Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Omega/F$ be a field extension and $K,E$ be two subfields of $\Omega/F$. Assume that $KE/F$ is a finite Galois.

I have a theorem in my lecture notes that claim $\text{Gal}(KE/E)\cong \text{Gal}(K/K\cap E)$, while it is easy to see that $KE/F$ is Galois $\implies KE/E$ is Galois I can not understand why $K/K\cap E$ is Galois (it is separable since $KE/F$ is, but I don't see why it's normal).

Why $K/K\cap E$ is Galois ?

share|improve this question
    
I think you need to assume that $K$ is Galois over $K\cap E$, in which case you get the isomorphism... –  Arturo Magidin Jul 20 '12 at 21:27

1 Answer 1

up vote 1 down vote accepted

You need to assume that $K$ is Galois over $K\cap E$; if that is the case, then the isomorphism drops out of the Galois correspondence and the Isomorphism Theorems.

We can replace $F$ with $K\cap E$, so that we are in the following situation:

  • $KE/F$ is finite Galois;
  • $K\cap E = F$.

If $G=\mathrm{Gal}(KE/F)$, let $M$ be the subgroup corresponding to $K$ and $N$ the subgroup corresponding to $E$. Then $M\cap N=\{e\}$ (since $KE$ is the field "on top"), and $\langle M,N\rangle = G$.

Now, $K$ is Galois over $F$ if and only if $M$ is normal in $\langle M,N\rangle$.

In particular, if $K$ is Galois over $F$, then $M\triangleleft \langle M,N\rangle$, so $\langle M,N\rangle = MN$. Thus, $N\cap M$ is normal in $N$, and by the isomorphism theorems we have that $$\frac{G}{M} =\frac{MN}{M} \cong \frac{N}{N\cap M} \cong N.$$ Now, $\frac{G}{M}\cong\mathrm{Gal}(K/F)$; and $N=\mathrm{Gal}(KE/E)$; so we get the isomorphism if $K$ is Galois.

For an example showing that the given conditions do not imply that $K$ is Galois over $K\cap E$, let $F=\mathbb{Q}$, $K=\mathbb{Q}[\sqrt[3]{2}]$, and $E=\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive cubic root of unity; then $K\cap E=\mathbb{Q}$. Then $KE$ is the splitting field of $x^3-2$ over $\mathbb{Q}$, hence is Galois. Even though $KE$ is Galois over $E$, $K$ is not Galois over $\mathbb{Q}$, so you cannot have the claimed isomorphism.

share|improve this answer
    
Thank you very much! –  Belgi Jul 20 '12 at 21:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.