Timeline for Dimension of a general partial derivative of a linear subspace of polynomials
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Nov 8, 2022 at 16:36 | comment | added | Zach Teitler | I'm not sure if it's salvageable. Sure, if the basis elements of $U$ have leading terms involving $x_1$, then what I wrote earlier is fine, but that's not guaranteed. For example if $n=4$ and $U$ is spanned by $\{x_2 x_3, x_1 x_3, x_2^2, x_1 x_2, x_1^2\}$ then it is Borel-fixed but missing the monomial $x_1 x_4$. So... I don't know. Worse: I'm now questioning whether we have enough generality to get the Borel-fixed property while still fixing $\partial = \partial/\partial x_1$. My hunch is that it's okay but I'm not sure. | |
Nov 7, 2022 at 21:36 | comment | added | Zach Teitler | Ah, there is something to think about here. A theorem of Galligo-Bayer-Stillman (Theorem 15.20 in Eisenbud's book) shows that the generic initial ideal of the ideal generated by $U$ is Borel-fixed. However a Borel-fixed collection of monomials in degree $d$ is not quite the same thing as a lex initial segment. | |
Nov 7, 2022 at 12:33 | vote | accept | Ben | ||
Nov 7, 2022 at 13:11 | |||||
Nov 7, 2022 at 5:55 | comment | added | Ben | I am confused about your first line... do you claim that $GL(n) \cdot U$ is Zariski open dense in the Grassmannian variety of $\dim(U)$-dimensional subspaces of the polynomial ring? This is false by dimension considerations. So I guess you mean that $GL(n) \cdot U$ is ``general enough" so that one of its elements admits such a basis? If so, why is this true? | |
S Oct 16, 2022 at 6:05 | history | suggested | Ben | CC BY-SA 4.0 |
In the last sentence, the inequality is only satisfied, not necessarily with equality.
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Oct 16, 2022 at 3:33 | vote | accept | Ben | ||
Nov 7, 2022 at 12:12 | |||||
Oct 16, 2022 at 3:07 | review | Suggested edits | |||
S Oct 16, 2022 at 6:05 | |||||
Oct 16, 2022 at 0:01 | history | answered | Zach Teitler | CC BY-SA 4.0 |