Weakened conditions for étale + X implies faithfully flat.

Let $F:R \to S$ be an étale morphism of rings. It follows with some work that $f$ is flat.

However, faithful flatness is another story. It's not hard to show that faithful + flat is weaker than being faithfully flat. An equivalent condition to being faithfully flat is being surjective on spectra.

The question: Is there any further condition we can require on an étale morphism that implies faithful flatness?

"Faithfully flat implies faithfully flat" or "surjective on spectra is equivalent to faithfully flat" do not count. The answer should in some way use the fact that the morphism is étale (or at least flat).

As you can see by the tag, all rings commutative, unital, etc.

Edit: Why faithfully flat is weaker than faithful + flat.

Edit 2: I resent the voting down of this question without accompanying comments as well as the voting up of the glib and unhelpful answer below. It's clear that some of you are in the habit of voting on posts based on the poster rather than the content, and I think that is shameful. There is nothing I can do because none of you has the basic decency to at least leave a comment. I am completely at your mercy. You've won. I hope it's made you very happy.

a.) Reading this post by Jim Borger

b.) Asking my commutative algebra professor in an e-mail

Which led me to believe (perhaps due to a flawed reading of said sources) that this was a harder question than it turned out to be.

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I replaced your $\acute{e}$s with é: the joys of an international keyboard layout! :) –  Mariano Suárez-Alvarez Feb 12 '10 at 2:09
(As a parenthetical aside, one of the standard definitions of an etale map is one which is [sometimes only locally] of finite type, flat and unramified. From this definition, etale implies flat takes little work!) It would be great if finite etale maps were surjective on spectra. This is not always true: e.g. $\pi_1: \mathbb{C} \times \mathbb{C} \rightarrow \mathbb{C}$ is finite etale but not faithfully flat. However, I think it becomes true under further useful hypotheses. More later, or perhaps this will jog someone else's memory. –  Pete L. Clark Feb 12 '10 at 2:45
Ah, I see. That's also a perfectly reasonable definition of "faithful." Should I add "tensoring with S over R never kills a non-zero module" as an answer, or is that your definition? –  Anton Geraschenko Feb 12 '10 at 3:30
Expanding on Pete's comment: if R --> S is injective, finite, and etale, then it is faithfully flat. (Pf: The map Spec S --> Spec R has dense image, by the injectivity, and has closed image, by finiteness, hence is surjective.) –  Emerton Feb 12 '10 at 3:49
Dear Harry, Anton's answer to this question is a textbook answer, available in any number of places (Matsumura certainly, Atiyah--MacDonald it seems based on Anton's answer, surely Eisenbud's tome, ... ). In the past you've criticised others for posting questions with easily available answers of this type. Perhaps others are voting down your question for the same reason. –  Emerton Feb 12 '10 at 5:04

The definition of $f:R\to S$ being faithfully flat that I first saw is that $S\otimes_R-$ is exact and faithful (meaning that $S\otimes_R M=0$ implies $M=0$). I'm not sure exactly what your definition of "faithfully flat" is, but it looks like you're happy with "flat and surjective on spectra." You get flatness for free from étaleness, so I'll show that the extra faithfulness condition implies surjectivity on spectra.

Upon tensoring with $S$, $f$ becomes $f\otimes_R id_S:S\cong S\otimes_R R\to S\otimes_R S$, given by $s\mapsto s\otimes 1$. This is injective since it is a section of multiplication $S\otimes_R S\to S$. By flatness of $S$, this shows that $S\otimes_R \ker f=0$, so $\ker f=0$. So I'll identify $R$ with a subring of $S$.

Let $\mathfrak p\subseteq R$ be a prime ideal. We wish to show that there is a prime $\mathfrak q\subseteq S$ such that $\mathfrak q \cap R=\mathfrak p$. Let $K$ be the kernel of the morphism $R/\mathfrak p\to S/\mathfrak p S$ of $R$-modules. Upon tensoring with $S$, this morphism becomes injective (as before, it's a section of the multiplication map $S/\mathfrak p S\otimes_R S/\mathfrak p S\to S/\mathfrak p S$), so by flatness of $S$, we have $S\otimes_R K=0$, so $K=0$. This shows that $\mathfrak p S \cap R=\mathfrak p$ (if the intersection were any larger, $K$ would be non-zero). So $\mathfrak p$ generates a proper ideal in the localization $(R\setminus \mathfrak p)^{-1}S$. Let $\mathfrak q\subseteq (R\setminus \mathfrak p)^{-1}S$ be a maximal ideal containing $\mathfrak p$. This corresponds to some prime ideal $\mathfrak q$ (slight abuse of notation to use the same letter) of $S$ which contains $\mathfrak p$ but does not intersect $R\setminus \mathfrak p$, so $\mathfrak q\cap R=\mathfrak p$.

I assume that you mean that $f:R\to S$ is faithful. –  Harry Gindi Feb 12 '10 at 4:13