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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

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    The evaluation at $1$.2011-09-14
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    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.2011-09-14
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    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.2011-09-14
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    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() at 1 and got it to work, thank you very much for your help.2011-09-14
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    @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.2011-09-14

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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.