2

Let $I\subseteq \mathbb C[x_1,\ldots,x_n,y_1,\ldots,y_m]$ be a prime ideal. Consider the variety $V:=V(I)$ in $\mathbb C^n\times \mathbb C^m$ it defines (it is irreducible.) Let $\pi$ be the projection of $V$ onto the second factor $\mathbb C^m$. For generic $y'\in \pi(V)$ is it true that the specialization $I(y')$ is radical?, where \begin{align} I(y'):=\{f(x,y'):f\in I\}\subseteq \mathbb C[x_1,\ldots,x_n]. \end{align}

Observe that this is not true for all $y\in \pi(V)$. Take for instance the radical ideal generated by the single polynomial $x^2+y$. Then the specialization $y=0$ yields the ideal generated by $x^2$ which is not radical.

Any help or reference is appreciated.

Cellardoor
  • 312
  • 1
    Maybe this reference will help: https://math.stackexchange.com/questions/168979/when-a-scheme-theoretical-fiber-is-reduced. – danneks Nov 22 '23 at 12:48
  • Thanks, but I'm not so familiar with scheme language. Do you know if that reference is saying what I'm asking? – Cellardoor Nov 22 '23 at 15:51
  • 1
    Yes, the spectrum of $\mathbb C[x_1, \ldots, x_n]/I(y')$ is isomorphic to the scheme-theoretic fiber of $V$ over $y'$, and this fiber is reduced iff $I(y')$ is radical. – danneks Nov 22 '23 at 16:54
  • Thank you. Do you happen to know of a source? The source given in the link you sent me was broken unfortunately. – Cellardoor Nov 22 '23 at 18:51
  • 1
    In EGA IV.9.7.7 it is proved that the property of a fiber being geometrically reduced is locally constructible on the base. It follows that if the fiber over the generic point has this property, then it holds on an open subset. – danneks Nov 23 '23 at 05:31
  • What I am still confused about is the description of the generic fiber as $\mathrm{Spec} K(A)\otimes_A B$, where in our case $A=\mathbb C[y]/I(\pi(V))$ and $B=\mathbb C[x,y]/I(V)$. This fiber does not depend on $y'\in \pi(V)$. In contrast, Hartshorne's definition of fiber of a point is $\mathrm{Spec} \mathbb C[y]/I(y=y') \otimes_A B$ (Chapter 2, Section 3.) I'm guessing that you are saying that the latter spectrum is isomorphic to $\mathbb C[x]/I(y')$, which is the ideal we want to be radical. How do I see that the two descriptions are the same? – Cellardoor Nov 29 '23 at 22:41
  • 1
    By generic fiber I mean the fiber over generic point of the base scheme (the closure of this point is the whole scheme). And the fiber over a point $y'$ belonging to an open subset of the base is usually called the general fiber. – danneks Nov 30 '23 at 05:50
  • Let me clarify: In the link you sent me, they refer to the fiber of the generic point, but in my context I want the fiber of a general point. And your claim is that by the EGA result, the fact that it holds for the generic point, means that it holds for a general point? Further, the scheme-theoretic fiber of $V$ over the general point $y'$ (as defined by Harthshorne) is the spectrum of $\mathbb C[x]/I(y')$? Thank you in advance. – Cellardoor Nov 30 '23 at 14:02
  • 1
    All is correct. If a locally constructible subset of an affine space (or any irreducible Noetherian scheme) contains the generic point, then it contains all points from some open subset. – danneks Nov 30 '23 at 14:36
  • In fact it holds for any irreducible scheme. – danneks Nov 30 '23 at 14:54
  • Finally, I should understand why the spectrum of $\mathbb C[x]/I(y')$ is isomorphic to the scheme-theoretic fiber of $V$ over $y'$. I don't immediately see this using Hartshorne's definition. Do you know of a source for this, or can you explain it? Thanks. – Cellardoor Nov 30 '23 at 21:03
  • 1
    See Vakil's notes Foundations of Algebraic Geometry, Exercise 10.2.B. And note that $\mathbb C[x, y]/(I, y-y') \cong \mathbb C[x]/I(y')$. The notes are available at http://math.stanford.edu/~vakil/216blog/. – danneks Dec 01 '23 at 10:55
  • Perfect, thanks! – Cellardoor Dec 04 '23 at 13:57

0 Answers0