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](http://mathworld.wolfram.com/FaithfullyFlatModule.html) 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.


Edit 3: To answer Emerton's comment, I asked here after:


a.) Reading this [post](http://sbseminar.wordpress.com/2009/08/06/algebraic-geometry-without-prime-ideals/#comment-6354) 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.



Weakened conditions for étale + X implies faithfully flat.
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 $\mathfrak q\cap R=\mathfrak p$.


See also exercise 16 of Chapter 3 of Atiyah-Macdonald.


