Obtaining invertible matrix from a separable field extension

Question

I was reading a proof in Bourbaki, Chapter VIII on Non-commutative algebra ($\S7$, n$^{\circ}2$, Proposition $3$ b)). In the proof, they claim the following result (translated from French) :

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Let $A/K$ be a separable field extension and $\Omega$ be an algebraic closure of $A$. Suppose $a_1,\cdots,a_n \in A$ are linearly independent over $K$. Then there exist $\sigma_1,\cdots,\sigma_n \in \mathrm{Aut}(\Omega/K)$ such that the matrix $(\sigma_i(a_j)) \in \mathrm{Mat}_{n \times n}(\Omega)$ is invertible.

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The author insists on separability and on the fact that $A/K$ does not have to be algebraic.

They give a reference (Chapitre V, $\S7$, n$^{\circ} 2$) but the section is about separable polynomials and contains nothing about the automorphism group $\mathrm{Aut}(\Omega/K)$. I've tried looking in the sections before and after but no success. Does anyone have a reference for this result? It feels like a well-known result but then again I can't think of anywhere to look. (Even though this has to do with separability, I don't even know how to prove it in characteristic zero.)

Answer 1

Here is a proof for the algebraic case (and it is probably not very hard to extend it to the general case, although my proof for that is still incomplete). So suppose that $A/K$ is algebraic, i.e. has finite degree.

By the primitive element theorem, there is a $\alpha\in K$ such that $A=K[\alpha]$. Let $M$ be the minimal polynomial of $\alpha$, and let $d=[A:K]$. Then the degree of $M$ is exactly $d$, and by separability the roots $\alpha_1=\alpha,\alpha_2,\ldots,\alpha_d$ of $M$ in $\Omega$ are distinct. For each $k$ , there is a unique field homomorphism $\tau_k : A \to \Omega$ sending $\alpha$ to $\alpha_k$ and acting as the identity on $K$, and this homomorphism may be (nonuniquely) extended to a field homomorphism $\sigma_k : \Omega \to \Omega$. We can find $a_{n+1},\ldots,a_d \in L$ such that $(a_1,a_2,\ldots,a_d)$ is a $K$-basis of $A$. Since the characters $\tau_1,\ldots,\tau_d$ are linearly independent by Dedekind's lemma, it follows that the matrix $M_1=(\tau_i(a_j)) \in \mathrm{Mat}_{d \times d}(\Omega)$ is invertible. It follows that the matrix $M_2$ consisting of the first $n$ columns of $M_1$ has rank exactly $n$, so at least one of the $n\times n$ minors of $M_2$ must be nonzero, and we are done.

Answer 2

The reference in the book I was reading was incorrect (and not off by just one or two pages!) but I ended up knowing where to look, thanks to Ewan Delanoy.

A proof is outlined in Bourbaki, Algèbre, Chapitre V, $\S$ 15, p. 120, Section 6, Theorem 4 (i) $\Longleftrightarrow$ (iii). He uses the notion of rationality of vector spaces on field extensions and things are not very trivial, but doable with a lot of patience (and knowledge of French!). The proof may seem short, but as any proof in Bourbaki, it assumes a lot of material, so if someone reads this answer and just wants to have a look, I am warning you : you might be missing a lot of material to read this proof, so be ready to read 10, 20, maybe 50 pages of material to fully understand it.

I have written down the proof myself for my own purposes (in English) in my commutative algebra notes. They are available on my academic webpage (see the Separable extensions section).