Field Extensions and their dimensions

There is a powerful theorem with respect to a field $F$ extensions and their dimensions.

$F<E<K \ \Rightarrow [K:F] = [K:E][E:F] $

This is analogous to the famous Lagrange's theorem with respect to groups. Is there any relationship between these two theorems.

Solution 1:

This is a great question. Both the degree formula for field extensions and Lagrange's theorem are special cases of a Lemma taking place in a symmetric monoidal category. Here is this Lemma for the case of abelian groups.

Lemma: Let $M$ be a free left $S$-module, $R \to S$ a homomorphism of rings, such that $S$ is free as a left $R$-module. Then the left $R$-module $M|_R$ is free. Specifically, if $B$ is an $S$-basis of $M$ and $C$ is an $R$-basis of $S$, then $C \cdot B$ is an $R$-basis of $M|_R$ with cardinality $|C \cdot B| = |C|\cdot |B|$.

Proof: The basis $B$ induces an isomorphism $M \cong S^{\oplus B}$, hence $M|_R \cong (S|_R)^{\oplus B}$, and $C$ induces an isomorphism $S|_R \cong R^{\oplus C}$. Thus, $M|_R \cong (R^{\oplus C})^{\oplus B} \cong R^{\oplus C \times B}$. By construction this maps $(c,b) \in C \times B$ to $c \cdot b \in M$. QED

The degree formula for field extensions is an immediate corollary.

Now notice that the proof of this Lemma is entirely formal. We don't need any elements at all. Therefore we can generalize this Lemma to arbitrary symmetric monoidal categories with coproducts which distribute over $\otimes$. Then $R,S$ are monoid objects, $M$ is a left $S$-module object, which is called free when it is a coproduct of copies of $S$. The same proof as above works.

Let us apply this to the cartesian category of sets ($\otimes=\times$ and $\oplus=\sqcup$). Then $R,S$ are monoids in the usual sense. Let us restrict to the case of two groups $H,G$ equipped with a homomorphism of groups $H \to G$, w.l.o.g. injective. Then $M$ is a $G$-set. By decomposing $M$ into orbits, we find that $M$ is a coproduct of copies of $G$, i.e. it is a free $G$-set. A basis consists of a system of representatives $B$ for the orbits. Similarly, we find that $G$ is a free $H$-set, a basis consists of a system of representatives $C$ for the right cosets of $H$ in $G$. The Lemma tells us that $C \cdot B$ is a system of representatives for the orbits of $M|_H$. In other words: $$[H/M] = [H/G] \cdot [G/M].$$ Applying this to a group $K$ equipped with an injective homomorphism $G \to K$, we obtain Lagrange's theorem $$[G/K] = [H/G] \cdot [G/K].$$ Right modules give the corresponding formula for left cosets.