Geometric interpretation of primitive element theorem?

Question

The primitive element theorem is a basic result about field extensions. I was wondering whether there are nice geometric ways to visualize it or think about it. Since field spectra are singletons, it has to be about the non-trivial automorphisms of points (I think), and I don't know how to think about it.

Answer 1

Reformulate the primitive element theorem as follows:

Let $\mathfrak m \subset k[X_1, \dotsc, X_n]$ be a maximal ideal, such that the obtained field extension is separable. For general $(a_1, \dotsc, a_n) \in k^n$, the map $$k[T] \to k[X_1, \dotsc, X_n]/\mathfrak m, T \mapsto a_1X_1 + \dotsb + a_nX_n$$ is surjective.

Geometrically, any closed point of $\mathbb A_k^n$ with separable residue field (i.e. a smooth point) can be realized as a closed point of $\mathbb A_k^1$.

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Another geometric interpretation is the following. To be precise it is a consequence, but I am not sure whether it is equivalent.

Let $X$ be an irreducible variety of dimension $n-1$ over a perfect field $k$. Then $X$ is birationally equivalent to a hypersurface in $\mathbb A_k^n$.

Let $K(X)$ be the function field of $X$. Pick a transdence basis, i.e. a finite extension $K(X)/k(x_1, \dotsc, x_{n-1})$. By the primitive element theorem we can write this extension as

$$K(X) = k(x_1, \dotsc, x_{n-1})[x_n]/(f(x_n)) = \operatorname{Frac} k[x_1, \dotsc, x_n]/(f(x_n))$$

, hence $X$ is birationally equivalent to the hypersurface given by $f=0$.

Answer 2

Here’s a geometric argument that has nothing to do with algebraic geometry. You may find it insufficiently rigorous, but the idea is certainly sound.

Consider a separable extension $K\supset k$. One consequence of separability is that there are only finitely many intermediate fields $E$, $K\supset E\supset k$. Consider the finitely many proper subfields, and look at their (set-theoretic) union — not their join, not their compositum, just their union. This is a union of finitely many proper $k$-subspaces of $K$, each of them of dimension less than the dimension $n=[K:k]$ of $K$ as a $k$-space. But certainly, if $k$ is infinite, you can’t fill up an $n$-dimensional $k$-space with finitely many subspaces of lower dimension. So there has to be an element $\alpha\in K$ that isn’t in any proper subfield of $K$. Then this $\alpha$ must therefore generate $K$ as a field.