Show that $M_{n}(R)/M_{n}(I)\cong M_{n}(R/I)$

Question

I have the following problem that I'm working on:

Let $R$ be a ring with multiplicative identity $1_{R}$ and let $I$ be a (two-sided) ideal of $R$. Let $n$ be a natural number and denote by $M_{n}(R)$ the ring of $n\times n$ matrices over $R$. Show that $$\frac{M_{n}(R)}{M_{n}(I)}\cong M_{n}(R/I).$$

I'm trying to apply the First Isomorphism Theorem. Here's my work so far:

Let $\Theta:M_{n}(R)\to M_{n}(R/I)$ be the map that carries a matrix with entries in $R$ to the same matrix, but with entries considered in $R/I$. In other words, if $S=(s_{ij})\in M_{n}(R)$, then $\Theta[(s_{ij})]=(\overline{s_{ij}})$, where $\overline{s_{ij}}=s_{ij}+I$ is the coset of $s_{ij}$ in $R/I$. Then:

• $\Theta$ is a morphism. If $S=(s_{ij})$ and $T=(t_{ij})$ are two matrices in $M_{n}(R)$, then $$\Theta(S+T)=\Theta[(s_{ij}+t_{ij})]=(\overline{s_{ij}+t_{ij}})=(\overline{s_{ij}})+(\overline{t_{ij}})=\Theta(S)+\Theta(T).$$ Similarly, $\Theta$ respects multiplication, since $$\Theta(ST)=\Theta[(s_{ik}t_{kj})]=(\overline{s_{ik}t_{kj}})=(\overline{s_{ik}})(\overline{t_{jk}})=\Theta(S)\Theta(T).$$
• $\mathrm{Ker}(\Theta)=M_{n}(I)$. Indeed, we have

\begin{split}S\in\mathrm{Ker}(\Theta)&\Leftrightarrow \Theta(S)=(\overline{s_{ij}})=(0_{R})_{n\times n} \\ &\Leftrightarrow \overline{s_{ij}}=\overline{0}\:\:\: \forall\:\: i,j\\ &\Leftrightarrow s_{ij}\in I\:\:\: \forall\:\: i,j\\ &\Leftrightarrow S\in M_{n}(I). \end{split}

Does my work look correct? Also, there are two things that I'm stuck on: showing that $\Theta$ is well-defined, and showing that $\Theta$ is surjective. Thanks in advance for any help or suggestions!

UPDATE: The 'multiplication' above doesn't make sense, because there is no definition on $k$. However, writing in the summation notation (that I should have done in the beginning), we have $$\Theta(ST)=\Theta\left[\left(\sum_{k=1}^{n}s_{ik}t_{kj}\right)\right]=\left(\overline{\sum_{k=1}^{n}s_{ik}t_{kj}}\right)=\left(\sum_{k=1}^{n}\overline{s_{ik}}\cdot\overline{t_{kj}}\right)=(\overline{s_{ij}})(\overline{t_{ij}})=\Theta(S)\Theta(T).$$

Answer 1

Everything looks good until the multiplication part. What is the $ij$ coefficient of $ST$ ? (certainly not $s_{ik}t_{kj}$ because there is no defined $k$ !)

Then in your proof of $\mathrm{Ker}\Theta = M_n(I)$, it should read $0_{R/I}$, not $0_R$.

As for the things you are stuck on:

$\Theta$ is well-defined; proof: you defined it. More seriously, $\Theta(S)_{ij} = \pi(s_{ij})$ with $\pi : R\to R/I$ the canonical projection is perfectly well-defined, why would there be an issue here ? As a "rule of thumb", well-definedness questions mostly appear when defininf maps out of a quotient, because such maps are usually defined using one representative of the equivalence class, when the image of the class should only depend on the class, not on a specific representative.

To show that $\Theta$ is surjective, take a matric $M\in M_n(R/I)$, and look at its $ij$ coefficient : $m_{ij} \in R/I$. What can you then say about $m_{ij}$ ? And thus about $M$ ?