Degree of composite field

Question

Let $E/K$ and $F/K$ be finite subextensions of $L/K$, denote $EF/K$ the composite subextension. Then $[EF:F]\leqslant[E:K]$ and $[EF:K]\leqslant[E:K]\cdot[F:K]$.

If we assume $E\cap F=K$, then will we get "="?

Answer 1

No. Trivial intersection does not guarantee that $[EF:K]=[E:K]\cdot [F:K]$.

A standard counterexample of this is $K=\Bbb{Q}$, $E=\Bbb{Q}(\root3\of2)$, $F=\Bbb{Q}(e^{2\pi i/3}\root3\of2)$. The elements adjoined to get $E$ or $F$ are both zeros of $p(x)=x^3-2$. Therefore we see that $[E:K]=3=[F:K]$. As three is a prime, neither $E/K$ nor $F/K$ has non-trivial intermediate extensions, so $[E\cap F:K]=1$ or $3$. Here $E$ is real and $F$ is not, so $3$ is ruled out, and $E\cap F=K$.

But, the compositum $EF$ is the splitting field of $p(x)$ over $\Bbb{Q}$, well known to be Galois with Galois group $S_3$. Therefore $[EF:K]=6<3\cdot3$.

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The condition that guarantees equality, $[EF:K]=[E:K]\cdot[F:K]$, is linearly disjoint extensions. A study group I was running once needed the basics so I wrote up a quick intro.

Pete L. Clark has better material on this theme.

• See his question, and
• Notes on field theory. You want Chapter 12 in particular.