text: End$(A)$ for $A=\mathbb{Z}$, $A=\mathbb{Q}$ and $A=\mathbb{R}$
  
  By End$(A)$ I mean the set of all ring endomorphisms of $A$.
  
  I would appreciate if you could check if my proof is correct.
  
  First, for $\mathbb{Z}$ we have
  
  $$\phi \colon \ \mathbb{Z} \longrightarrow \mathbb{Z} \text{ and } \phi(1)=:k \ne 0.$$
  
  We can prove by induction $\phi(n)=kn, \quad \forall n\in \mathbb{N}$
  
  Let $$P(n)\colon \ \phi(n)=kn$$
  
  We have,
  
  $$\phi(n+1)=\phi(n)+\phi(1)=nk+k=k(n+1)$$
  
  Therefore $$P(n)\Longrightarrow P(n+1)$$
  
  This is obviously true for $n=1$, and
  
  $$\phi(-n)=-\phi(n)=-kn \quad \forall n\in \mathbb{N}$$
  
  Therefore
  
  $$\phi(n)=kn, \quad \forall n\in \mathbb{Z}$$
  
  Also note that
  
  $$\phi(1\cdot n)=\phi(n) \iff k\cdot \phi(n)=\phi(n) \Longrightarrow k=1$$
  
  because $\phi(\mathbb{Z})\subseteq \mathbb{Z}$ is a subring of an integral domain, then it must be an integral domain itself.
  
  That means,
  
  $$\phi(n)=n \text{ and } \phi_0(n)=0, \quad \forall n\in \mathbb{Z} \text{ are the only endomorphisms on } \mathbb{Z}.$$
  
  Now let
  
  $$\psi \colon \ \mathbb{Q} \longrightarrow \mathbb{Q}$$
  
  and let's assume $\ker{(\psi})\ne \mathbb{Q}$
  
  then
  
  $$\psi(n)=n, \quad \forall n\in \mathbb{Z}$$
  
  and
  
  $$\psi(n^{-1})=(\psi(n))^{-1}=n^{-1}$$
  
  that means,
  
  $$\psi(q)=q \text{ and } \psi_0(q)=0, \quad \forall q\in \mathbb{Q} \text{ are the only endomorphisms on } \mathbb{Q}.$$
  
  Finally, let $f\colon \ \mathbb{R} \longrightarrow \mathbb{R}$
  
  Because $\mathbb{R}$ is a field and $\ker{(\mathbb{R})}$ is an ideal,
  
  $$\ker{(f)}=\{0\} \lor \ker{(f)}=\mathbb{R}$$
  
  Considering only the first case,
  
  $$f(x^2)=(f(x))^2\ge 0 \Longrightarrow f(x)\ge 0, \quad \forall x\ge 0$$
  
  and
  
  $$f(-x)=-f(x)<0, \quad \forall x>0$$
  
  Let $x_0\in \mathbb{R}$ and $x_1\ge0$
  
  then
  
  $$f(x_0+x_1)=f(x_0)+f(x_1)\ge f(x_0)$$
  
  Therefore $f$ is increasing (or non-decreassing, I'm not sure how you would say it in english).
  
  Finally,
  
  $$\forall x\in \mathbb{R}, \ \exists q_1,q_2 \in \mathbb{Q}\colon \ q_1\le q_2 \text{ and } q_1\le x \le q_2$$
  
  Because f is increasing, then
  
  $$f(q_1)\le f(x) \le f(q_2) \iff q_1\le f(x) \le q_2$$
  
  $$\Longrightarrow 0\le f(x) -x\le 0 \iff f(x)=x, \quad \forall x\in \mathbb{R}$$
  
  Therefore,
  
  $$f(x)=x \text{ and } f_0(x)=0, \quad \forall x\in \mathbb{R} \text{ are the only endomorphisms on } \mathbb{R}$$
  
  I'm mostly not sure about the $\phi(1)=1$ part.
  
  Thank you
  
  • I don't see anything wrong. By the way, $\phi(1)=\phi(1\cdot1)=\phi(1)^2$. So either $\phi(1)=1$ or $\phi(1)=0$, cases that are distinguished by what you are assuming $\ker\phi$ to be. And also, "increasing" is the right term to use above, since you are assuming that $\ker f=0$, which means $f(x_1)\neq0$. Mar 14, 2013 at 6:18
  • Thank you, just wanted to make sure :) Mar 14, 2013 at 6:25
  • After thinking about it, I dont think it's accurate to say $\phi(1)$ is either $1$ or $0$ in general. For example, consider the ring homomorphism $\phi \colon \ \mathbb{R} \longrightarrow$ M$_2(\mathbb{R}),\quad \phi(x)= \left( \begin{array}{cc} x & 0 \\ 0 & 0 \\ \end{array} \right)$. In here we have $\phi(1)=\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right) \ne \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) = Id$, but it is the identity of $\phi(\mathbb{R})$. Mar 14, 2013 at 16:03
  • Unless of course, by $\phi(1)=1$ you meant the identity of $\phi(A)$ on a $\phi \colon \ A \longrightarrow B$ ring homomorphism and not necesarly of $B$ Mar 14, 2013 at 16:11
  • Yes I wasn't talking about general rings. I meant specifically for the rings in your question, which all embed in $\mathbb{R}$. In $\mathbb{R}$, the only solutions to $x^2=x$ are $0$ and $1$. Mar 14, 2013 at 17:49
explanation: The document is a detailed mathematical proof that explores the set of all ring endomorphisms (End$(A)$) for the sets of integers ($\mathbb{Z}$), rational numbers ($\mathbb{Q}$), and real numbers ($\mathbb{R}$). The author provides a step-by-step proof for each case and asks for verification of the correctness of the proof. The document also includes a discussion and a response to a comment from another user, addressing a specific part of the proof.
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distilabel_metadata.content: End$(A)$ for $A=\mathbb{Z}$, $A=\mathbb{Q}$ and $A=\mathbb{R}$
  
  By End$(A)$ I mean the set of all ring endomorphisms of $A$.
  
  I would appreciate if you could check if my proof is correct.
  
  First, for $\mathbb{Z}$ we have
  
  $$\phi \colon \ \mathbb{Z} \longrightarrow \mathbb{Z} \text{ and } \phi(1)=:k \ne 0.$$
  
  We can prove by induction $\phi(n)=kn, \quad \forall n\in \mathbb{N}$
  
  Let $$P(n)\colon \ \phi(n)=kn$$
  
  We have,
  
  $$\phi(n+1)=\phi(n)+\phi(1)=nk+k=k(n+1)$$
  
  Therefore $$P(n)\Longrightarrow P(n+1)$$
  
  This is obviously true for $n=1$, and
  
  $$\phi(-n)=-\phi(n)=-kn \quad \forall n\in \mathbb{N}$$
  
  Therefore
  
  $$\phi(n)=kn, \quad \forall n\in \mathbb{Z}$$
  
  Also note that
  
  $$\phi(1\cdot n)=\phi(n) \iff k\cdot \phi(n)=\phi(n) \Longrightarrow k=1$$
  
  because $\phi(\mathbb{Z})\subseteq \mathbb{Z}$ is a subring of an integral domain, then it must be an integral domain itself.
  
  That means,
  
  $$\phi(n)=n \text{ and } \phi_0(n)=0, \quad \forall n\in \mathbb{Z} \text{ are the only endomorphisms on } \mathbb{Z}.$$
  
  Now let
  
  $$\psi \colon \ \mathbb{Q} \longrightarrow \mathbb{Q}$$
  
  and let's assume $\ker{(\psi})\ne \mathbb{Q}$
  
  then
  
  $$\psi(n)=n, \quad \forall n\in \mathbb{Z}$$
  
  and
  
  $$\psi(n^{-1})=(\psi(n))^{-1}=n^{-1}$$
  
  that means,
  
  $$\psi(q)=q \text{ and } \psi_0(q)=0, \quad \forall q\in \mathbb{Q} \text{ are the only endomorphisms on } \mathbb{Q}.$$
  
  Finally, let $f\colon \ \mathbb{R} \longrightarrow \mathbb{R}$
  
  Because $\mathbb{R}$ is a field and $\ker{(\mathbb{R})}$ is an ideal,
  
  $$\ker{(f)}=\{0\} \lor \ker{(f)}=\mathbb{R}$$
  
  Considering only the first case,
  
  $$f(x^2)=(f(x))^2\ge 0 \Longrightarrow f(x)\ge 0, \quad \forall x\ge 0$$
  
  and
  
  $$f(-x)=-f(x)<0, \quad \forall x>0$$
  
  Let $x_0\in \mathbb{R}$ and $x_1\ge0$
  
  then
  
  $$f(x_0+x_1)=f(x_0)+f(x_1)\ge f(x_0)$$
  
  Therefore $f$ is increasing (or non-decreassing, I'm not sure how you would say it in english).
  
  Finally,
  
  $$\forall x\in \mathbb{R}, \ \exists q_1,q_2 \in \mathbb{Q}\colon \ q_1\le q_2 \text{ and } q_1\le x \le q_2$$
  
  Because f is increasing, then
  
  $$f(q_1)\le f(x) \le f(q_2) \iff q_1\le f(x) \le q_2$$
  
  $$\Longrightarrow 0\le f(x) -x\le 0 \iff f(x)=x, \quad \forall x\in \mathbb{R}$$
  
  Therefore,
  
  $$f(x)=x \text{ and } f_0(x)=0, \quad \forall x\in \mathbb{R} \text{ are the only endomorphisms on } \mathbb{R}$$
  
  I'm mostly not sure about the $\phi(1)=1$ part.
  
  Thank you
  
  • I don't see anything wrong. By the way, $\phi(1)=\phi(1\cdot1)=\phi(1)^2$. So either $\phi(1)=1$ or $\phi(1)=0$, cases that are distinguished by what you are assuming $\ker\phi$ to be. And also, "increasing" is the right term to use above, since you are assuming that $\ker f=0$, which means $f(x_1)\neq0$. Mar 14, 2013 at 6:18
  • Thank you, just wanted to make sure :) Mar 14, 2013 at 6:25
  • After thinking about it, I dont think it's accurate to say $\phi(1)$ is either $1$ or $0$ in general. For example, consider the ring homomorphism $\phi \colon \ \mathbb{R} \longrightarrow$ M$_2(\mathbb{R}),\quad \phi(x)= \left( \begin{array}{cc} x & 0 \\ 0 & 0 \\ \end{array} \right)$. In here we have $\phi(1)=\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right) \ne \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) = Id$, but it is the identity of $\phi(\mathbb{R})$. Mar 14, 2013 at 16:03
  • Unless of course, by $\phi(1)=1$ you meant the identity of $\phi(A)$ on a $\phi \colon \ A \longrightarrow B$ ring homomorphism and not necesarly of $B$ Mar 14, 2013 at 16:11
  • Yes I wasn't talking about general rings. I meant specifically for the rings in your question, which all embed in $\mathbb{R}$. In $\mathbb{R}$, the only solutions to $x^2=x$ are $0$ and $1$. Mar 14, 2013 at 17:49
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a detailed mathematical proof that explores the set of all ring endomorphisms (End$(A)$) for the sets of integers ($\mathbb{Z}$), rational numbers ($\mathbb{Q}$), and real numbers ($\mathbb{R}$). The author provides a step-by-step proof for each case and asks for verification of the correctness of the proof. The document also includes a discussion and a response to a comment from another user, addressing a specific part of the proof.
  
  Label: question-and-answer