In Raynaud's "Anneaux locaux henseliens," a proof is given of the following fact: Let $R \to S$ be a finite type morphism of rings, $\mathfrak{q} \in \mathrm{Spec} S$, $\mathfrak{p} $ the inverse image of $\mathfrak{q}$ in $R$, such that $R \to S$ is locally unramified at $\mathfrak{q}$. Then $S$ is isomorphic near $\mathfrak{q}$ to a quotient of a standard etale algebra (i.e. of the form $(R[x]/P)_Q$ where the localization is such that $P'$ is a unit).

In Raynaud's book, Zariski's Main Theorem is first used to reduce to the case of $S$ finite over $R$. The part I don't understand is the reduction to the case of $S$ generated by one element over $R$, where $R$ is local.

Here is the argument, as I understand it. Let the maximal ideal of $R$ be $\mathfrak{m}$ and those of the semi-local ring $S$ be $\mathfrak{q}, \mathfrak{q}_1, \dots, \mathfrak{q}_k$. Choose $x \in S$ generating the residue extension $S/\mathfrak{q}/R/\mathfrak{m}$ (possible by the primitive element theorem, separability following from structure theory of unramified algebras over fields) and in the other maximal ideals $\mathfrak{q}_i$. Consider the algebra $C = R[x] \subset S$ and the ideal $\mathfrak{r} = \mathfrak{q} \cap C$. Raynaud claims on p.53 that $\mathfrak{q}$ is the only maximal ideal of $S$ lying over $\mathfrak{r}$.

I don't understand the claim. Suppose we have an extension of number fields $L/K$ which splits completely at some prime. Let $B/A$ be the ring of integers, and let $\mathfrak{p}$ be the prime of $K$ which splits completely. Then the extension $A_{\mathfrak{p}} \to B_{\mathfrak{p}}$ is unramified and $B_{\mathfrak{p}}$ is a finite $A_{\mathfrak{p}}$-module. However, since the residue extensions are all trivial, we can take $x$ in the above paragraph to be zero. It is, however, not the case that only one prime of $B_{\mathfrak{p}}$ lies over $\mathfrak{p}$, unless the extension is trivial. What am I missing?


closed as no longer relevant by Akhil Mathew, Andrés E. Caicedo, Yemon Choi, Daniel Moskovich, Pete L. Clark Jan 2 '11 at 23:43

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ It suffices to impose $x$ to be non-zero in $S/\mathfrak{q}$. $\endgroup$ – Qing Liu Jan 2 '11 at 22:17
  • $\begingroup$ OK, I think I see this argument. If another prime lay above $\mathfrak{r}$, then $x$ would lie in this prime, and so in $\mathfrak{r}$. But it cannot, as it does not lie in $\mathfrak{q}$. $\endgroup$ – Akhil Mathew Jan 2 '11 at 23:08
  • 1
    $\begingroup$ @Qing Liu: Thanks for pointing that out, by the way! $\endgroup$ – Akhil Mathew Jan 2 '11 at 23:12
  • 3
    $\begingroup$ The question has been closed at the request of the author. $\endgroup$ – Pete L. Clark Jan 2 '11 at 23:44
  • 1
    $\begingroup$ Please do not delete it! $\endgroup$ – Mariano Suárez-Álvarez Jan 3 '11 at 1:58