Galois extension

  • Thread starter math_grl
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This stuff is killing me...

Let [tex]K \leq M \leq L[/tex] be fields such that L is galois over M and M is galois over K. We can extend [tex]\phi \in G(M/K)[/tex] to an automorphism of L to show L is galois over K.

I need help filling in the details in why exactly L is galois over K.
 

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  • #2
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Something like this, I think: let [tex][M:K]=a[/tex] and [tex][L:M]=b[/tex]. Since [tex]M/K[/tex] is Galois, there are [tex]a[/tex] automorphisms of [tex]M[/tex] that fix [tex]K[/tex]. For the same reason, there are [tex]b[/tex] automorphisms of [tex]L[/tex] that fix [tex]M[/tex]. So there are [tex]ad[/tex] automorphisms of [tex]L[/tex] that fix [tex]K[/tex]. Since [tex]ad[/tex] is also the degree of [tex]L/K[/tex], it's Galois.

If you're using Dummit and Foote, check out theorems 13.8 and 13.27.
 

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