I am currently working through Hungerford’s Algebra book. Question 2 on page 156 states:
Let $\operatorname{Mat}_n R$ be the ring of $n \times n$ matrices over a ring $R$. Then for each $n \geq 1$. $$ (\operatorname{Mat}_n R)[x] \cong \operatorname{Mat}_n R[x] $$ $$ (\operatorname{Mat}_n R)[[x]] \cong \operatorname{Mat}_n R[[x]]. $$
Now, I know how to prove these statements by creating maps and proving the isomorphisms by hand, but this is a fairly tedious task, as one can see by link. I figured that with the language of category theory I would be able to prove these statements relatively painlessly using some universal properties. My problem is that I am fairly new to category theory and don’t quite know how to write an explicit proof in this language.
I would appreciate some help by either showing a simple proof of the statement using category theory, showing how one should go about thinking about a proof using category theory, telling me why there is no proof in category theory, or why a proof would be even more tedious than the one method I mentioned above.
This can be done but it's a little indirect. More generally the following is true:
This implies the first result by taking $k = \mathbb{Z}, S = \mathbb{Z}[x]$ (since for $R$ a ring we have $R[x] \cong R \otimes \mathbb{Z}[x]$). I'll discuss the second result after.
Proof. $M_n(R)$, abstractly, is the endomorphism ring of $R^n$ as a right $R$-module. This gives that
$$\begin{align*} M_n(R \otimes_k S) &\cong \text{End}_{R \otimes_k S}( (R \otimes_k S)^n ) \\ &\cong \text{Hom}_{R \otimes_k S}((R \otimes_k S)^n, (R \otimes_k S)^n) \\ &\cong \text{Hom}_{R \otimes_k S}(R^n \otimes_k S, R^n \otimes_k S) \\ &\cong \text{Hom}_R(R^n, R^n \otimes_k S) \\ &\cong \text{Hom}_R(R^n, R^n) \otimes_k S \\ &\cong M_n(R) \otimes_k S. \\ \end{align*}$$
This requires a little elaboration. On the third line we use that $(-) \otimes_k S$ commutes with finite direct sums. On the fourth line we use a tensor-hom adjunction. The real content of this argument is on the fifth line; this is not a general property of Hom but it is true that for (bi)modules we have
$$\text{Hom}(P, Q \otimes R) \cong \text{Hom}(P, Q) \otimes R$$
(all three of $P, Q, R$ are bimodules here in full generality but I am too lazy to decorate everything appropriately) if $P$ is finitely generated projective, and in this case $P$ is finite free. The proof for the finite free case just requires writing $P$ as a direct sum of rank $1$ free modules and using that everything in sight commutes with finite direct sums.
This still requires a little elaboration: what the above argument actually shows a priori is that there is a natural map one can write down, which is an isomorphism of $k$-modules. But since it is also a ring homomorphism it is also an isomorphism of $k$-algebras. $\Box$
Now, this doesn't directly imply the second result, because it is not true that $R[[x]] \cong R \otimes \mathbb{Z}[[x]]$ (this is a nice exercise; they're already not isomorphic for $R = \mathbb{Q}$), so we can't just take $S = \mathbb{Z}[[x]]$.
However, by taking $S = \mathbb{Z}[x]/x^k$ we have a compatible family of isomorphisms $M_n(R[x]/x^k) \cong M_n(R)[x]/x^k$ for all $k$ and then we can take the cofiltered limit, then use the fact that the functor $R \mapsto M_n(R)$ is representable and so commutes with limits.