3
$\begingroup$

Let $K$ and $L$ be extensions of $F$. Show that $KL$ is Galois over $F$ if both $K$ and $L$ are Galois over $F$. Is the converse true?

I should show $|Gal(KL/F)|=[KL:F]$ and for this I am using $|Gal(K/F)|=[K:F], |Gal(L/F)|=[L:F] $ and $[KL:F]\leq[K:F][L:F]$, and I get to the next but I do not know what else to do, could someone help me please? Is there a counterexample to the converse? Thank you in advance.

$\endgroup$
3
  • 1
    $\begingroup$ The converse is easy to disprove, what non-Galois extension do you know ? And (a separable finite extension) $K/F$ is Galois iff there is a collection of elements $\alpha_1,\ldots,\alpha_n$ closed under $F$-conjugaison (other roots of the $F$-minimal polynomial) such that $K = F(\alpha_1,\ldots,\alpha_n)$. If $K/F$ and $L/F$ are Galois then $ KL = F(\ldots)$ $\endgroup$
    – reuns
    Oct 16, 2017 at 15:55
  • $\begingroup$ @reuns $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}$ $\endgroup$
    – user425181
    Oct 16, 2017 at 15:59
  • 2
    $\begingroup$ Your proof for the direct statement is not OK as it is. $\endgroup$
    – orangeskid
    Oct 16, 2017 at 16:01

1 Answer 1

Reset to default
4
$\begingroup$

To prove that the converse is not true, consider the extension $\mathbb{Q}(\sqrt[3]{2},\zeta_{3})$ which is a Galois extension of $\mathbb{Q}$, but $\mathbb{Q}(\sqrt[3]{2})$ is not! (Edit: Take $K=\mathbb{Q}(\sqrt[3]{2})$ and $L=\mathbb{Q}(\zeta_{3})$ and $\mathbb{F}=\mathbb{Q})$ As for the direct statement, we might use splitting fields. Now, if we want to use splitting fields, if $K$ is a splitting field over $F$ for $f(x)$ and $L$ is a splitting field of $g(x)$ over $F$, then $KL$ is a splitting field of...

$\endgroup$
1
  • 1
    $\begingroup$ Mention separability (or characteristic zero), otherwise those are only normal extensions $\endgroup$
    – reuns
    Oct 16, 2017 at 16:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .