Composite of Galois extensions is Galois, a particular proof.

Question

This has been asked over and over again on math.stackexchange and I will ask it again.

Let $L_1/K$ and $L_2/K$ be finite Galois extensions of $K$ inside a common field, then $L_1L_2/K$ is a finite Galois extension.

I'm interested in one common proof of that fact. It goes like this:

$L_1L_2/K$ is finite so it suffices to prove that $L_1L_2$ is the splitting field of a separable polynomial over $K$. $L_i$ is the splitting field of a separable polynomial $f_i$ over $K$. Then $L_1L_2$ is the splitting field for the product of $f_1$ and $f_2$ with common factors only used once.

However, to me it seems that this only works when (the product of) common factors belong to $K[X]$ and I cannot think of a reason why this would be guaranteed (except e.g. when $L_1\cap L_2 = K$).What am I missing, or is this a bogus proof?

Answer 1

Consider, for instance, $L_1 = \Bbb Q(\sqrt2)$ and $L_2 = \Bbb Q(\sqrt3)$ over $K = \Bbb Q$. Then $L_1$ is the splitting field of $f_1(x) = x^2-2$ and $L_2$ is the splitting field of $f_2(x) = x^2-3$. The field $L_1L_2$ is the splitting field of $f_1f_2$ over $K$. Nothing about this construction gives you a polynomial outside of $K[x]$, and it may be applied quite generally, even when $f_1$ and $f_2$ have factors in common.

If it turns out that $f_1$ and $f_2$ have roots in common, then they have a non-1 $\gcd$. But that $\gcd$ is still an element of $K[x]$, as for instance the Euclidean algorithm applies, and doesn't take you out of $K[x]$.

So "the product of $f_1$ and $f_2$ with common factors only used once" is given by $$ \frac{f_1f_2}{\gcd(f_1, f_2)}\in K[x] $$

Answer 2

We can prove it's separable and normal independently.

As $L_1/K$ is Galois, $L_1$ is the splitting field of a separable polynomial $f(x)\in K[x] \subseteq L_2[x]$, the roots of $f(x)$ generate $L_1$ over $K$, so it generate $L_1L_2$ over $L_2$. In other words, $L_1L_2$ is the splitting field of a separable polynomial $f(x)$ over $L_2$, so $L_1L_2/L_2$ is Galois, hence separable.

Since $L_2/K$ is separable and separable is transitive, we know that $L_1L_2/K$ is separable.

Then we prove it's normal by your argument, no need to worry whether it contains multiple factors.

Answer 3

Lemma:

Let $L_1$ be the splitting field of $f_1(x)$ over $K$.

Let $L_2$ be the splitting field of $f_2(x)$ over $K$.

Then I claim that $L_1 L_2$ (the compositum of the two fields, as subfields of the algebraic closure of $K$) is the splitting field of $f_1(x) f_2(x)$.

proof Since $L_1 L_2$ contains $L_1$ it is clear that $f_1(x)$ splits in this field. Same with $f_2(x)$. So the polynomial $f_1(x) f_2(x)$ will also split completely in $L_1 L_2$.

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This implies that $L_1 L_2 / K$ is a normal extension.

The polynomial $f_1(x) f_2(x)$ may have multiple roots if $f_1$ and $f_2$ share some roots. In that case the polynomial would not be seperable. But as discussed in comments that can be factored out and $L_1 L_2$ is the splitting field of that factored out polynomial, which is seperable. So it is a Galois extension. But note that if $f_1$ and $f_2$ are irreducible polynomials then they cannot share any common factors, so in that case $f_1(x) f_2(x)$ would be seperable.

Note that only one of the two fields needs to be Galois for the compositum to be Galois, but proving this would take a bit more.