I'm looking for an integral morphism $f\colon X \rightarrow Y$ between schemes $X$ and $Y$ which has a fiber consisting of infinitely many points. If $f$ were locally of finite type, then $f$ would be finite and thus would have finite fibers. An example must thus be integral but not locally of finite type.

  • $\begingroup$ There exists domains whose integral closure in its fraction field is not finitely generated as a module. These are pathological examples and if my memory serves me, some examples can be found in Nagata's local rings. $\endgroup$
    – Mohan
    Sep 13, 2016 at 19:34

1 Answer 1


For a finite field $k$, take $ {k[x]}/{(x^2)}$. Then this is integral over $k$, and has only finitely many elements, so you find a polynomial killing all of them. Then this polynomial also kills every element of $\prod _{i \in \mathbb{N}} k[x]/(x^2)$ thus this is integral over $k$, but the fibre of the inclusion is infinite as the spectrum of $\prod _{i \in \mathbb{N}} k[x]/(x^2)$ is infinite.

Another not so intuitive example is the map $\bar{\mathbb{Q} }\to \bar{\mathbb{Q} }\otimes_{\mathbb{Q}} \bar{\mathbb{Q} }$ induced by base changing the inclusion of $\mathbb{Q}$ into $\bar{\mathbb{Q}}$. This is integral as the inclusion is, and base changing does not change that, and does not have finite fibres as $\bar{\mathbb{Q} }\otimes _{\mathbb{Q}} \bar{\mathbb{Q} }$ is infinite.

  • $\begingroup$ Cool, thanks! Sadly, I do not have enough reputation to up-vote this, but it was exactly what I was looking for. $\endgroup$
    – Lisa
    Sep 16, 2016 at 22:18
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    $\begingroup$ Is there a reason here you used an infinite product of $k[x]/(x^2)$ and not an infinite product of $k$? It seems that the map from an infinite discrete space to $\operatorname{Spec} k$ would also work. $\endgroup$
    – C.D.
    Aug 2, 2020 at 0:00

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