Surjective ring homomorphism from $M_n(R)$ to $M_n(R/I)$ where $R$ is a ring and $I$ is an ideal for R?

Question

I'm looking for such a surjective homomorphism. I was thinking of starting from the canonical surjection from $R$ to $R/I$ and induce one but somehow I get stuck... Can you help me?

Thank you very much!

Answer 1

If $n > 0$, we have a functor $M_n : \operatorname{Ring} \to \operatorname{Ring}$ where $M_n(R)$ is the ring of $n \times n$ matrices with entries in $R$. If $f : R \to S$ is a ring homomorphism, then $M_n(f) : M_n(R) \to M_n(S)$ is a ring homomorphism defined by $$M_n(f)\Bigg( \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn}\end{bmatrix}\Bigg) = \begin{bmatrix} f(a_{11}) & \cdots & f(a_{1n}) \\ \vdots & & \vdots \\ f(a_{n1}) & \cdots & f(a_{nn})\end{bmatrix}.$$

In our specific case, we let $R$ be a ring and $I$ an ideal, so we can consider the ring $R/I$ and the quotient homomorphism: $f : R \to R/I$ defined by $f(r) = \overline r = r + I$. Notice that our homormorphism is clearly surjective.

Let's pick an arbitrary element in $M_n(R/I)$: $$\begin{bmatrix} \overline {a_{11}} & \cdots &\overline{ a_{1n}} \\ \vdots & & \vdots \\ \overline{a_{n1}} & \cdots & \overline{a_{nn}}\end{bmatrix} = \begin{bmatrix} a_{11} + I & \cdots & a_{1n} + I \\ \vdots & & \vdots \\ a_{n1} + I & \cdots & a_{nn} + I\end{bmatrix}$$ where we can choose the obvious representatives: for $\overline{a_{ij}}$ we pick $a_{ij} \in R$ and notice that $f(a_{ij}) = \overline{a_{ij}} = a_{ij} + I$.

It should be clear that $$\begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn}\end{bmatrix}$$ is an element such that $$M_n(f)\Bigg(\begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn}\end{bmatrix}\Bigg) = \begin{bmatrix} \overline {a_{11}} & \cdots &\overline{ a_{1n}} \\ \vdots & & \vdots \\ \overline{a_{n1}} & \cdots & \overline{a_{nn}}\end{bmatrix}.$$

Hence the surjective homomorphism you requested should be $M_n(f)$ where $f$ is the canonical quotient map.