# Tychonoff's theorem for products of finite discrete topologies?

I need the following specific version of Tychonoff's theorem:

Suppose $\{X_\alpha\}_\alpha$ is a collection of finite sets endowed with discrete topologies, then $\prod_\alpha X_\alpha$ is compact.

I wonder whether there's any direct proof for this specific version, since it's pervasive in mathematics (other than general topology), say the theory of pro-finite groups where infinite Galois theory is based. Many existence theorem is deduced from compactness.

I wonder whether there are more direct proofs (but Zorn's lemma is allowed) for this special case. I don't like base many existence theorems (especially some existence of a specific morphism between extension fields) on the black box of Tychonoff's theorem.

For example, suppose $A$ is an integrally closed domain with quotient field $F$, and $K/F$ is a normal extension. Let $B$ be the integral closure of $A$ in $K$. Given a prime ideal $P\subsetneq A$, we want to show that $\operatorname{Gal}(K/F)$ acts transitively on prime ideals lying over $P$. We can first prove it for finite extension $K/F$, and deduce the general case from this special case by compactness of $\operatorname{Gal}(K/F)$, see Serre's Local Algebra, say. However, I want to see how this Galois element is constructed without appealing to Tychonoff's theorem.

Any help is welcome. Thanks!

• What would you consider a (more) direct proof? Note that this statement is equivalent to the Boolean Prime Ideal Theorem. – user642796 Dec 17 '15 at 20:20
• @ArthurFischer In some sense, the proof of existence of maximal ideals for commutative rings is very intuitive, and existence of algebraic closure, and existence of extension to splitting fields. I need to use it to extract something. I'll edit my post to include an example. – Yai0Phah Dec 17 '15 at 20:35
• This is an exercise in futility. Tychonoff's theorem is equivalent to Zorn's lemma. – Matt Samuel Dec 18 '15 at 6:43

Since the Cantor set $K$ is a countable product of $\{0,1\}$, the compactness of $\prod X_\alpha$ for $X_\alpha=\{0,1\}$ implies the compactness of the product for $X_\alpha=K$. Now there is a continuous map from $K$ onto $[0,1]$, so we get the result for $X_\alpha=[0,1]$. And any compact Hausdorff space can be embedded in a (high-dimensional) cube, hence if $X_\alpha$ is compact Hausdorff then $\prod X_\alpha$ is a closed subset of a compact Hausdorff space.