Timeline for Is the tensor product of regular rings still regular
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| May 23, 2011 at 14:38 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
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| Dec 23, 2010 at 23:22 | comment | added | Ying Zhang | (continuation of above)Then, to show $G(I^{\cdot})$ is also an injective(acyclic?) sequence of $B\otimes B$ modules, I think we need sth like $M$ to be a flat $B\otimes B$ module, which is not always true, since we choose $M$ arbitrarily. Maybe there are some ways to get around with this, or that I was trying to show sth not necessary. Anyway, the core question can be stated very clearly: is the composition of two functors each of whose derived functor has finite homological dimension also the derived functor for itself with finite homological dimension? | |
| Dec 23, 2010 at 23:15 | comment | added | Ying Zhang | @Tom Goodwillie: Thanks very much, it is much clearer now. I was still a little worried about "the derived functors of the composition is the composition of the derived functors". I was trying to show the finiteness directly. Suppose for an arbitrary $A\otimes B$ module M, find an injective resolution $I^{\cdot}$, which are also (at least acyclic) $A$ and $B\otimes B$ modules, could this always be achieved? Then, we have finite homological dimension when applying the functor $G$. | |
| Dec 23, 2010 at 4:15 | comment | added | Tom Goodwillie | For a $B\otimes B$-module $X$, a $B\otimes B$-linear map $B\to X$ corresponds precisely to an element $x\in X$ satisfying the condition that for every $b\in B$ we have $(b\otimes 1)x=(1\otimes b)x$. Taking $X$ to be the group of $A$-module maps $M\to N$ and defining the module structure by $((b_1\otimes b_2)f)(m)=b_1f(b_2m)$, we find that this condition becomes $f(bm)=bf(m)$, or in other words, $f$ is not just $A$-linear but $(A\otimes B)$-linear. | |
| Dec 23, 2010 at 3:30 | history | edited | Tom Goodwillie | CC BY-SA 2.5 |
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| Dec 22, 2010 at 21:43 | comment | added | Ying Zhang | Now if we assume all the rings and tensor products involved are Noetherian, could you pls provide some more details for your argument? E.g. do you mean $Hom_{A\otimes_C B}(M,-)=Hom_{B\otimes_C B}(B, Hom_A(M,-))$ for any $A\otimes_C B$ module N? (This looks to me like an adjointness argument, but I don't know how to show it...) Besides, I guess there is a typo, first you used $Hom_A(M,-)$, then in the rightmost of the line below it is $B-Mod$, (well this line is not an exact sequence, is it?) If these however are standard facts provided somewhere, I would appreciate any reference. | |
| Dec 22, 2010 at 21:32 | comment | added | Ying Zhang | Thanks, but I have a confusion. Here by tensor you always mean over a fixed base ring C, right? I think we need some assumptions: I only know "regularity $\Leftrightarrow$ finiteness of $Ext$-dimension" when the ring is Noetherian. (Is it true for non Noetherian?) In general if the rings A, B and base ring C is Noetherian, $A\otimes_C B$ may not be Noetherian, see mathoverflow.net/questions/29764/pushouts-of-noetherian-rings | |
| Dec 22, 2010 at 3:12 | history | answered | Tom Goodwillie | CC BY-SA 2.5 |