Let \(K\) be a field of characteristic zero and \(L\) a Galois extension of \(K\). Let \(f\) be an irreducible polynomial in \(K[x]\) of degree 7 and suppose \(f\) has no zeroes in \(L\). Show that \(f\) is irreducible in \(L[x]\).
Solution
1. Let \(\alpha\) be a root of \(f\) in an algebraic closure of \(L\), and set\[ E=K(\alpha). \]Then\[ [E:K]=7. \]
Since \(f\) is irreducible over \(K\) and has degree \(7\), it is the minimal polynomial of \(\alpha\) over \(K\). Therefore \[ [E:K]=\deg f=7. \]
2. If \(f\) were reducible over \(L\), then it would be reducible over some finite Galois intermediate extension\[ K\subseteq M\subseteq L. \]
A nontrivial factorization of \(f\) in \(L[x]\) uses only finitely many coefficients from \(L\). Let \(N/K\) be the finite extension generated by those coefficients. Since \(L/K\) is Galois, hence algebraic, normal, and separable, the normal closure over \(K\) of the finite extension \(N/K\) is a finite Galois extension \(M/K\) contained in \(L\). The same factorization is then defined over \(M\), so \(f\) is reducible in \(M[x]\).
3. For every finite Galois intermediate extension \(M/K\) inside \(L\),\[ [M(\alpha):M]=[E:E\cap M]. \]
This is the standard degree formula for a compositum when one of the two finite extensions is Galois over the base: since \(M/K\) is Galois and \(E/K\) is finite, \[ [EM:M]=[E:E\cap M]. \] Here \(EM=M(\alpha)\) because \(E=K(\alpha)\).
4. For every such \(M\),\[ E\cap M=K. \]
1. Hence\[ [E\cap M:K]\in\{1,7\}. \]If this degree were \(7\), then \(E\cap M=E\), so \(E\subseteq M\subseteq L\). In particular \(\alpha\in L\), contradicting the hypothesis that \(f\) has no zero in \(L\). Therefore\[ [E\cap M:K]=1, \]so \(E\cap M=K\).
5. Hence for every finite Galois \(M/K\) contained in \(L\),\[ [M(\alpha):M]=7. \]
Combine 3 and 4: \[ [M(\alpha):M] =[E:E\cap M] =[E:K] =7. \] Thus the minimal polynomial of \(\alpha\) over \(M\) has degree \(7\). Since it divides the degree-\(7\) polynomial \(f\), it must equal \(f\). Therefore \(f\) is irreducible over \(M\).
6. The polynomial \(f\) is irreducible in \(L[x]\).
If \(f\) were reducible over \(L\), 2 would give a finite Galois intermediate field \(M\) over which \(f\) is reducible, contradicting 5.