Prove that $M_{n}(R)$ is a profinite ring and that $GL_{n}(R)$ is a profinite group. 
Let $R$ be a profinite commutative ring with an identity element, and let $n$ be a positive integer. Write $M_{n}(R)$ for the ring of $n \times n$ matrices and $GL_{n}(R)$ for the group of invertible $n \times n$ matrices over $R$. Then $M_{n}(R)$ has a natural topology defined by regarding it as $R^{n^{2}}$ with the product topology. Show that $M_{n}(R)$ is a profinite ring and that $GL_{n}(R)$ is a profinite group.

This question is taken directly from the exercises in chapter 1 of Profinite Groups by John Wilson. I'm working on this question for my dissertation topic, however this question is not assessed in any way, I'm just trying to get a better understanding of the subject by attempting some exercises.
I don't really know where to start with this question. I'm effectively punching above my weight here because I've had to learn a fair amount of things by myself ahead of my course. Any help would be appreciated.
 A: Well it's a few months down the line since I asked this and I think I have a reasonable understanding of the topic at this point, though feel free to correct me if I slip up.
$R$ is given to be a profinite commutative ring with an identity element. Then $R$ is the inverse limit of some inverse system of finite rings $R_{i}$, where $i$ is an element of a directed set $I$. Label the projection maps from $R \rightarrow R_{j}$ and $R_{j} \rightarrow R_{i}$ by $\psi_{j}$ and $\varphi_{ij}$ respectively. We don't need to know what they are, only that the universal property holds for this system, that is $\varphi_{ij} \circ \psi_{j} = \psi_{i}$ whenever $i \leq j$.
Now we can conjecture that similarly we have $\varprojlim_{i \in I}M_{n}(R_{i}) = M_{n}(R)$. We want to construct morphisms such that the universal property holds for the inverse system $(M_{n}(R_{i}), \alpha_{ij})$. Now, define $$\theta_{j}\colon M_{n}(R) \rightarrow M_{n}(R_{j}); a_{xy} \mapsto \psi_{j}(a_{xy}), \\
\alpha_{ij}\colon M_{n}(R_{j}) \rightarrow M_{n}(R_{i}); a_{xy} \mapsto \varphi_{ij}(a_{xy})$$
for each $1 \leq x, y \leq n$, where the $a_{xy}$ represent the coefficients of the respective matrix. Since $R$ is a profinite ring, we can conclude that both $\theta_{j}$ and $\alpha_{ij}$ are continuous and surjective since $\psi_{j}$ and $\varphi_{ij}$ are, and it's clear from this construction that the universal property holds:
$$\alpha_{ij} \circ \theta_{j}(a_{xy})=\alpha_{ij} \circ\psi_{j}(a_{xy}) = \varphi_{ij} \circ \psi_{j}(a_{xy}) = \psi_{j}(a_{xy}) = \theta_{i}(a_{xy}) \; \text{for all}\; i \leq j$$
by the universal property of the first inverse limit. Hence we have the desired result. I imagine a similar approach would yield an answer for the second part too.
