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## Does this isomorphism between Galois groups hold for transcendental extensions?

Suppose $L/K$ and $M/K$ are algebraic extensions of a field $K$, such that $L\cap M=K$, and $L/K$ is a normal extension. It is well-known that, with these conditions, we have:

$$\text{Gal}(L/K)\cong \text{Gal}(LM/M).$$

However, suppose we now complicate matters by specifying that $M/K$ (and hence also $LM/L$) is not algebraic but transcendental ($L/K$ remaining normal). In this case, am I right in thinking that this identity still holds?

If not, can anyone provide a counterexample, or is there an obvious reason why this doesn't work? What if we specify that $K$ has characteristic zero?

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What do you mean by a transcendental normal extension? – Qiaochu Yuan Oct 10 at 3:26
Oh dear, I do apologise...I appear to have mixed up my $M$'s and my $L$'s. As it was the question made no sense, so sorry for wasting your time. Serves me right for posting questions so late at night... I have edited the question - hopefully it is clear now. – MrB Oct 10 at 12:46

## 1 Answer

The answer is yes, no matter what the characteristic is. Since $L/K$ is not assumed to be separable, I believe that $\text{Gal}(L/K)$ denotes the group of $K$-automorphisms of $L$. Let $K'$ be the fixed field of this group. By the usual properties of linearly disjoint extensions, we know that the fields $L$ and $K'M$ are linearly disjoint over $K'$, so they intersect in $K'$. The usual translation theorem of Galois theory (see e.g. Lang's Algebra, there is no algebraicity assumption on $LM/K'M$) shows that the homomorphism $\text{Gal}(LM/K'M)\to\text{Gal}(L/K')$ (restriction to $L$) is bijective. Obviously $\text{Gal}(LM/M)\to\text{Gal}(L/K)$ is injective, and the claim follows from noting that $\text{Gal}(L/K)=\text{Gal}(L/K')$ and $\text{Gal}(LM/M)\ge\text{Gal}(LM/K'M)$.

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