Let $$ M \supset M_1 \supset \ldots \supset M_n \supset \ldots \text{ and } N \supset N_1 \supset \ldots \supset N_n \supset \ldots$$ be exhaustive decreasing filtrations of modules over a commutative integral ring $R$. I would like to know if there is any way to define a filtration over $M \otimes_R N$ such that the graduation $Gr_k(M \otimes_r N)$ verifies $$Gr_k(M \otimes_r N)=\oplus_{i+j=k} M_i/M_{i+1}\otimes_RN_j/N_{j+1}.$$ I know it is possible to do so in the case of finite dimensional vector spaces by defining $$(M \otimes_R N)_k=\sum_{i+j=k}M_i \otimes N_j.$$ So if anyone has an idea or a reference where something like could be written, I would appreciate. Thanks for reading me.
$\begingroup$
$\endgroup$
10
-
2$\begingroup$ If this were possible, it would in particular imply that $[M \otimes_R N] = \sum_{i,j} [\operatorname{Gr}_i M \otimes \operatorname{Gr}_j N]$ in the Grothendieck group of $R$. But this is not always true; for example let $R$ be a Dedekind domain admitting a prime ideal $\mathfrak p \subseteq R$ of order $>1$ in the class group, and let $M_0 \supsetneq M_1 \supsetneq M_2$ be $R \supseteq I \supseteq 0$ and $N_0 \supsetneq N_1$ be $R/\mathfrak p \supseteq 0$. $\endgroup$– R. van Dobben de BruynCommented Feb 23, 2021 at 23:22
-
2$\begingroup$ Then $M \otimes N \cong R/\mathfrak p$ and $M_0/M_1 \otimes N \cong R/\mathfrak p \cong M_1/M_2 \otimes N$. Since $K_0(R) \cong \operatorname{Cl}(R) \oplus \mathbf Z$ (taking $R/\mathfrak p$ to $(\mathfrak p,0)$ and $R$ to $(0,1)$), we see that the two do not coincide. (There might be a positive statement in general if you use the derived tensor product $M \otimes_R^{\mathbf L} N$.) $\endgroup$– R. van Dobben de BruynCommented Feb 23, 2021 at 23:23
-
$\begingroup$ Thanks for the answer. I didn't thought about such examples in the first place. Is there any chance it would be true if we add the condition that M and N are flat ? $\endgroup$– Pierre21Commented Feb 24, 2021 at 10:20
-
$\begingroup$ You can make a similar example where $N_0 \supsetneq N_1 \supsetneq N_2$ is also $R \supset I \supset 0$. I think a better condition would be that $M_i/M_{i+1}$ and $N_i/N_{i+1}$ are flat, and then the construction you suggest should probably work. $\endgroup$– R. van Dobben de BruynCommented Feb 24, 2021 at 19:10
-
1$\begingroup$ I know this is an old question, but this may help: arxiv.org/pdf/1306.1359.pdf $\endgroup$– FunctionOfXCommented Dec 13, 2022 at 0:41
|
Show 5 more comments