Timeline for reduced ⊗ reduced = reduced; what about connected?
Current License: CC BY-SA 3.0
25 events
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| May 23, 2020 at 17:55 | comment | added | user20948 | @HeinrichD It looks like a constructive proof of (1) is possible. Indeed, we can assume that $A,B$ are finitely generated over $k$, and then it might be possible to start with Gröbner basis. | |
| Dec 14, 2016 at 17:57 | history | edited | darij grinberg | CC BY-SA 3.0 |
improve latex and add link
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| Sep 18, 2016 at 11:44 | answer | added | HeinrichD | timeline score: 6 | |
| Sep 18, 2016 at 8:12 | comment | added | darij grinberg | @HeinrichD: no. All I've learned from this thread is that commutative algebra is hard... | |
| Sep 18, 2016 at 7:42 | comment | added | HeinrichD | @darij: Even if $A,B$ are fields, I don't see a constructive proof either. Almost all proofs in commutative algebra that I've learnt are terribly non-constructive. Regarding (2), have you found some counterexample? | |
| Sep 16, 2016 at 19:20 | comment | added | darij grinberg | @HeinrichD: No, unfortunately. | |
| Sep 16, 2016 at 19:02 | comment | added | HeinrichD | @darij: Have you meanwhile found a constructive proof of (1), avoiding the use of prime ideals? | |
| Aug 18, 2012 at 10:15 | history | edited | darij grinberg | CC BY-SA 3.0 |
now that was some bullshit
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| Feb 25, 2010 at 22:54 | history | edited | Charles Stewart |
edited tags
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| Feb 25, 2010 at 15:33 | answer | added | Qing Liu | timeline score: 7 | |
| Dec 21, 2009 at 22:44 | answer | added | Hideyuki Kabayakawa | timeline score: 1 | |
| Dec 20, 2009 at 22:16 | comment | added | darij grinberg | Ah, okay. Maybe this can be rescued by restating the condition that $k$ be algebraically closed in terms of schemes, but anyway this is just a rewording of my question. | |
| Dec 20, 2009 at 21:48 | history | edited | Anton Geraschenko | CC BY-SA 2.5 |
edited title; edited title
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| Dec 20, 2009 at 21:46 | comment | added | Georges Elencwajg | @darij : my comment was a counter-example to Mike's assertion " you're asking if the product of connected affine schemes is connected (which is true)" . | |
| Dec 20, 2009 at 21:20 | answer | added | Georges Elencwajg | timeline score: 5 | |
| Dec 20, 2009 at 20:35 | answer | added | user2035 | timeline score: 10 | |
| Dec 20, 2009 at 20:03 | comment | added | darij grinberg | That's why I edited my above post to require $k$ to be algebraically closed (for (3)). Is it still wrong then? | |
| Dec 20, 2009 at 19:50 | comment | added | Georges Elencwajg | It is NOT true that the product of two connected affine schemes is connected. For example the affine scheme X=Spec(Q(i)) is connected since it has only one point, but the product of X with itself is equal to the sum (=coproduct) of two copies of X. Topologically it is a discrete space with two ponts, hence not connected. | |
| Dec 20, 2009 at 19:29 | comment | added | darij grinberg | I meant the assertion that the product of connected affine schemes is connected (I assume that you mean the fibred product, which is not a topological product - or am I mistaken here?). What you just showed is that the tensor product of rings corresponds to the fibred product of the corresponding affine schemes over the trivial scheme - but I know that. | |
| Dec 20, 2009 at 19:25 | comment | added | Mike Skirvin | Since we're talking about affine schemes, my assertion above is equivalent to showing that the tensor product of two rings is a push-out in the category of commutative rings. So you just need to show that, given any ring $R$ and maps $A \to R$ and $B \to R$ which agree on $k$, this induces a map $A \otimes_k B \to R$ making the obvious diagram commute. This latter fact is a pretty straightforward exercise in ring theory. | |
| Dec 20, 2009 at 19:08 | comment | added | darij grinberg | Thanks, but how do you prove this true fact? | |
| Dec 20, 2009 at 19:08 | history | edited | darij grinberg | CC BY-SA 2.5 |
@(3): note to self: first think, then post; deleted 2 characters in body
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| Dec 20, 2009 at 19:04 | comment | added | Mike Skirvin | Tensor product of rings corresponds to pull-backs of affine schemes. So in question (3) you're asking if the product of connected affine schemes is connected (which is true). | |
| Dec 20, 2009 at 19:02 | history | edited | darij grinberg | CC BY-SA 2.5 |
added 57 characters in body
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| Dec 20, 2009 at 18:55 | history | asked | darij grinberg | CC BY-SA 2.5 |