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I'm trying to understand the proof of Proposition 2.2.10 in Gille-Samuely's book "Central simple algebras and Galois cohomology" (2nd edition), and I believe it has an error.

Here's the setup (which I'll try to write in such a way that you won't need the book to follow this). Let $k$ be an infinite field and let $D$ be a central division algebra over $k$ of degree $n$. We then have $D \otimes_k \overline{k} \cong M_n(\overline{k})$. We can identify $M_n(\overline{k})$ with $\mathbb{A}^{n^2}_{\overline{k}}$. Let $U \subset \mathbb{A}^{n^2}_{\overline{k}}$ be the open set consisting of matrices whose characteristic polynomial is separable.

What the authors now want to argue is that $U$ contains a point of $D$, i.e. a $k$-rational point. Their argument for this seems fishy: they say that we can identify $D$ with $\mathbb{A}^{n^2}_k \subset \mathbb{A}^{n^2}_{\overline{k}}$, which makes no sense! The $k$-structure we are considering is not the obvious one.

How can I fix this?

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    $\begingroup$ Note: The book has two editions. $\endgroup$
    – darij grinberg
    Commented Dec 8, 2020 at 23:56
  • $\begingroup$ @darijgrinberg: I'm reading the second edition (so I guess if I'm interpreting it correctly this was not corrected in the revision!). $\endgroup$
    – Rita
    Commented Dec 8, 2020 at 23:59
  • $\begingroup$ Just meant this should probably be said in the post :) I agree -- this argument looks fishy. $\endgroup$
    – darij grinberg
    Commented Dec 9, 2020 at 0:16
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    $\begingroup$ Even if the identification is not the usual one, this shouldn't matter, right? The standard copy of $\Bbb A^{n^2}_k\subset \Bbb A^{n^2}_{\overline{k}}$ is dense because $k$ is infinite, and then the image of $D$ in the standard $\Bbb A^{n^2}_k$ is of the same dimension, so it must be dense too. So the image of $D$ in $\Bbb A^{n^2}_{\overline{k}}$ is dense, and must intersect $U$. $\endgroup$
    – KReiser
    Commented Dec 9, 2020 at 0:27
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    $\begingroup$ If $x_1,\ldots,x_{n^2}$ is a basis for $D$, we could identify $D\otimes \bar{k}$ with $\mathbb{A}^{n^2}_{\bar{k}}$ via $\sum x_i\otimes\lambda_i\mapsto (\lambda_i)_i$ (this identification is not the same as the one coming from an isomorphism $D\otimes \bar{k}\simeq M_n(\bar{k})$). With this identification, it is clear that $D$ is identified with $\mathbb{A}^{n^2}_k$, so $D\cap U\neq\varnothing$. $\endgroup$
    – Julian Rosen
    Commented Dec 9, 2020 at 3:22

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