# Showing a Ring of endomorphisms is isomorphic to a Ring

Im trying to show that $\mathrm{End}(\langle \mathbb{Z},+\rangle)$ is naturally isomorphic to $\langle \mathbb{Z},+,\cdot\rangle$, but I'm not quite sure which ring homomorphism to use.

Thank you

• The evaluation at $1$. – Pierre-Yves Gaillard Sep 14 '11 at 16:59
• If you are not sure which ring homomorphism to use, it is because probably you donot know why the statement is true. One way to see why it holds is to make lots and lots of examples of endomorphisms of $\mathbb Z$: when you have produced lots of them, you will see a pattern. Use that pattern to construct your isomorphism. – Mariano Suárez-Álvarez Sep 14 '11 at 17:02
• Pieree-Yves and Mariano have told you what you need to know. I don't want to give the solution away, but you should not lose sight of the fact that (Z,+) is a cyclic group. – Chris Leary Sep 14 '11 at 17:11
• Oh yes i see now, by doing the examples i saw that what matters is where 1 is sent to by each homomorphism, and then i made my ring isomorphism as the evaluation of each element of End(<Z,+>) at 1 and got it to work, thank you very much for your help. – Chris Birkbeck Sep 14 '11 at 17:37
• @Chris: You might want to consider posting your proof as an answer; you can then get feedback on writing it properly, and then this question will not appear as "unanswered" on the website. – Arturo Magidin Sep 14 '11 at 18:07

OK, so let me denote the elements of End$(⟨\mathbb{Z},+⟩)$ by $\theta_{i}$ , where we have $\theta_{i}(1)=i$ for some $i \in \mathbb{Z}$. Also I define the evaluation homomorphism as $\alpha(\theta_{i})=\theta_{i}(1)$. Now we can see that $\alpha:$End$(⟨\mathbb{Z},+⟩)$ $\rightarrow$ $⟨\mathbb{Z},+,\cdot⟩$, and i need to show that in fact $\alpha$ is a ring isomorphism.

First ill show its a ring homomorphism, to do this, observe that$\alpha(\theta_{i} + \theta_{j})=(\theta_{i} + \theta_{j})(1)=\theta_{i}(1) + \theta_{j}(1)=\alpha(\theta_{i})+\alpha(\theta_{j})$.

And also that $\alpha(\theta_{i}\theta_{j})=\theta_{i}(\theta_{j}(1))=\theta_{i}(j)=ij=\alpha(\theta_{i})\alpha(\theta_{j})$, so it is a ring homomorphism.

Now i just need to show its one-to-one and onto.

one-to-one: if $\alpha(\theta_{i})=\alpha(\theta_{j})$, then $\theta_{i}(1)=\theta_{j}(1)$, and since the homomorphisms are determined by what values they send the generators to, we must have $\theta_{i}=\theta_{j}$, i.e they are the same homomorphism. Hence $\alpha$ is one-to-one.

Onto: now $\forall x \in \mathbb{Z}$, we have $\theta_{x}$ as defined above being the homomorphism sending $1$ to $x$, so $\alpha(\theta_{x})=x$, so we can see that $\alpha$ is obviously onto.