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Let $R \rightarrow S$ be a faithfully flat (unital) homomorphism of commutative rings. Does it follow that the corresponding map of topological spaces $\Spec S \twoheadrightarrow \Spec R$ is open?

If not, is there a counterexample where both $R$ and $S$ are finitely generated over the same algebraically closed field?

EDIT: The second question was answered in a comment, but I'm still interested in the first part.

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    If $R \to S$ is flat and of finite presentation, then $\operatorname{Spec}(f)$ is open (see \[Stacks, Tag [00I1](http://stacks.math.columbia.edu/tag/00I1)\]). The idea is that Chevalley's theorem implies the image of $\operatorname{Spec}(S)$ in $\operatorname{Spec}(R)$ is constructible, and flat maps satisfy going down, hence the image is stable under generization. – 2017-02-23
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    Oh right; of course. Well that answers the second part of my question. Any ideas on the first part? For that, if I remove "faithfully", then a counterexample is given by embedding any non-field integral domain into its fraction field. – 2017-02-23
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    As you said, $R\rightarrow\operatorname{Frac}R$ is not faithful. But $R\rightarrow R\times\operatorname{Frac}R$ is, and $\operatorname{Spec}(R\times\operatorname{Frac}R)=\operatorname{Spec}R\coprod\operatorname{Spec}\operatorname{Frac}R\rightarrow\operatorname{Spec}R$ is not open. – 2017-02-23
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    @Roland Great; thanks! If you want to post this as an answer, I'll accept it. – 2017-02-23
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    ..though now I wonder if there's a counterexample that is a local homomorphism of local rings. – 2017-02-23

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