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
[Math] Showing a Ring of endomorphisms is isomorphic to a Ring
abstract-algebraring-theory
Best Answer
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.