Timeline for Constructing a free resolution of a $\mathbb Z[x_1,\dotsc,x_n]$-module using a related free resolution of a $\mathbb Q[x_1,\dotsc,x_n]$-module
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2022 at 13:28 | vote | accept | Pranay Gorantla | ||
| Apr 10, 2022 at 12:43 | comment | added | Pranay Gorantla | No problem. Thanks for all your help! | |
| Apr 10, 2022 at 10:23 | comment | added | Aurora | As now you let resolution UP TO changing generators, so one can try to find an ideal in $R'$ whose any minimal generating set with integer coefficients, generates an ideal in $R$ of projective dimension more than that of the original ideal. I expect that such an example exists (when lifting problems from equicharacteristic 0 to prime characteristic, the classical technique is to invert some integers to be able to have that some properties such as pdim is preserved). But finding such as example requires more work, I can't spend more time; My goal was to trying to be of some help anyways. | |
| Apr 10, 2022 at 2:45 | comment | added | Pranay Gorantla | As I mentioned in the post, I always start with $M'$, a submodule of a free $R'$-module, then find $A_0$ associated with the minimal generating set of $M'$ with least cardinality, and only then define $M = \operatorname{im} A_0$ over $R$. | |
| Apr 10, 2022 at 2:43 | comment | added | Pranay Gorantla | Let $N' = \mathbb Q$ and consider its minimal free resolution over $R' = \mathbb Q[x_1,x_2,x_3]$ given by the Koszul complex. Let $\epsilon: R' \rightarrow N'$ be the rightmost map, and $A_0 = \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix}$ be the one to the left of it. Then, $M'=\operatorname{im} A_0$ over $R'$ is a submodule of $R'$. I define $M = \operatorname{im} A_0$ over $R = \mathbb Z[x_1,x_2,x_3]$. In this case, both sequences are exact with the same $A_i$'s of the Koszul complex. | |
| Apr 10, 2022 at 2:33 | comment | added | Aurora | I mean you can take the ideal of $M$ instead of $M$ I defined and the same reasoning works, just the projective dimensions would be $3$ and $2$ instead of $4$ and $3$. | |
| Apr 10, 2022 at 2:20 | comment | added | Aurora | No problem, take the image of the first matrix! | |
| Apr 10, 2022 at 2:18 | comment | added | Pranay Gorantla | In my question, I mentioned that $M’$ is a submodule of a free $R’$-module. But in your answer $M’ = \mathbb Q$, which is not a submodule of a free $R’$-module. | |
| Apr 10, 2022 at 2:15 | history | edited | Aurora | CC BY-SA 4.0 |
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| Apr 10, 2022 at 1:56 | comment | added | Aurora | I modified my answer upon your new question. | |
| Apr 10, 2022 at 1:54 | history | edited | Aurora | CC BY-SA 4.0 |
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| Apr 10, 2022 at 1:48 | history | edited | Aurora | CC BY-SA 4.0 |
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| Apr 10, 2022 at 0:59 | comment | added | Pranay Gorantla | I upvoted your answer but haven't accepted it yet. Technically, you answered my original question and it is still relevant to the new question. If you think it's better for the new question (which is a weaker version of the old question) to be a new post, I can accept your answer and create a new post for the new question. | |
| Apr 10, 2022 at 0:35 | comment | added | Pranay Gorantla | In both the counterexamples above, we could find a different choice of $A_1$'s that worked because $\mu_R(\ker A_0) = \mu_{R'}(\ker A_0)$, i.e., the minimal cardinalities of generating sets of $\ker A_0$ over $R$ and $R'$ are the same. I modified my question asking if they are always equal. | |
| Apr 10, 2022 at 0:18 | comment | added | Pranay Gorantla | While both of them are valid counterexamples to my question 1, we know that there is a choice of $A_1$ for which we do get sequences which are exact over both $R'$ and $R$. For instance, in your counterexample with $A_0 = \begin{pmatrix} x_1 - 2x_2 & x_1 - 2x_3 & x_1 \end{pmatrix}$, we can choose $A_1 = \begin{pmatrix} -x_1 + 2x_3 & x_3 & -x_1 + 2x_3 \\ x_1 - 2x_2 & -x_2 & -2x_2 \\ 0 & x_2 - x_3 & x_1 - 2x_3 \end{pmatrix}$ and $A_2 = \begin{pmatrix} x_1 \\ -x_1+2x_3 \\ x_1-2x_2 \end{pmatrix}$. (I found them on Macaulay2.) This gives an exact sequence over both $R'$ and $R$. | |
| Apr 10, 2022 at 0:11 | comment | added | Pranay Gorantla | Thanks for the counterexample! After seeing your answer, I came up with another (perhaps simpler) counterexample. Say $n=2$ and $A_0 = \begin{pmatrix} x_1 & x_2 \end{pmatrix}$, but $A_1 = \begin{pmatrix} 2x_2 \\ -2x_1 \end{pmatrix}$. (This is a minor modification of the example in the OP.) Then, over $R'$, this still gives an exact sequence but over $R$ it doesn't because $\begin{pmatrix} x_2 \\ -x_1 \end{pmatrix} \in \ker A_0$ over $R$ but it is not contained in $\operatorname{im}A_1$ over $R$. | |
| Apr 9, 2022 at 18:12 | history | edited | LSpice | CC BY-SA 4.0 |
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| Apr 9, 2022 at 17:49 | history | edited | Aurora | CC BY-SA 4.0 |
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| Apr 9, 2022 at 17:29 | history | answered | Aurora | CC BY-SA 4.0 |