If K $\subset$ L, K Galois over F, L is a splitting field of f(x) over F, can we say L Galois over F

Asked Mar 04 '15 at 16:41

Active Mar 04 '15 at 16:41

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Let $F$ be a field, $f(X) \in F[X]$ an irreducible polynomial of degree $n$ over $F$, $L$ a splitting field of $f(X)$ over $F$, and $\alpha \in L$ a root of $f(X)$. If $K$ is any Galois extension of $F$ contained in $L$, Can we draw the conclusion that $L$ is Galois over $F$?

This is part of a problem from Dummit&Foote(14.2 28). The hint of the problem says that If $H$ is the subgroup of the Galois group of $L$ over $F$ corresponding to $K$ then blablabla. I solve the problem via following the hint. However, the hint presumes that $L$ is Galois over $F$ and I use this result frequently in my proof. However, I am not sure that the statement $L$ is Galois over $F$ is true. Can anyone give any-counter example or hint how to prove the statement?

asked Mar 04 '15 at 16:41

scitamehtam

There is something wrong with your problem statement. The field $K$ is not involved at all. – bzc Mar 04 '15 at 16:45
@ Michael Burr Char not necessarily be zero – scitamehtam Mar 04 '15 at 16:47
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Do you mean either of these? Splitting field implies Galois extension or Galois extension is transitive – Michael Burr Mar 04 '15 at 16:48
@Brandon Carter if K plays no role, then I think I can't directly draw the conclusion that L is Galois over F since f(x) is not necessarily separable. – scitamehtam Mar 04 '15 at 16:51
@Michael Burr I am sorry that none of them are equivalent to the statement.~ – scitamehtam Mar 04 '15 at 16:52
$K$ plays no role since $K=F$ is Galois over $F$. – Gregory Grant Mar 04 '15 at 16:52
@Gregory Grant great idea! Thanks a lot! so the statement is false? – scitamehtam Mar 04 '15 at 16:53
Yeah since it's not true that a splitting field is necessarily Galois. – Gregory Grant Mar 04 '15 at 16:57

If K $\subset$ L, K Galois over F, L is a splitting field of f(x) over F, can we say L Galois over F

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