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(everything below is defined over an algebraically closed field)

Let $D$ be a (smooth) surface, and let $X \subset T \times D$ be a flat family of curves on $D$, where $T$ is irreducible. Let $E$ be a curve on $D$. Fix some $t_0 \in T$ such that $X_{t_0}$ does not contain $E$, the intersection $X_{t_0} \times_D E$ is some not necessarily reduced closed subscheme of $D$ of dimension 0.

Is it true then that there exists an open dense subset $U \subset T$, such that $X_t \times_D E$ are all isomorphic for different $t \in U$? Does one actually need to assume $D$ to be projective? smooth? Is the previous statement true for an affine $D$?

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I don't understand the role played by $E$ here. Given a family of curves, can't you always choose an $E$ and a $t_0$ so that your hypothesis is satisfied ? Also, consider the moduli space ($=T$) of elliptic curves with a linear rigidification; above $T$ you have a family of elliptic curves ($=X$) and an embedding $X\to T\times{\bf P}^2$, where $D={\bf P}^2$, given by the Weierstras equations. Clearly, there is no open subset of $T$ above which all the fibres are geometrically isomorphic. This seems to show that the answer to your question is no. – Damian Rössler Jun 24 '12 at 15:18
Sorry I just realized that I completely misunderstood you question. Forget my comment. – Damian Rössler Jun 24 '12 at 15:40
up vote 3 down vote accepted

Probable counterexample: take $D=\mathbb{A}^2$ (or $\mathbb{P}^2$, if you prefer), $T=\mathbb{A}^1$, and let $X_t$ be the union of the four lines $x=0$, $y=0$, $x=y$, and $x=ty$. It is well known that $t$ is "almost" determined by the isomorphism class of $D_t$, as the cross-ratio of the four tangents at the origin. ("Almost" refers to permutations of the four lines, but only finitely many $t$'s give rise to isomorphic $D_t$'s). My claim is that this remains true if you replace $D_t$ by a sufficiently big infinitesimal neighbourhood of the origin (presumably the one given by the ideal $(x,y)^5$ works).

Assuming this, you have a counterexample as soon as $E$ contains the neighbourhood in question, i.e. is sufficiently singular at the origin.

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Thank you, this answers my question. – Dima Sustretov Jun 25 '12 at 17:16

Let me just flesh out Laurent's answer. $X_t$ is the vanishing set of $f_t=xy(x-y)(x-ty)$, an equation of degree $4$. Let $g$ be any homogeneous polynomial of degree $5$, and let $E$ be the vanishing set of $G$. Then the intersection of $E$ and $X_t$ has a unique singular point, the origin, whose maximal ideal is $(x,y)$. Thus, the local ring at the origin, modulo the fifth power of the maximal ideal, is determined by the isomorphism class of the intersection.

This local ring is $k[x,y]/(f_t,x^5,x^4y,x^3y^2,x^2y^3,xy^4,y^5)$. There is a unique-up-to-scaling degree $4$ homogeneous polynomial on the Zariski tangent space that vanishes in this ring, and that polynomial is $f_i$. Degree-$4$ homogenous determine $4$-point subsets of $\mathbb P^1$ determine elliptic curves. The $j$-invariant of that curve is a nonconstant function of $t$ and is determined solely by the isomorphism type of the intersection.

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