Let R be a ring, show that $(M_n(R))[x]\cong M_n(R[x])$.
Given a matrix $A\in M_n(R[X])$, $(A)_{ij}=\sum_0^nb_k^{ij}x^k$ (where $b_k^{ij}$ is the coefficent of the k'th power on the polynomial).
If I define $A_k\in M_n(R)$ such that $(A_k)_{ij}=b_k^{ij}$, and let $n>0$ such that we define $(X^n)_{ij}=\begin{cases}0&\text{if } i\neq j\\x^n&\text{if }i=j \end{cases}$.
Then $(A_kX^k)_{ij}=\sum_{r=0}^n(A_k)_{ir}(X^k)_{rj}=\sum_{r=0}^nb_k^{ir}(X^k)_{rj}=b_k^{ij}x^k$. Therefore $(A)_{ij}=\sum_{k=0}^n(A_kX^k)_{ij}$ which implies $A=\sum_{k=0}^nA_kX^k$.(In the case $k=0$ $A_0$ is well defined so there's not trouble there)
I have two questions: 1) how do I show that the $X^k$ I defined is the one in the expansion of a polynomial in $(M_n(R))[x]$.
2) If the above holds is this injection well defined? I can't seem to prove it.
Edit: as the comment below says, $R$ may not contain a multiplicative identity.
You forgot that the polynomials in $A$ can be of degree higher (or lower) than $n$, so here a little correction:
Let $A=\left(\sum_{k=0}^{n_{ij}} a_k^{ij}x^k\right)_{ij} \in M_n\left(R[x]\right)$ where $n_{ij}\in\mathbb{N}_0$, $a_k^{ij}\in R$ with the convention of $a_0^{ij}x^0 :=a_0^{ij}$ (this is just a convenience notation-wise as we (maybe) don't have a unit in $R$).
We then define: $$A_k:=(a_k^{ij})_{ij} \text{ for } 0\le k \le \max \{n_{ij}\}\text{ such that } a_k^{ij}=0 \text{ for all } k>n_{ij}$$ and $$X^k:=\left(\begin{cases}x^k&i=j\\0&\text{else}\end{cases}\right)_{ij} \text{ for } k>0$$ For $k>0$ we then find that: \begin{align} A_kX^k &=(a_k^{ij})_{ij}\left(\begin{cases}x^k&i=j\\0&\text{else}\end{cases}\right)_{ij}\\ &=\left(\sum_{r=1}^n a_k^{ir} \begin{cases}x^k&r=j\\0&\text{else}\end{cases}\right)_{ij}\\ &=(a_k^{ij}x^k)_{ij} \end{align} Introducing $$A_0X^0:=A_0$$ we indeed see: \begin{align} \sum_{k=0}^{\max\{n_{ij}\}} A_kX^k &=A_0+\sum_{k=1}^{\max\{n_{ij}\}} A_kX^k\\ &=A_0+\sum_{k=1}^{\max\{n_{ij}\}}(a_k^{ij}x^k)_{ij}\\ &=A_0+\left(\sum_{k=1}^{\max\{n_{ij}\}} a_k^{ij}x^k\right)_{ij}\\ &=(a_0^{ij})_{ij}+\left(\sum_{k=1}^{n_{ij}} a_k^{ij}x^k\right)_{ij}\text{ as } a_k^{ij}=0\text{ for }k>n_{ij}\\ &=\left(\sum_{k=0}^{n_{ij}} a_k^{ij}x^k\right)_{ij}\text{ again with }a_0^{ij}x^0:=a_0^{ij}\\ &=A \end{align} If we from now on use $Y$ as the formal variable in $M_n(R)[\,\cdot\,]$ (to avoid confusion between the variable power $X^k$ and the matrix $X^k$) we can define: \begin{align} \phi: M_n(R[x])&\longrightarrow M_n(R)[Y]\\ A&\longmapsto \sum_{k=0}^{\max\{n_{ij}\}}A_kY^k \end{align} One can easily proof that for $0,A,B\in M_n(R[X])$ we have $\phi(0)=0$ and $\phi(A+B)=\phi(A)+\phi(B)$.
Additionally (also using $\phi(A+B)=\phi(A)+\phi(B)$ and the convention $MY^0:=M$ for all $M\in M_n(R)$ ): \begin{align} \phi(AB) &=\phi\left(\left(\sum_{k=0}^{n_{ij,A}}a_k^{ij}x^k\right)_{ij}\left(\sum_{l=0}^{n_{ij,B}} b_l^{ij}x^l\right)_{ij}\right)\\ &=\phi\left(\left(\sum_{r=1}^n\sum_{k=0}^{n_{ij,A}}\sum_{l=0}^{n_{ij,B}}a_k^{ir}\,b_l^{rj}\,x^{k+l}\right)_{ij}\right)\\ &=\sum_{k=0}^{\max\{n_{ij,A}\}}\sum_{l=0}^{\max\{n_{ij,B}\}} \phi\left(\left(\sum_{r=1}^na_k^{ir}\,b_l^{rj}\,x^{k+l}\right)_{ij}\right)\\ &=\sum_{k=0}^{\max\{n_{ij,A}\}}\sum_{l=0}^{\max\{_{ij,B}\}} \left(\sum_{r=1}^na_k^{ir}\,b_l^{rj}\right)_{ij}Y^{k+l}\\ &=\sum_{k=0}^{\max\{n_{ij,A}\}}\sum_{l=0}^{\max\{_{ij,B}\}} (a_k^{ij})_{ij}\,(b_l^{ij})_{ij}Y^{k+l}\\ &=\left(\sum_{k=0}^{\max\{n_{ij,A}\}}(a_k^{ij})_{ij}Y^k\right)\left(\sum_{l=0}^{\max\{n_{ij,B}\}} (b_l^{ij})_{ij}Y^l\right)\\ &=\phi\left(\sum_{k=0}^{\max\{n_{ij,A}\}}(a_k^{ij})_{ij}X^k\right)\phi\left(\sum_{l=0}^{\max\{n_{ij,B}\}} (b_l^{ij})_{ij}X^l\right)\\ &=\phi(A)\phi(B) \end{align} Given that $\phi(I=\text{diag}(1,1,\dots,1))=I$ if $R$ is unitary we see that $\phi$ indeed is a morphism from $M_n(R[X])$ to $M_n(R)[X]$ whether there is $1\in R$ or not.
Whats left to show is, that $\phi$ is bijective. For the moment I'm out of time. If you need any more help, I'll gladly come back :)