, 720-721.
 People vs. Berdin, G.R. No. 137598, November 28, 2003 at 16-17, citing People vs. Bolivar, 352 SCRA 438 (2001); Article 14, par. 16, Revised Penal Code.
 People vs. Ampo-an, G.R. No. 75366, July 4, 1990, 187 SCRA 173, 189; People vs. Manzanares, G.R. No. 82696, September 8, 1989, 177 SCRA 427.
 The crime was committed before Republic Act No. 7659 which took effect on January 1, 1994.
 People vs. Berdin, supra at 18, citing People vs. Joey Manlansing y Ambrosio and Mario Manlansing y Ambrosio, G.R. Nos. 131736-37, March 11, 2002.
 People vs. Abrazaldo, G.R. No. 124392, February 7, 2003.
 People vs. Solamillo, supra, citing People vs. Marquez, 380 SCRA 561 (2002); People vs. Yatco, 379 SCRA 432 (2002); People vs. Matic, 377 SCRA 314 (2002).
 People vs. Berdin, supra at 19, citing People vs. Pacina, 338 SCRA 195, 215-216 (2000).
 G.R. No. 137842, August 23, 2001, 363 SCRA 621, cited in People vs. Berdin, supra at 19.