We were discussing base change in algebraic geometry in the class today but unfortunately I did not quite get it because there were few examples. So I am trying to think of some on my own.

Let us choose an algebraically closed field $K$. Take the projective nodal cubic over $K$. Its normalization is the projective line. If we get rid of a point in the preimage of the node, we get a map from the affine line to the projective nodal cubic that is bijective on points. In fact, it is a homeomorphism if we consider varieties as schemes.

Any base change of this map is going to be both injective and surjective so the map is universally bijective. It can not be universally closed because then the affine line would be a complete variety. Thus there is a base change of the map that is not a homeomorphism. What is an example of such base change?

  • $\begingroup$ What about the base-change of your morphism $\mathbb{A}^1\to X$ along itself with $X$ your projective nodal cubic? $\endgroup$ – Ariyan Javanpeykar Jul 23 at 14:58
  • $\begingroup$ @AriyanJavanpeykar I am not sure. Which closed subset will have non-closed image? $\endgroup$ – user690882 Jul 23 at 17:47

Let's say $X$ is the projective nodal cubic, $g:\mathbb{P}^1\to X$ is the normalization, and $f:\mathbb{A}^1\to X$ is your bijective restriction. Then I claim the base change of $f$ along $g$ is not closed. Indeed, note that the fiber product $\mathbb{A}^1\times_X\mathbb{P}^1$ is a closed subscheme of $\mathbb{A}^1\times\mathbb{P}^1$ consisting of the diagonal $\Delta$ in $\mathbb{A}^1\times\mathbb{A}^1$ together with the point $(x,\infty)$, where $x\in\mathbb{A}^1$ is the point that maps to the node. In particular, $\Delta$ is closed in $\mathbb{A}^1\times_X\mathbb{P}^1$. But the projection of $\Delta$ to $\mathbb{P}^1$ is just $\mathbb{A}^1$, which is not closed in $\mathbb{P}^1$.

  • $\begingroup$ is it true that $\mathbb{A}^1\times_X \mathbb{P}^1$ is the disjoint union of the diagonal (isomorphic to the affine line) and the point (which, if we are over an algebraically closed field, is $\mathrm{Spec}\:K$)? In particular, it is not connected? $\endgroup$ – user690882 Jul 25 at 6:45
  • $\begingroup$ That's right. So it's $\mathbb{P}^1$ except that the point at $\infty$ has been "split off", and the base change of $f$ is the map to $\mathbb{P}^1$ that "glues it back on". $\endgroup$ – Eric Wofsey Jul 25 at 6:57
  • $\begingroup$ I think $f$ is a connected morphism in the sense of EGA IV, partie 2, Def. 4.5.5 (I am not entirely sure about the fiber over the generic point). Then Prop. 4.5.7 seems to predict that $\mathbb{A}^1\times_X\mathbb{P}^1$ is connected (because $\mathbb{P}^1$ is connected and $g$ is a finite morphism so universally closed). Who is wrong: my interpretation of the situation, your answer, or EGA IV? $\endgroup$ – user690882 Jul 27 at 12:16
  • $\begingroup$ 4.5.7 requires $f$ to be universally closed (or one of the other conditions), not $g$. $\endgroup$ – Eric Wofsey Jul 27 at 13:24

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