If \frac{K}{F} is Galois, and the coefficients of the monic

Pamela Meyer

Pamela Meyer

Answered question

2022-01-13

If KF is Galois, and the coefficients of the monic fK[x] generates K, then σGal(KF)σfF[x]
How does one see that gF[x]?

Answer & Explanation

Jenny Bolton

Jenny Bolton

Beginner2022-01-14Added 32 answers

Step 1
Let KF be a galois extension then
KGal(KF)={xF:σx=x, σGal(KF)}=F()
Thus, pick some σGal(KF). We have
σf=στGal(KF)τf=τGal(KF)τf=f
because the map
Gal(KF)Gal(KF),τστ
induces a permutation of Gal(KF) leaving the product unchanged. Hence applying any σGal(KF) has left the coefficients of f fixed, thus by () they must lie within F.
scoollato7o

scoollato7o

Beginner2022-01-15Added 26 answers

Step 1
If you want go gain some intuition, consider
F=Q
andK=Q[2]
f=2x+1
then
g=(2x+1)(2x+1)=2x2+1Q[x]
The proof is straightforward. Any σGal(KF)
permutates F and hence leaves g unchanged. And by KF is Galois, the only elements in K that are invariant under Gal(KF) are elements of F, thus gF[x].
Pick a root α of f, then α is also a root of g. If g is irreducible over F, then
[F(α):F]=deg(g)=deg(f)|Gal(KF)|=deg(f)[K:F]
And
[F(α):F][K(α):F]=[K(α):K][K:F]
[K(α):K][F(α):F][K:F]=deg(f)
Since α is a root of f, we also have [K(α):K]deg(f), hence [K(α):K]=deg(f) and f must be the minimal polynomial of α over K hence irreducible.
Note that we don't even need the coefficients of f generate K.

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