Hartshorne, Algebraic Geometry, Exercise III.9.1 asks one to prove
A flat morphism $f : X \to Y$ of finite type of Noetherian schemes is open, i.e., for every open subset $U \subseteq X$, $f(U)$ is open in $Y$.
So far as I can tell this is essentially equivalent to the going down theorem, which only needs the hypothesis of flatness. Are the Noetherian and finite-type conditions actually needed here?
Best Answer
Let me just say the finite type hypothesis is 100% necessary; otherwise the result is false. Consider the inclusion $k[t] \to k(t)$. A module over a PID is flat iff it is torsion-free, so the induced map on spec here is flat. But $\operatorname{Spec} k(t)\to \operatorname{Spec} k[t]$ is not open!