This exercise is the 5.3.9 of Liu's famous book about algebraic geometry.

Let Y be a normal, locally Noetherian, integral scheme, and let $f : X \mapsto Y$ be a projective dominant morphism with $X$ integral. (a) Let us suppose $f$ is finite. Show that $f$ is purely inseparable if and only if the extension $K(Y ) \mapsto K(X)$ is purely inseparable. Moreover, if $f$ is purely inseparable, it will be a homeomorphism.

(b) Let us suppose that the algebraic closure of $K(Y )$ in $K(X)$ is purely inseparable over $K(Y )$. Show that the fibers of $X \mapsto Y$ are geomet- rically connected.

The answer of (a) is Ok. I'm looking for (b). Here is what I've done :

It's enough to check that $\mathcal{O}_{Y} \mapsto f_{*}\mathcal{O}_{X}$ is an isomorphism by the Corollary $3.17.$ page $201$. We can suppose $X = Spec(A), Y = Spec(B)$ affine with $A, B$ integral noetherian and $B$ integrally closed and $f$ is defined by a morphism of ring $\phi : B \mapsto A$. It's enough to see $\phi$ is an isomorphism. We know it's injective by dominance. Since the morphism is projectif (hence proper) the elements of $A$ are integral over $B$ (Proposition $3.18.$ page $105$) so they are in the algebraic closure of $K(Y)$ in $K(X)$. By hypothesis, we can show their minimal polynom are of the form $X^{p^{d}} - b$ with $b \in B$.

Here, I don't know what to do. If b would have a $p-$root, I could conclude but I have any hypothesis on the field $K(X)$.

I precise the link given above only answer to (a) and not to (b) (the question I'm looking for). Is their a possibility to re open the conversation to accept answer?

  • $\begingroup$ If you want a proof of (a) or a details about $X^{p^{d}} - b$ tell me... $\endgroup$
    – Analyse300
    Sep 3, 2023 at 17:02
  • $\begingroup$ Please use Mathjax for your whole post… I’m not sure how you can assume that $X$ and $Y$ are affine in (b), unless you keep the assumption that $f$ is finite. $\endgroup$
    – Aphelli
    Sep 3, 2023 at 17:06
  • $\begingroup$ 1. Please look around MSE to see if your question has been answered before prior to posting. 2. Please use MathJax for all of your post, including the statement of the problem. 3. Your reduction is not correct - it will not always be the case that $\mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism (consider Spec of any purely inseparable field extension, for instance). $\endgroup$
    – KReiser
    Sep 3, 2023 at 17:11
  • $\begingroup$ Hello KReiser, thank you for your answer. I already looked and I don't find satisfying things and I find your link. The link only answer to (a). $\endgroup$
    – Analyse300
    Sep 4, 2023 at 7:10
  • $\begingroup$ I'm going to edit the first line with MathJax, sorry. $\endgroup$
    – Analyse300
    Sep 4, 2023 at 7:13

1 Answer 1


We'll first reduce to the case when $X\to Y$ is finite: by Stein factorization, we may write $f:X\to Y$ as $X\stackrel{g}{\to} Y'\stackrel{h}{\to} Y$ where $g$ is projective with $g_*\mathcal{O}_X \cong \mathcal{O}_{Y'}$ (and thus geometrically connected fibers) and $h:Y'\to Y$ is finite. If we can show the fibers of $h$ are singletons, we'll have demonstrated the claim.

Now consider the fields $K(Y)$, $K(Y')$, $K(X)$, and $K(Y)^{alg}$, the algebraic closure of $K(Y)$ in $K(X)$. We know that $K(Y)\subset K(Y')\subset K(X)$ from the sequence of dominant morphisms $X\to Y'\to Y$, and we know that $K(Y)\subset K(Y')$ is a finite extension of fields, so we have that $K(Y')\subset K(Y)^{alg}$, too. Since every subextension of a purely inseparable extension is purely inseparable, this means $K(Y)\subset K(Y')$ is a purely inseparable field extension, so now we apply (a) and we're done.

  • $\begingroup$ Thank you very much KReiser, it works. I'll use $@-reply next time (I didn't knew this tool). Thanks again. $\endgroup$
    – Analyse300
    Sep 10, 2023 at 8:11

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