Proof.
Let \alpha \in E be a primitive element. Let P be the minimal polynomial of \alpha over F. Let E \subset M be a splitting field for P over E, so that P(x) = (x - \alpha )(x - \alpha _2) \ldots (x - \alpha _ n) over M. For ease of notation we set \alpha _1 = \alpha . Next, let E/K/F be a subextension. Let Q be the minimal polynomial of \alpha over K. Observe that \deg (Q) = [E : K]. Writing Q = x^ d + \sum _{i < d} a_ i x^ i we claim that K is equal to L = F(a_0, \ldots , a_{d - 1}). Indeed \alpha has degree d over L and L \subset K. Hence [E : L] = [E : K] and it follows that [K : L] = 1, i.e., K = L. Thus it suffices to show there are at most finitely many possibilities for the polynomial Q. This is clear because we have a factorization P = QR in K[x] in particular in E[x]. Since we have unique factorization in E[x] there are at most finitely many monic factors of P in E[x].
If F is a finite field (equivalently E is a finite field), then E/F has a primitive element by the discussion in Section 9.18. Next, assume F is infinite and there are at most finitely many proper subfields E/K/F. List them, say K_1, \ldots , K_ N. Then each K_ i \subset E is a proper sub F-vector space. As F is infinite we can find a vector \alpha \in E with \alpha \not\in K_ i for all i (a vector space can never be equal to a finite union of proper subvector spaces; details omitted). Then \alpha is a primitive element for E over F.
Having established the equivalence of (1) and (2) we now turn to the final statement of the lemma. Choose an algebraic closure \overline{F} of F. Enumerate the elements \sigma _1, \ldots , \sigma _ n \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F}). Since E/F is separable we have n = [E : F] by Lemma 9.12.11. Note that if i \not= j, then
V_{ij} = \mathop{\mathrm{Ker}}(\sigma _ i - \sigma _ j : E \longrightarrow \overline{F})
is not equal to E. Hence arguing as in the preceding paragraph we can find \alpha \in E with \alpha \not\in V_{ij} for all i \not= j. It follows that |\mathop{\mathrm{Mor}}\nolimits _ F(F(\alpha ), \overline{F})| \geq n. On the other hand [F(\alpha ) : F] \leq [E : F]. Hence equality by Lemma 9.12.11 and we conclude that E = F(\alpha ).
\square
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