I need to prove that rings $End(\mathbb{Z}^{n})$ and $M_{n}(\mathbb{Z})$ are isomorphic.
To start with, I let $A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots& & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \in M_{n}(\mathbb{Z})$ (so that all the entries $a_{ij} \in \mathbb{Z}).$
Then, I defined $\Gamma(A): \mathbb{Z}^{n} \to \mathbb{Z}^{n}$ to be such that $\Gamma(A) \begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots& & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}\begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix}$, and checked that $\Gamma(A)$ is a group homomorphism by considering $\begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix}, \begin{pmatrix} w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{pmatrix} \in \mathbb{Z}^{n}$ and showing that $\Gamma(A)\left[ \begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix} + \begin{pmatrix} w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{pmatrix} \right] = \Gamma(A)\begin{pmatrix} z_{1} \\ z_{2} \\ \vdots \\ z_{n} \end{pmatrix} + \Gamma(A)\begin{pmatrix} w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{pmatrix}$
So, now, since $\Gamma(A)$ is a group homomorphism from $\mathbb{Z}^{n}$ into itself, $\Gamma(A) \in End(\mathbb{Z}^{n})$, the ring of endomorphisms on $\mathbb{Z}^{n}$.
I'm not sure how to use what I just did, though, to show that an isomorphic map exists between $End(\mathbb{Z}^{n})$ and $M_{n}(\mathbb{Z})$, or even if it helps at all. Could somebody please help me with this proof?
Thank you.
Maybe this wasn't the best first attempt, but I am honestly very confused, and constructive criticism is the best kind of criticism.
This is just a sketch. The map $$\Gamma: M_n(\mathbb{Z}^n)\rightarrow End(\mathbb{Z}^n), A\mapsto \Gamma(A) $$ is the desired isomorphism. It is easy to check that $\Gamma(A+B)=\Gamma(A)+\Gamma(B)$ as well as $\Gamma(AB)=\Gamma(A)\circ \Gamma(B)$, which means that $\Gamma$ is a ring homomorphism. To see the injectivity, let $e_i$ denote the $i$-th unit vector and note that $Ae_i$ is the $i$-th column of $A$. Thus, if $\Gamma(A)=0$, we also have $0=\Gamma(A)(e_i)=Ae_i$ for all $i$, so $A=0$.
The surjectivity remains. If $f$ is an endomorphism, let $A_f$ be the matrix with $f(e_i)$ as $i$-th column. Use the linearity of $f$ to show that $\Gamma(A_f)(x)=f(x)$ for all $x\in \mathbb{Z}^n$, that is $\Gamma(A_f)=f$.