Timeline for Singling out irreducible components
Current License: CC BY-SA 4.0
7 events
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Apr 27, 2021 at 6:42 | comment | added | D. Dona | Here's the example needing $\gg D^{2}$: it's a generalization of Elkies's answer in here. Fix $k$, and take $x^{k-1}y-z$ and $y^{k}-z^{k-1}$: they define a variety $V$ in $\mathbb{A}^{3}$ that is the union of $L=\{y=z=0\}$ and of the curve $C$ parametrized as $(t,t^{(k-1)^2},t^{k(k-1)})$. Now let $P$ be a polynomial vanishing on $C$ and not on $L$: if $P$ doesn't vanish on $L$, it must have a monomial $x^r$ inside, and if $P=0$ on $C$ then $r\gg k^2$, because any other monomial with $y,z$ becomes $t^s$ with $s\geq\min\{(k-1)^2,k(k-1)\}\gg k^2$. | |
Apr 26, 2021 at 19:21 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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Apr 26, 2021 at 16:34 | history | edited | H A Helfgott | CC BY-SA 4.0 |
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Apr 26, 2021 at 16:16 | comment | added | H A Helfgott | Daniele Dona has just shown me an example demonstrating that degree $\gg D^2$ can be needed. | |
Apr 26, 2021 at 14:07 | comment | added | H A Helfgott | I'd also be happy, but it would be the same question! | |
Apr 26, 2021 at 14:02 | comment | added | user178109 | What happens if you don't take the intersection with $V$ i.e. ask for a variety $W$ defined by equations of degree $O(nD)$ such that $V_0^+$ is the only top-dimensional component? | |
Apr 26, 2021 at 13:07 | history | asked | H A Helfgott | CC BY-SA 4.0 |