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.

  • $\begingroup$ I know it is related to the following theorem: Let R,S,T be division ring s.t. $R\subset S\subset T$ then $dim_R T=(dim_S T)(dim_R S)$ I hope this helps you. $\endgroup$
    – Lotus
    Feb 5, 2014 at 3:10
  • $\begingroup$ You pretty much listed all their similarities already. Are you thinking there is a theorem that they are both corollaries of? $\endgroup$
    – rschwieb
    Feb 5, 2014 at 3:13
  • 2
    $\begingroup$ The connection you may be looking for is Galois theory (if you haven't already learned it yet). This tells you why, for Galois extensions, these theorems are basically the same $\endgroup$
    – zcn
    Feb 5, 2014 at 4:41
  • 2
    $\begingroup$ I've added the tags "group theory" and "field theory" for the obvious reasons, and the tag "category theory" because of my answer. In my opinion, category theory is the key tool in order to understand the connections between different fields of mathematics. $\endgroup$ Feb 5, 2014 at 11:36
  • $\begingroup$ I love the turn this question took. This is how you learn God Damn it...:) $\endgroup$
    – Quester
    Feb 6, 2014 at 22:07

1 Answer 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.

  • $\begingroup$ simply beautiful! $\endgroup$ Feb 5, 2014 at 11:47
  • $\begingroup$ Thank you. I always wonder if such generalizations of algebra to monoidal categories are known / well-known / written down somewhere? I've developed this kind of algebra for quite some time now, and don't know any references except perhaps for the special case of Tannakian categories. $\endgroup$ Feb 6, 2014 at 0:12

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