My questions refer to Exercise 5.3.9 from Liu's "Algebraic Geometry" (page 208):
I have two questions:
How to show in (a) that
$K(Y) \to K(X)$ pure inseparable implies $f$ purely inseparable
My considerations:
As @Laurent Moret-Bailly stated in his comment below it is neccessary to assume that $X$ dominates $Y$ (otherwise the extension $K(X)/K(Y)$ don't make any sense since in such case this map in genereal would not exist).
By definition $f: X \to Y$ is called purely inseparable if $f$ injective and if for every $x ∈ X$, the extension of residue fields $k(f(x)) → k(x)$
is a purely inseparable extension.
We obtain following diagram
$$
\require{AMScd}
\begin{CD}
O_{Y,f(x)} @>{}>> O_{X,x} \\
@VVV @VVV \\
K(Y)=O_{Y, \eta_Y} @>{}>> K(X)=O_{X, \eta_X}
\end{CD}
$$
By finiteness & purely separateness we have $K(X) = K(Y)[a^{1/n}]$ for appropriate $a \in K(Y)$. The question is how this strucure can be "transfered" to $O_{Y,f(x)}/m_{f(x)}=k(f(x)) \to k(x)= O_{X,x}/m_x$?
Honestly, I have no idea.
Best Answer
A caution: it's not necessarily true that $K(X)=K(Y)[a^{1/n}]$ - for instance, $\Bbb F_2(x,y)\subset \Bbb F_2(x,y)[x^{1/2},y^{1/2}]$ is a purely inseparable finite extension which is not generated by a single element.
Now on to the real problem. The first thing to do is to note that we may reduce to the case of both $X,Y$ affine - making the reduction to $Y$ affine is clear, and then the fact that finite morphisms are affine implies the reduction for $X$. So we may write $X=\operatorname{Spec} A$ and $Y=\operatorname{Spec} B$ for $A,B$ both noetherian integral domains and $B$ integrally closed (by normality). $f$ finite implies that $A$ is a finite $B$-module, so now we can pick a finite set of generators $\{z_i\}_{i\in I}$ for $A$ as a $B$-algebra.
Lemma: Let $R$ be an integrally closed domain and $F=Frac(R)$. Let $F\subset K$ be a finite field extension. Then for any $\alpha\in K$ integral over $R$, the minimal polynomial $m(x)$ of $\alpha$ is actually in $R[x]$.
Proof: As $\alpha$ is integral, there's a monic polynomial $f\in R[x]$ with $f(\alpha)=0$. But by the inclusion $R[x]\subset F[x]$, $f$ is also a polynomial in $F$ which vanishes on $\alpha$. So it's divisible by $m(x)$, and we may write $f=gm$. As all roots of $m$ are roots of $f$, all roots of $m$ are integral over $R$. As the coefficients of $m$ are the elementary symmetric polynomials in these roots, the coefficients of $m$ are again integral over $R$. As $R$ is integrally closed, these coefficients are actually in $R$, and thus we've shown that $m(x)\in R[x]$. $\blacksquare$
By definition of a purely inseparable extension, every element of $Frac(A)$ has a minimal polynomial over $Frac(B)$ of the form $x^{p^n}-b$. Applying the lemma above to the $z_i$ which are integral over $B$, we see that each $z_i$ satisfies the relation $x^{p^{n_i}}-b_i$ for $b_i\in B$ and $n_i\in \Bbb Z_{> 0}$.
To show that $f$ is purely inseparable, it suffices to compute $A/\mathfrak{m}$ for any maximal ideal $\mathfrak{m}\subset B$ and to show it's a local ring with residue field a purely inseparable extension of $B/\mathfrak{m}$ (this suffices because it shows that the fibers are all single points, which implies $f$ injective). But this is clear - for every $b_i\notin \mathfrak{m}$, we adjoin a $p^{n_i}$th root of something in the field $B/\mathfrak{m}$, and for every $b_i\in\mathfrak{m}$, we get some nilpotents in $A/\mathfrak{m}$. So $A/\mathfrak{m}$ is a local ring with maximal ideal consisting of the nilpotents we added, and we're done.