I have a question concerning the following exercise in Hartshorne:

The inclusion of $k[s]$ into $k[s,t]/(s-t^2)$ induces a morphism of the corresponding affine schemes $X\to Y$. The exercise itself is about calculating the fibres of this morphism.

However I asked myself, how the geometric generic fibre looks like and I'm stuck at the following point:

Let $\xi_Y$ be the generic point of $Y$. Then the geometric generic fibre is the $X_{\xi_Y} = X \times_Y Spec (\overline{k(s)}) = Spec (k[s,t]/(s-t^2) \otimes_{k[s]} \overline{k(s)})$

So at the end I have just the algebraic question, what this last tensor product looks like.

Further question: As already in this seemingly easy example, the geometric fibre is very hard to grasp, I would like to know, if there is a general strategy, how to deal with geometric fibres, especially for the non-closed points or in case of non-algebraically closed base-fields (or no base-field at all).



Note $\sqrt{s}\in\overline{k(s)}$ a square root of $s$. Then the polynomial $s-t^2$ splits as $(\sqrt{s}-t)(\sqrt{s}+t)$. So $$k[s,t]/(s-t^2)\otimes_{k[s]}\overline{k(s)}=\overline{k(s)}[t]/(\sqrt{s}-t)(\sqrt{s}+t)\simeq\overline{k(s)}\times\overline{k(s)}$$ the last equality following from the Chinese remainder theorem.

This shows that the geometric generic fiber consists of two points with residue field isomorphic to $\overline{k(s)}$.

In general, I think of the geometric generic fiber as a generic geometric fiber : in other words it looks like how most of the other fibers look like.

Indeed, if $k$ is algebraically closed, the fiber over a point $s=a$ consists of two points except at $s=0$. So a general fiber consists of two points, and so does the geometric generic fiber.

  • $\begingroup$ so the tensor product commutes with the algebraic closure of $k(s)$? Is this obvious? This was my problem, but maybe I just didn't look at it close enough $\endgroup$ – kesa Mar 22 '16 at 22:11
  • $\begingroup$ For your last comment (this is also how it is stated in Hartshorne), I believe this is not true in characteristic 2, where you need that $t^2 - a$ is separable (i.e. $a$ has two different square roots, else one ends up with the same situation as for $a = 0$) Please correct me if I'm wrong. $\endgroup$ – kesa Mar 22 '16 at 22:13
  • $\begingroup$ @kesa In general $A[x_1,\dots,x_n]/I\otimes_A B=B[x_1,\dots,x_n]/I$. This is the commutation I did with $A=k[s]$ and $B=\overline{k(s)}$. $\endgroup$ – Roland Mar 23 '16 at 7:57
  • $\begingroup$ @kesa For your second comment, you are right : in characteristic 2 the fiber over $s=a$ consist of a single point with multiplicity 2 exactly like the case $a=0$. But note that this is also the case in the geometric generic fiber : you get the single point $t=\sqrt{s}$ with multiplicity 2, so it still holds that the generic geometric fiber looks like a general geometric fiber. $\endgroup$ – Roland Mar 23 '16 at 8:01
  • $\begingroup$ Thanks for your answers, I really should have thought about this algebraic fact, now it seem so obvious :D $\endgroup$ – kesa Mar 27 '16 at 9:40

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