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?
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.