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Jan 17, 2019 at 20:01 vote accept user45397
Jan 17, 2019 at 20:01 comment added user45397 @JasonStarr Thank you. I understand now.
Jan 17, 2019 at 19:40 comment added Jason Starr Laurent Moret-Bailly already answered your question. I am just repeating his second counterexample, since you seem not to have understood his point. Let $X$ be $\mathbb{A}^1_k$ with coordinate $s$, let $A$ be $k[t]$, and let $Y$ be the closed subscheme with defining ideal $\langle s^2,st \rangle$. The fiber $Y_t$ is an effective Cartier divisor in $X\times\{t\}$ for every $t$ in $\text{Spec}\ A$. However, $Y$ is not flat over $\text{Spec}\ A$.
Jan 17, 2019 at 19:02 history edited user45397 CC BY-SA 4.0
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Jan 17, 2019 at 18:51 history edited user45397 CC BY-SA 4.0
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Jan 17, 2019 at 17:43 history edited user45397 CC BY-SA 4.0
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Jan 17, 2019 at 17:31 answer added Laurent Moret-Bailly timeline score: 2
Jan 17, 2019 at 17:10 history asked user45397 CC BY-SA 4.0