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Dec 9, 2020 at 22:28 comment added Kimball I guess Julian answered your question, but IIRC, I believe this book has errata posted online.
Dec 9, 2020 at 3:38 comment added Rita @JulianRosen: Ah, that makes a lot of sense. Thanks!
Dec 9, 2020 at 3:22 comment added Julian Rosen 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 $D\cap U\neq\varnothing$.
Dec 9, 2020 at 2:29 history edited Toby Bartels CC BY-SA 4.0
typo, clarify edition (per comments)
Dec 9, 2020 at 1:38 comment added Rita @KReiser: I'm not sure I follow. Why must $D$ lie in the standard $\mathbb{A}_k^{n^2}$?
Dec 9, 2020 at 0:27 comment added KReiser 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 $U$.
Dec 9, 2020 at 0:16 comment added darij grinberg Just meant this should probably be said in the post :) I agree -- this argument looks fishy.
Dec 8, 2020 at 23:59 comment added Rita @darijgrinberg: I'm reading the second edition (so I guess if I'm interpreting it correctly this was not corrected in the revision!).
Dec 8, 2020 at 23:57 review First posts
Dec 9, 2020 at 2:29
Dec 8, 2020 at 23:56 comment added darij grinberg Note: The book has two editions.
Dec 8, 2020 at 23:54 history asked Rita CC BY-SA 4.0