Timeline for How to prove these two rings are not isomorphic
Current License: CC BY-SA 2.5
8 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 8, 2010 at 20:07 | comment | added | Georges Elencwajg | Tom, I'm happy that our little misunderstanding has been cleared. Cheers. | |
| Jul 8, 2010 at 19:40 | comment | added | Tom Church | The point I was missing is that an isomorphism between the original rings would give an isomorphism over $\mathbb{C}$ between $\text{SL}_n\mathbb{C}$ and $\mathbb{A}^{n^2-1}$, which would certainly imply that their $\mathbb{C}$--points are homeomorphic. My confusion was that I thought you were arguing that $\mathbb{C}[x_1,\ldots,x_{n^2-1}]$ and $\mathbb{C}[y_1,\ldots,y_{n^2}]/(\text{det}=1)$ are non-isomorphic as rings, which as far as I can tell doesn't follow from your argument, rather than as $\mathbb{C}$-algebras. (Of course this isn't what was asked.) | |
| Jul 8, 2010 at 19:31 | comment | added | Georges Elencwajg | @Tom Church: I am baffled at your claim that my argument proves that $\mathbb R$ and $\mathbb R^2$ are isomorphic : it certainly does nothing of the sort . Also, Serre's example is irrelevant in this context: he starts with a scheme defined over a quadratic number field, which has two embeddings into $\mathbb C$ . Here we have schemes defined over $\mathbb Z$ and I don't even know what you mean by "conjugate" . For the rest, Robin has given a very clear explanation: thanks Robin! | |
| Jul 8, 2010 at 17:17 | comment | added | Robin Chapman | This is all quite straightforward: if we have two affine subvarieties $X$ and $Y$ of complex affine space, and a variety isomorphism $f:X\to Y$, then $f$ is given by a bunch of polynomials so it is continuous as a map on the complex points. The same holds for $f^{-1}$, so the sets of complex points of $X$ and $Y$ are homoemorphic as topological spaces. | |
| Jul 8, 2010 at 15:39 | comment | added | Tom Church | Your argument seems to also prove that $\mathbb{R}$ and $\mathbb{R}^2$ are not isomorphic as groups (they're not homeomorphic!). And there are varieties which are conjugate but whose $\mathbb{C}$-points are not homeomorphic (I know Serre constructed a non-singular projective surface which is an example, not sure about affine examples). Can you explain how to get the reduction to the classical topology (without going through étale cohomology or something like that)? | |
| Jul 8, 2010 at 15:23 | history | edited | Georges Elencwajg | CC BY-SA 2.5 |
Added "h" to OP's name
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| Jul 8, 2010 at 12:20 | history | answered | Georges Elencwajg | CC BY-SA 2.5 |