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Oct 8, 2018 at 8:38 vote accept Raffaele C
Oct 6, 2018 at 10:51 answer added Jason Starr timeline score: 5
Oct 5, 2018 at 12:22 comment added Jason Starr No, that is not true. Let $X$ be $\text{Spec}\ k[x,y]/\langle xy \rangle.$ Let $S$ be $\text{Spec}\ k[\epsilon]/\langle \epsilon^2 \rangle.$ Let $I$ be $\langle x,y-\epsilon\rangle.$ Then $\mathcal{O}_{X\times S}/\mathcal{I}^2$ is a direct sum of two copies of $\mathcal{O}_S$ (generated by $1$ and $y$) plus one summand $\mathcal{O}_S/\epsilon \mathcal{O}_S$ (generated by $x$). It is not $\mathcal{O}_S$-flat.
Oct 5, 2018 at 12:05 history edited Raffaele C CC BY-SA 4.0
more precise setting
Oct 5, 2018 at 10:43 history asked Raffaele C CC BY-SA 4.0