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I would like to ask some questions about algebraic geometry:

  • Can someone provide an example of a morphism of schemes with finite fibers such that under a base change it doesn't have finite fibers? I guess it is related with the infinite extension of fields but I cannot find a precise example.

  • When is $\operatorname{Spec}(\mathbb{F}_p[x,y]/(xy^2-m))$ an irreducible scheme, for $m\in \mathbb{Z}$?

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1 Answer 1

up vote 3 down vote accepted

If $K \subset L$ is an extension of fields, then Spec $L \to$ Spec $K$ certainly has finite fibres: there is only one fibre, and it is a single point. Now consider the base-change over $L$, i.e. Spec $L\otimes_K L \to $ Spec $L$. This again has only one fibre: is it finite?

If $L$ is finite over $K$, the answer is yes, since in that case $L\otimes_K L$ is an Artinian $L$-algebra.

But in general the answer is no; e.g. if $L = K(x)$, then $L\otimes_K L$ has infinite Spec.

This is why in EGA quasi-finite is defined as having finite fibres and being of finite type; with this definition it is stable under base-change.


If $p \not\mid m$ then $\mathbb F_p[x,y]/(xy^2 - m) = \mathbb F_p[y,y^{-1}]$ (with $x$ being identified with $my^{-2}$), and hence is an integral domain. Thus its Spec is irreducible; indeed, it is $\mathbb A^1 \setminus \{0\}$.

If $p \mid m,$ then $\mathbb F_p[x,y](xy^2 - m) = \mathbb F_p[x,y]/(xy^2)$, and so its Spec is the union of two lines, one of them doubled, and so is neither reduced nor irreducible.

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Dear Matt, great answer! The scheme $Spec(\mathbb C(x)\otimes_\mathbb C \mathbb C(y))$ is the one-dimensional scheme obtained by removing from the plane all closed points and all generic points of vertical or horizontal lines, right? –  Georges Elencwajg May 23 '12 at 12:20
    
Dear Georges, Thanks for the kind words, and yes, that's a good way to describe Spec $(\mathbb C(x)\otimes_{\mathbb C} \mathbb C(y))$. Best wishes, –  Matt E May 23 '12 at 14:54

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