Suppose that $(R, m) \subseteq (S,n)$ is a finite extension of normal local domains that is:

  • etale on the punctured spectrum
  • not flat / etale at the origin
  • and such that the residue fields $R/m = S/n$ are equal.

Question: Does anyone know of such examples where $S$, when viewed as an $R$-module, has more than 1 free $R$-summand?

Note, that if we remove the second condition, then we can have lots of summands (since $S$ will be free if flat, and flat + etale in codim 1 implies etale by one of the standard purity of the branch locus theorems).

Likewise if we remove the residue field condition, then it's very easy to construct such examples (do an extension with 1 summand and then extend the residue field).

  • 1
    $\begingroup$ There are no examples arising as homogeneous coordinate rings, $\Gamma_*(Y)\to \Gamma_*(X)$, (localized at the maximal ideal of the vertex) associated with a finite 'etale morphism $f:X\to Y$ and an ample invertible sheaf $\mathcal{O}(1)$ on $Y$ (use the pullback to $X$ to define the coordinate ring). Every $R$-summand of $S/R$ corresponds to a direct summand of $f_*\mathcal{O}_X/\mathcal{O}_Y$ that is isomorphic to $\mathcal{O}(d)$. Via the perfect trace pairing, there is a summand $\mathcal{O}(-d)$. One of $d$ and $-d$ is nonnegative, so that $H^0(X,\mathcal{O}_X)$ is greater than $1$. $\endgroup$ – Jason Starr Apr 13 '16 at 11:27
  • $\begingroup$ Excellent! (I secretly hope they don't exist in general). $\endgroup$ – Karl Schwede Apr 13 '16 at 15:43

Ok, so I realized how to do this. There is always at most 1 summand and here's why.

Setting: $(R, m) \subseteq (S, n)$ is a finite local extension of normal Noetherian domains.

Consider the trace map $T \in \text{Hom}_R(S, R)$. Since $R \subseteq S$ is etale in codimension 1, it is easy to see that $T$ generates $\text{Hom}_R(S, R)$ as an $S$-module (in general, viewed as a section, it corresponds to the ramification divisor).

Suppose for a contradiction that $S = R^{\oplus 2} \oplus M$ for some $R$-module $M$. Consider the two projection maps $\pi_1, \pi_2 : S \to R$ onto the two factors we identify. Both of these are $S$-multiples of $T$. Say $\pi_1(\bullet) = T(x \bullet)$ and $\pi_2(\bullet) = T(y \bullet)$.

Now, it is easy to see / well known (let me know if people want a proof) that $T(n) \subseteq m$. It follows that $\pi_i(n) \subseteq m$ as well. We have a map $\pi_1 \oplus \pi_2 : S \to R^{\oplus 2}$. By hypothesis, this map is surjective. But if it's surjective, it will remain surjective after modding out by $n$ and $m$. So we have a $k$-linear surjective map $$k = S / n \to (R/m)^{\oplus 2} = k^2$$ which is obviously impossible.


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