# Show that $K_1K_2/k$ is Galoisienne if $K_1$ and $K_2$ are Galoisienne.

Let $k$ a field and $K_1/K$ and $K_2/k$ two Galois extension.

1) Show that $K_1K_2/k$ is a Galois extension.

2) Let $\varphi: Gal(K_1K_2/k)\longrightarrow Gal(K_1/k)\times Gal(K_2/k)$ defined by $$\varphi(\sigma )=(\sigma |_{K_1},\sigma |_{K_2}).$$ Show that $\varphi$ is a group homomorphism.

My Attempts

1) Hard question.

$\bullet$ For the separability, I guess that I have to prove that $$|Hom_k(K_1K_2,k^{alg})|=[K_1K_2:k^{alg}].$$ (I denote $Hom_A(B,C)=\{\varphi:B\to C\mid \varphi|_A=id \}$).

Q1) Sorry, but I can't do it. Could you help ?

$\bullet$ For the the fact that the extension is normale, if $\{\alpha_i\}$ is a base pf $K_1$ and $\{\beta_j\}$ a base of $K_2$, then $\{\alpha_i\beta_j\}$ is a base of $K_1K_2$. Let $x\in K_1K_2$. Then, there is $(x_{ij})_{ij}$ s.t. $$x=\sum_{i,j}x_{ij}\alpha_i\beta_j\underset{\varphi\in Hom_k(K_1K_2,k^{alg})}{\implies} \varphi(x)=\sum_{ij}x_{ij}\varphi(\alpha_i)\varphi(\beta_j).$$ Since $K_i/k$ is Galois, $\varphi|_{K_i}\in K_i$ and thus $$x_{ij}\varphi(\alpha_i)\varphi(\beta_j)\in K_1K_2\implies \varphi\in Aut_k(K_1K_2).$$ Therefore, it's normale.

Q2) Is it correct ?

2) Q3) I don't really know what to show, that $\varphi(\sigma _1\sigma _2)=\varphi(\sigma _1)\varphi(\sigma _2)$ ? It's all ?

• Are you restricting to the finite case? – Lubin Dec 1 '15 at 19:23
• The English adjective is just "Galois," although I do admire the poetry of the French term here. – Qiaochu Yuan Dec 1 '15 at 22:07

## 1 Answer

I don’t think you’ve chosen the most efficient way of looking at things. When you have a Galois extension $K\supset k$, it may not help to talk about the Galois group.

In this case, let’s deal with normality first. I’ll take as definition of whether $K\supset k$ is normal the condition that whenever you have a $k$-morphism $\sigma$ of $K$ into an algebraically closed field $\Omega\supset K$, it is necessarily the case that $\sigma(K)\subset K$. Now consider two normal extensions $K_1$ and $K_2$ of $k$, both in $\Omega$. A $k$-morphism $\sigma$ defined on $K_1K_2$ necessarily sends each $K_i$ into itself, therefore sends the compositum into itself as well.

For separability, there are many definitions, and which definition you choose will affect the difficulty of your proof. I choose the definition that an extension $K\supset k$ is separable if $K$ may be generated over $k$ by a set of elements $\{\alpha_i\}$ separable over $k$: this means that for each $i$, the minimal $k$-polynomial for $\alpha_i$ has distinct roots. Now if $\{\alpha_i\}$ is a set of separable generators for $K_1$ and $\{\beta_i\}$ is similarly a set for $K_2$, then the union of these two sets of generators is a set of generators for $K_1K_2$.

• A little bit late @Rick, but I hope that my comment is helpful. I think Lubin's answer is understandable, but the definition of separability is not really a standard one. So I will give another hint for the proof of separability: You can write any element in $K_1K_2$ by using finitely many elements in $K_1$ and $K_2$. If you adjoin them to $k$, you get a separable extension. Hence any element of $K_1K_2$ is separable. – Daniel Bernoulli Dec 25 '15 at 9:21