Proof of primitive element theorem in Lang's *Algebra*

Question

I am reading the proof of the Primitive Element Theorem from Lang's Algebra. This is from the 3rd edition, on page 243.

Theorem 4.6. (Primitive Element Theorem) Let $E$ be a finite extension of a field $k$. There exists an element $\alpha \in E$ such that $E = k(\alpha)$ if and only if there exists only a finite number of fields $F$ such that $k \subset F \subset E$. If $E$ is separable over $k$, then there exists such an element $\alpha$.

We additionally assume that $k$ is an infinite field. I am stuck at a particular point in the proof of the forward implication, that is, if there is a primitive element, then there are only finitely many intermediate fields. Here is how the proof in the book for this part goes:

Let $E = k(\alpha)$ and $f(X) = \mathrm{Irr}(\alpha,k,X)$. Then, there is a map from the set of intermediate fields to the set of polynomials in $E[X]$ that divide $f$, given by $$F \mapsto g_F,$$ where $k \subset F \subset E$, and $g_F(X) = \mathrm{Irr}(\alpha,F,X)$. Clearly, if we show that this map is injective, then we will have shown that the number of intermediate fields is finite.

To show that the map is injective, we proceed as follows:

Let $F_0$ be the subfield of $F$ generated over $k$ by the coefficients of $g_F(X)$. Then $g_F$ has coefficients in $F_0$ and is irreducible over $F_0$ since it is irreducible over $F$. Hence the degree of $\alpha$ over $F_0$ is the same as the degree of $\alpha$ over $F$. Hence $F = F_0$.

I am stuck at the last line. I am unable to show that $$[F(\alpha):F] = [F_0(\alpha):F_0] \implies F = F_0.$$ I've tried to manipulate the expression on the left-hand side, to show that $[F:F_0] = 1$, but somehow with no success. What I could do so far is just: $$ \begin{align} [F(\alpha):F_0] &= [F(\alpha):F_0(\alpha)] [F_0(\alpha):F_0]\\ [F(\alpha):F_0] &= [F(\alpha):F] [F:F_0]\\ \therefore\ [F(\alpha):F_0(\alpha)] &= [F:F_0]. \end{align} $$ I'm sure that I'm only missing something quite elementary. Can someone help me show this implication?

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EDIT: As @YumekuiMath points out in the comments, I have forgotten to use the fact that $E = k(\alpha)$. The rest is now easy from @DonAntonio’s hint.

Answer 1

All you need here is the well known theorem from linear algebra "If $\;V\;$ is a finite dimensional linear space over a field $\;F\;$ and $\;W\;$ is a subspace of $\;V\;$ such that $\;\dim W=\dim V\;$ , then $\;W=V\;$ " ...

Answer 2

For the sake of completeness, I am expanding on @DonAntonio's hint and @YumekuiMath's comment under the accepted answer.

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Since $k \subset F, F_0 \subset E$ and $E = k(\alpha)$, we have $F(\alpha) = E = F_0(\alpha)$. We have already shown that $[F:F_0] = [F(\alpha):F_0(\alpha)]$. The RHS is just $[E:E] = 1$, so $F = F_0$.

Alternatively, observe that $g_F(X)$ is irreducible over $F_0$, since it is irreducible over $F \supset F_0$. Hence, $[E:F_0] = [E:F] = \deg g_F$. Thus, $$ [E:F_0]=[E:F][F:F_0] \implies [F:F_0] = 1. $$