Let \(F\) be a field, and let \(K, L\) be finite dimensional extensions of \(F\) inside a common overfield. Suppose \([K : F]\) and \([L : F]\) are relatively prime. Prove that \([KL : F] = [K : F][L : F]\). Here \(KL\) is the compositum of \(K\) and \(L\).
Give an example of fields \(F, K, L\) such that \(K \cap L = F\) and \([KL : F] < [K : F][L : F]\).
Solution
1. For any two finite extensions in a common overfield,\[ [KL:F]\leq[K:F][L:F]. \]
Put \(m=[K:F]\) and \(n=[L:F]\), and choose an \(F\)-basis \(\ell_1,\ldots,\ell_n\) of \(L\) with \(\ell_1=1\). Inside the common overfield, let \[ S=K\ell_1+\cdots+K\ell_n. \] This is a \(K\)-vector space of dimension at most \(n\), containing both \(K\) and \(L\). For every \(i,j\), the product \(\ell_i\ell_j\) is an \(F\)-linear combination of the basis elements. Consequently \(S\) is closed under multiplication, and it contains \(1\).
For nonzero \(s\in S\), multiplication by \(s\) is an injective \(K\)-linear map \(S\to S\), since the ambient ring is a field. Finite dimension makes this map surjective. In particular, \(st=1\) for some \(t\in S\), so \(S\) is a field. It contains \(K\) and \(L\) and is contained in their compositum, hence \(S=KL\). Therefore \([KL:K]\leq n\), and the tower law gives \([KL:F]=[KL:K]m\leq mn\).
2. Coprimeness forces equality, proving part (a).
1. Thus \([KL:F]=mn\).
3. For part (b), take two different cubic subfields of thesplitting field of \(x^3-2\).
Let \(\alpha=\sqrt[3]{2}\) be real and let \(\zeta\) be a primitive cube root of unity. Inside \(\mathbb C\), set \[ F=\mathbb Q,\qquad K=\mathbb Q(\alpha),\qquad L=\mathbb Q(\zeta\alpha). \] Both displayed generators have minimal polynomial \(x^3-2\), which is irreducible by Eisenstein’s criterion at \(2\) [@DF04]. Thus \([K:\mathbb Q]=[L:\mathbb Q]=3\). The fields are different because \(K\subseteq\mathbb R\), whereas \(\zeta\alpha\) is nonreal.
The degree \([K\cap L:\mathbb Q]\) divides \(3\) by the tower law. If it were \(3\), then \(K\cap L=K=L\), contrary to distinctness. Hence it is \(1\) and \(K\cap L=\mathbb Q\).
The compositum contains \(\zeta=(\zeta\alpha)/\alpha\), so \[ KL=\mathbb Q(\alpha,\zeta)=K(\zeta). \] The element \(\zeta\) satisfies \(x^2+x+1\) and does not belong to the real field \(K\), so \([K(\zeta):K]=2\). Consequently \[ [KL:F]=2\cdot3=6<9=[K:F][L:F]. \] This verifies both conditions in part (b).