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