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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
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@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
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2 Answers 2

up vote 5 down vote accepted

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
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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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