Projective limit of finite rings is compact

Question

In Serre's book Local Fields it mentions that the projective limit of finite rings is compact. How can it be proven?

Answer 1

Tychonoff's theorem says the product of infinitely many compact topological spaces is a compact topological space.

Let $(I,\leq)$ be a directed set, and let $R_i$ be a directed system of compact $T_1$ topological rings with continuous homomorphisms $f_{ij}:R_i\to R_j$ when $j\leq i$ such that $f_{ij}\circ f_{jk}=f_{ik}$ and $f_{ii}=\mathrm{id}_{R_i}$.

Let $X=\prod_{i\in I} R_i$. Then the inverse limit is the subspace of $X$ defined as:

$$R_{\infty}=\{(r_i)_{i\in I}\in X\mid \forall j\leq i\in I:f_{ij}(r_i)=r_j\}$$

$X$ is compact by Tychonoff, so we only need to prove that $R_{\infty}$ is closed in $X$ to get that $R_{\infty}$ is compact.

But if $\pi_i:X\to R_i$ is the natural projection, then for each $j\leq i\in I$ we define a continuous map: $\rho_{ij}:X\to R_j$ defined as:

$$\rho_{ij}(x)=\pi_{j}(x)-f_{ij}(\pi_i(x))$$

This map is continous since $f_{ij},\pi_i,\pi_j$ are continuous and subtraction is continuous in a topological ring.

But this means $\rho_{ij}^{-1}(\{0\})$ is closed in $X$, and we see that:

$$R_{\infty}=\bigcap_{j\leq i\in I} \rho_{ij}^{-1}(\{0\})$$

So $R_{\infty}$ is the intersection of closed subsets of $X$ and hence is closed in $X$ and hence compact.

The only reason we need $T_1$ is to ensure that $\{0\}$ is closed in each $R_i$. But in the original case, where $R_i$ are finite with the discrete topology, this is always true.

Tychonoff requires the axiom of choice (well, only the Boolean Prime Ideal theorem, which is slightly weaker.) But if $I$ is nice - say, $(\mathbb N,\leq)$ or otherwise countable - we can be more constructive, especially with the $R_i$ finite.

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Aside: Any topological ring $R$ with $\{0\}$ closed is actually Hausdorff, since the diagonal in $R\times R$ is $m^{-1}(\{0\})$ where $m(x,y)=x-y$ is continuous.