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Please give an example of a flat family {X_t} of closed subschemes of PP^n such that the family of projective cones of X_t is not a flat family in PP^{n+1}.

I still could not find...

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Why do you believe there is such a thing? – Steven Landsburg Aug 22 '11 at 14:34

As an example, let $X_t$ be three points in $\mathbb{P}^2$, at positions $(1:0:0)$, $(0:1:0)$ and $(1:1:t)$. This is certainly a flat family. Notice that the points are colinear when $t=0$ and not for $t \neq 0$.

For $t \neq 0$, the homogenous ideal of $X_t$ is $$\langle xz-txy, yz-txy, z^2-t^2 xy \rangle$$

Let $Y$ be the closure of $\langle xz-txy, yz-txy, z^2-t^2 xy \rangle$ in $\mathbb{A}^3 \times \mathbb{A}^1$. If I haven't made any errors, the ideal of $Y$ is $$\langle xz-txy, yz-txy, z^2-t^2 xy, xy(x-y) \rangle.$$ Since I defined $Y$ as the closure of a flat family over a smooth one dimensional base, it is flat. For $t \neq 0$, the fiber $Y_t$ is the cone on $X_t$.

However, at $t=0$, the fiber $Y_0$ is not the cone on $X_0$. They are the same away from the origin, but $Y_0$ has an extra nonreduced bit. Specifically, since $X_0$ lies on the line $z=0$, the cone on $X_0$ has $z$ in its defining ideal, while $Y_0$ has nothing in degree $1$. You can picture $Y_0$ as three coplanar lines, plus a nonreduced tangent vector pointing out the the $z=0$ plane; the cone on $X_0$ doesn't have that extra nonreduced piece.

What may be confusing you is that there isn't some non-flat family here for me to point to. The family $Y_t$, which exists, is flat, but its fibers are not the cones on the $X_t$. There is no closed subfamily of $\mathbb{A}^3 \times \mathbb{A}^1$ whose fibers are the cones on the $X_t$. This is the general picture whenever you have a flat family in $\mathbb{P}^N$ which does not lift to a flat family of projective cones.

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I just noticed you asked for a failure in $\mathbb{P}^{n+1}$, not $\mathbb{A}^{n+1}$. You can take the projective closure of this example to get that; the interest is all at the origin. – David Speyer Aug 22 '11 at 15:16
Here is a more precise (related) question: Let $R$ be a finitely generated $\mathbb{N}$-graded ring. If $\operatorname{Proj}(R)$ is flat over $\operatorname{Spec}(R_0)$, does this imply that $\operatorname{Spec}(R)$ is flat over $\operatorname{Spec}(R_0)$? – Nicholas Proudfoot Jun 20 '15 at 21:09
@NicholasProudfoot Well, there are stupid counter-examples. Set $R = k[t, \epsilon]/(t \epsilon, \epsilon^2)$, with $t$ in degree $0$ and $\epsilon$ in degree $1$. Then $\mathrm{Spec}(R)$ is not flat over $\mathrm{Spec}(R_0)$, but $\mathrm{Proj}(R)$ is empty. There are probably ways to add hypotheses to rule out that sort of silliness, though. – David Speyer Jun 20 '15 at 23:33

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