Timeline for Integrally closed factor rings and projective modules
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jul 10, 2013 at 17:09 | history | edited | Yemon Choi | CC BY-SA 3.0 |
Ann was upright before the LaTeX; it should remain upright *with* the LaTeX
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| Jul 10, 2013 at 17:07 | history | edited | Andrew Stacey | CC BY-SA 3.0 |
Fixed maths rendering
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| Nov 10, 2009 at 18:53 | comment | added | Greg Stevenson | I haven't thought hard enough about this point to know if it is a problem - but it isn't a problem in the sense that the local hypothesis is never really used. Projective is sufficient in the above - a projective module is still supported everywhere and our colimit is filtered so the flatness argument still goes through. I haven't had a chance to think further about the non-domain case yet. | |
| Nov 10, 2009 at 11:29 | comment | added | Jose Capco | There is another possible flaw in the argument (things that get really dangerous, when one say..) "we can reduce" to the case R is local. You see, in the initial argument (supposing that it's not necessarily integral domain) you made use of some "a" being essential over R.. but if we go deeper into the "we can reduce" thing, we are localizing at every closed point of Spec R, but I'm not sure if even here essentiality can be lifted. We are making use of something in the hypothesis that is not necessarily a local property | |
| Nov 8, 2009 at 1:38 | comment | added | Greg Stevenson | Yes, you are right - so whether or not this method works depends on whether or not one can choose an essential lift I suppose. One could possibly do this using a norm argument. | |
| Nov 8, 2009 at 0:48 | history | edited | Greg Stevenson | CC BY-SA 2.5 |
had an error
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| Nov 7, 2009 at 23:55 | comment | added | Jose Capco | Your conclusion with the last diagram isn't necessarily true.. You have a commutative diagram with the lower horizontal \bar R --> \bar R[\bar b] being essential. You cannot conclude that R --> R[b] is essential from that, even if the diagram is commuting. | |
| Nov 7, 2009 at 23:39 | comment | added | Jose Capco | Well.. I think one can show S=R without much fancy methods. If R[a] is an integral extension of R with a belonging to the quotient field of R, then R[a] is also essential (we assumed R is a domain here) and by hypothesis free (we assumed R is local). This can only be if R=R[a] since a is a fraction from R, and so there is an r in R nonzero such that ra=b is in R.. so a and 1 are linearly dependent (assuming a≠0, since otherwise theres nothing to prove) | |
| Nov 7, 2009 at 23:15 | history | edited | Greg Stevenson | CC BY-SA 2.5 |
added argument to deal with injectivity
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| Nov 7, 2009 at 22:23 | comment | added | Greg Stevenson | I'll edit in that part of the argument - I was a bit slack saying how this works one really uses the fact that S over R is actually faithfully flat not just flat | |
| Nov 7, 2009 at 17:31 | comment | added | Jose Capco | ok.. I'm a bit lost here. How did you used flatness to show that yS∩R = yR .. I know yS=yR@S but who does this show yR=yS∩R?? If your proof is correct, then the conjecture would at least hold for an integral domain. I hope the question isn't so silly. | |
| Nov 7, 2009 at 12:37 | comment | added | Jose Capco | Nevertheless, your proof does use some arguments that is interesting to me (like using flatness of S to argue that yS∩R = yR). I think it is providing me something to work on. Regarding essentiality, its a general term in Category theory: Let C be a categry and then a monomorphim a-->b (here we see injectivity!) is said to be essential iff for any morphism b-->c such that a-->b-->c is a monomorphism one has b-->c is a monomorphism. For Crings this is equivalent to saying that any non-zero ideal of the overring, from a commutative ring extension of R, restricts to a nonzero ideal of R. | |
| Nov 7, 2009 at 10:30 | comment | added | Greg Stevenson | It is probably not quite what you are after then. I guess it is reasonable to restrict the notion of essential to injective morphisms in which case a different argument would be required or a counterexample might exist (I interpreted over-ring as algebra, requiring injectivity did occur to me but I figured essential should make sense in the same generality as integral). | |
| Nov 7, 2009 at 10:07 | comment | added | Jose Capco | I am about to digest your proof. But I thought to give a comment first. You wrote in the beginning that b is essential over R.. b is supposed to be belonging to an over-ring of R. And as far as I know essentiality requires injectivity... I thought it was clear that b is supposed to belong to an over-ring of R. You could mean one element that belongs to the product of factor over minimal primes whose projection to R/P is b, but which of this element preserves essentiality and integrality for both R and R/P? more comments later | |
| Nov 7, 2009 at 0:49 | history | answered | Greg Stevenson | CC BY-SA 2.5 |