# Obtaining invertible matrix from a separable field extension

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) :

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.

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.)

• Is your extension $A/K$ finite ? – Souvik Dey May 29 '17 at 6:15
• @SouvikDey : It is not even assumed algebraic. You can define separable extensions in the case where the transcendence degree might be positive. A proof in the finite case might help (I know enough to work things out in the arbitrary case, I guess). – Patrick Da Silva May 30 '17 at 0:09

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.

• You only dealt with the case where $A/K$ is finite. If $A/K$ is algebraic but of infinite degree, let $E = K(a_1,\cdots,a_n)$, which is of finite degree because $A/K$ is algebraic. Since $E$ and $A$ have the same algebraic closure, by applying the result to the finite case, you can find the required automorphisms. – Patrick Da Silva May 30 '17 at 23:19
• I am still stuck in the arbitrary case. Again, I take $E = K(a_1,\cdots,a_n)$. By classical results on separable extensions and separating transcendence bases, the subset $\{a_1,\cdots,a_n\}$ contains a separating transcendence base. I pick it to be $a_{m+1},\cdots,a_n$ so that $E/K(a_{m+1},\cdots,a_n)$ is separable algebraic. By the result on the finite case, we find $\sigma_1,\cdots,\sigma_m \in \mathrm{Aut}(\Omega/K(a_{m+1},\cdots,a_n))$ which has corresponding matrix with non-zero determinant. This forces $\sigma_i(\alpha_j) = \alpha_j$ for $m < j \le n$. So I have $n$ rows of the matrix. – Patrick Da Silva May 30 '17 at 23:21
• With your help I managed to find Bourbaki's proof in his reference (it was about two chapters later but in the same book~; I couldn't find it since the reference was wrong, but your proof gave me the understanding to know where to look!) I'll post a proof when I am done reading. – Patrick Da Silva May 31 '17 at 4:14
• Yeah no this proof is way too long and intricate. Won't post that here, I'll just quote the reference. However, his proof is very different in flavor (and incredibly more complex), so it's nice to have a proof in the algebraic case that works much better! Thanks for that! – Patrick Da Silva Jun 2 '17 at 2:01

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).