Example demonstrating that to have finite fibers is not stable under base change?  I would like to ask some questions about algebraic geometry:


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*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 $m\in \mathbb{Z}$?
 A: 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.
