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.

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

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.

"tensoring with S over R never kills a non-zero module"as an answer, or is that your definition? $\endgroup$ – Anton Geraschenko Feb 12 '10 at 3:30