If KF is Galois, and the coefficients of the monic f∈K[x] generates K, then ∏σ∈Gal(KF)σf∈F[x] How does one see that g∈F[x]?

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If KF is Galois, and the coefficients of the monic f∈K[x] generates K, then ∏σ∈Gal(KF)σf∈F[x]  How does one see that g∈F[x]?

Jenny Bolton · Accepted Answer

Step 1Let KF be a galois extension thenKGal(KF)={x∈F:σx=x, ∀σ∈Gal(KF)}=F(⋆)Thus, pick some σ∈Gal(KF). We haveσf′=σ∏τ∈Gal(KF)τf=∏τ∈Gal(KF)τf=f′because the mapGal(KF)→Gal(KF),τ↦στinduces a permutation of Gal(KF) leaving the product unchanged. Hence applying any σ∈Gal(KF) has left the coefficients of f′ fixed, thus by (⋆) they must lie within F.

scoollato7o · Accepted Answer

Step 1If you want go gain some intuition, considerF=QandK=Q[2]f=2x+1theng=(2x+1)(−2x+1)=−2x2+1∈Q[x]The proof is straightforward. Any σ∈Gal(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 g∈F[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&#039;t even need the coefficients of f generate K.

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

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