# Galois group of the product of two fields being the product of Galois groups?

I'm just wondering if the following theorem is reasonable and whether the proof is makes sense or not? Also I have an application of it which I was trying to do earlier but I made a lot of mistakes so I would be very grateful if anyone would give some comments on this in case I have some very strong misunderstandings about Galois theory.

Theorem Let $E_1$ and $E_2$ be Galois field extensions of $F$ with trivial intersection ($E_1 \cap E_2 = F$), then $E_1 E_2$ has Galois group $\text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.

Proof

• $E_1$ and $E_2$ are splitting fields (over $F$) of some polynomials $f_1$ and $f_2$ so $E_1 E_2$ is the splitting field of $f_1 f_2$ over $F$ and therefore it is Galois and we are justified using the "Gal" notation.

• Let $\sigma_1 \in \text{Gal}(E_1/F)$ and $\sigma_2 \in \text{Gal}(E_2/F)$ then by the fundamental theorem of Galois theory we can lift these both into elements of $\text{Gal}(E_1 E_2/F)$. The claim is that $\sigma_1 \sigma_2 = \sigma_2 \sigma_1$, to see this consider an element $\alpha \in E_1$, we have both $\sigma_2 \alpha = \alpha$ and $\sigma_2 \sigma_1 \alpha = \sigma_1 \alpha$ because it is invariant with respect to $\sigma_2$, but we can combine these equalities to get $\sigma_1 \sigma_2 \alpha = \sigma_2 \sigma_1 \alpha$.

• Both groups $\text{Gal}(E_i/F)$ ($i = 1,2$) are included in $\text{Gal}(E_1 E_2/F)$ and the degrees say there are no more elements, furthermore since they commute we can conclude $\text{Gal}(E_1 E_2/F) = \text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.

So hopefully that theorem is, if not correct, fixable.. and can be used to prove that $\mathbb Q (\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_n})$ has Galois group $C_2^m$ ($m \le n$) over $\mathbb Q$?

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When you say that $E_1, E_2$ have trivial intersection, do you mean in a fixed algebraic closure of $F$? (One cannot speak of the intersection of two sets without specifying a set containing both of them as subsets.) Also, it is not obvious that this condition holds in the situation you want to apply it to. –  Qiaochu Yuan Apr 27 '11 at 16:13
@Qiaochu, the condition either holds or it doesn't: If it doesn't then we are not performing a field extension. –  quanta Apr 27 '11 at 16:28

Yes, your theorem and its proof are correct. This is Theorem 1.14 in Chapter 6 of Lang's Algebra.

Your application to $\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n})$ is also correct - by applying Theorem 1.14 repeatedly, one gets that $\text{Gal}(\mathbb{Q}(\sqrt{a_1},\ldots,\sqrt{a_n})/\mathbb{Q})$ injects into $\prod_{i=1}^n\text{Gal}(\mathbb{Q}(\sqrt{a_i})/\mathbb{Q})\cong C_2^n$.

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I can't think of what the correct term is, but I wouldn't say that $\sqrt2$ and $\sqrt3$ are algebraically independent: if $f(x,y)=x^2+y^3-5$, then $f(\sqrt2,\sqrt3)=0$. –  Mariano Suárez-Alvarez Apr 27 '11 at 14:57
Hmm, you're right, I was using that term without thinking about it. Is linear disjointness what we're after? –  Zev Chonoles Apr 27 '11 at 15:03

The term you are looking for is compositum. If $K$ is a field and $E_1, E_2 \subset \bar{K}$ are two extensions of $K$, then their compositum $E_1 E_2$ is the subfield of $\bar{K}$ generated by $E_1$ and $E_2$. Note that unless $E_1, E_2$ are normal extensions the compositum is not uniquely specified by what $E_1, E_2$ look like as abstract extensions; you really do need to specify an embedding in some bigger field, which might as well be the algebraic closure for algebraic extensions.

The compositum of a Galois extension and another extension is Galois, so in particular if $E_1, E_2$ are both Galois then $E_1 E_2$ is Galois. Now there is an obvious embedding $\text{Gal}(E_1 E_2 / K) \to \text{Gal}(E_1 / K) \times \text{Gal}(E_2 / K)$ whose image is precisely the subgroup of elements which fix $E_1 \cap E_2$ (which is also Galois). When $E_1 \cap E_2 = K$ the conclusion follows.

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