Let $X$ be a smooth proper curve over a field $k$, with function field $K$. Let $L$ be a finite separable tamely ramified extension of $K$, and let $Y$ be the normalization of $X$ in $L$. Is $Y\rightarrow \text{Spec }k$ smooth?

This seems to be true if $k$ is perfect, though I'd appreciate if someone could explain what might go wrong if $k$ isn't perfect.


1 Answer 1


No, but we can exactly characterize all counterexamples: $Y$ is non-smooth over $k$ at precisely those non-etale points of $Y \rightarrow X$ at which the residue field is not separable over $k$ (or equivalently the image point in $X$ has residue field not separable over $k$, since $f$ is separable between all residue fields at closed points by the tameness hypothesis).

For example, if ${\rm{char}}(k)=p>0$ and $c \in k - k^p$ then for any $e>1$ not divisible by $p$ the normalization of $\mathbf{P}^1_k$ in the everywhere-tame extension $k(t)[y]/(y^e - (t^p-c))$ of $k(t)$ is not smooth at the unique point over $t=c^{1/p}$ in $X$.

We first observe that any regular $k$-scheme $Z$ of finite type with pure dimension 1 (such as $Y$) is $k$-smooth near any closed $z \in Z$ such that $k(z)/k$ is separable. Indeed, the completed stalk $R:=\widehat{O}_{Z,z}$ has a (unique) compatible structure of $k(z)$-algebra by Hensel's Lemma, and is a dvr by regularity, so $R=k(z)[\![t]\!]$ as $k$-algebras. Thus, $\widehat{\Omega}^1_{R/k}$ is free of rank 1 over $R$, yet that is the completed stalk of $\Omega^1_{Z/k}$ at $z$, so $\Omega^1_{Z/k}$ is a line bundle near $z$, giving $k$-smoothness there. So $Y$ is $k$-smooth for perfect $k$ with $L/K$ an arbitrary finite extension.

Now consider $k$ that is imperfect but assume $L/K$ is tame relative to all closed points of $X$, and let $p = {\rm{char}}(k)>0$. First note that it is harmless to make a preliminary scalar extension to $k_s$ and pass to compatible connected components of $X_{k_s}$ and $Y_{k_s}$ since (i) the formation of the normalization of a normal noetherian $k$-algebra in a finite etale extension of its total ring of fractions commutes with separable extension on $k$, (ii) this retains the tameness hypothesis without any effect on ramification indices (easy local consideration in ramification theory), (iii) this has no effect on whether or not smoothness holds over the ground field, nor on whether or not a given residue field extension is separable.

Thus, now all residue fields at closed points on $X$ are separably closed (being of finite degree over $k=k_s$), so for any closed $y \in Y$ and its image $x \in X$ the finite extension $k(y)/k(x)$ (which is separable by tameness) is trivial. This will be convenient for local calculations.

It is necessary and sufficient that for any closed $y \in Y$ and $B := \widehat{O}_{Y,y}$, the $B$-module $\widehat{\Omega}^1_{B/k}$ (naturally identified with the completed stalk of $\Omega^1_{Y/k}$) is free of rank 1. If $x \in X$ is the image of $y$ and $A := \widehat{O}_{X,x}$ then $\widehat{\Omega}^1_{A/k}$ is a free $A$-module of rank 1. The subtlety is that we cannot easily describe $A$ as a $k$-algebra when $k(x)/k$ is not separable, and so we cannot easily describe a basis of the free $A$-module $\widehat{\Omega}^1_{A/k}$ of rank 1. For example, if $\pi$ is a uniformizer of $A$ then ${\rm{d}}\pi$ can fail to be a basis since it may even vanish: if $X = \mathbf{P}^1_k$ (with standard coordinate $t$) and $x = (t^p-c) \in {\rm{A}}^1_k$ for $c \in k - k^p$ then $\pi := t^p-c$ is a uniformizer of $A$ yet clearly ${\rm{d}}\pi=0$.

The $k$-smoothness of $X$ at $x$ ensures that $\widehat{\Omega}^1_{A/k}$ is free of rank 1 over $A$. By tameness at $y$ and the triviality of $k(y)/k(x)$, we have $B = A[y]/(y^e-\pi)=A[\![y]\!]/(y^e-\pi)$ as $A$-algebras for some uniformizer $\pi \in A$ and some $e >0$ not divisible by $p={\rm{char}}(k)$. Thus, $$\widehat{\Omega}^1_{B/k} = ((B \otimes_A \widehat{\Omega}^1_{A/k}) \oplus B {\rm{d}}y)/(ey^{e-1}{\rm{d}}y-{\rm{d}}\pi).$$
This is clearly free of rank 1 over $B$ when $e=1$ (as we know it must be, since then $Y \rightarrow X$ is etale at $y$). Suppose instead that $e>1$. Since $p \nmid e$, the obstruction to freeness as a module of rank 1 over the dvr $B$ is now exactly whether or not ${\rm{d}}\pi$ is a basis of $\widehat{\Omega}^1_{A/k}$.

This basis property holds whenever $k(x)/k$ is separable, or equivalently (since $k_s=k$) whenever $k(x)=k$, since in such cases $A = k[\![\pi]\!]$. But that basis property fails whenever $k(x)/k$ is not separable. Indeed, if it is a basis then for the finite flat (!) extension of complete dvr's $R := k[\![\pi]\!] \rightarrow A$ we have $$\Omega^1_{A/R} = \widehat{\Omega}^1_{A/R} = {\rm{coker}}(A \otimes_R \widehat{\Omega}^1_{R/k} \rightarrow \widehat{\Omega}^1_{A/k}) = 0,$$ making $A$ finite etale over $R$, which contradicts that the residue field extension is not separable.


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