Restriction of flat morphism to preimage of open subscheme.

Question

Let $f:Y \rightarrow X $ be a flat morphism of locally ringed space, i.e. the canonical map $f^\#:\mathcal{O}_{X,f(p)}\rightarrow \mathcal{O}_{Y,p}$ is a flat ring map for every $p \in Y$.

Then we can consider for $U \subset X$ open the map of locally ringed spaces $f_{|f^{-1}(U)}:f^{-1}(U) \rightarrow U$. I thought that the restriction will also be flat as we have for every $p \in f^{-1}(U)$ the identities $\mathcal{O}_{f^{-1} (U),p}\cong \mathcal{O}_{Y,p}$ and $\mathcal{O}_{U,f(p)}\cong \mathcal{O}_{X,f(p)}$. Hence $f_{|f^{-1}(U)}^\#:\mathcal{O}_{U,f(p)}\rightarrow \mathcal{O}_{f^{-1} (U),p}$ is flat as $f^\#:\mathcal{O}_{X,f(p)}\rightarrow \mathcal{O}_{Y,p}$ is flat for every $p \in f^{-1}(U)$.

But then I found this discussion Restriction of flat morphism and got confused. So what is the mistake in the argumentation above?

Answer 1

Everything is correct in your argument. In fact flat morphisms of schemes are stable under base change.

In the linked question the OP restricted also the domain of a flat morphism to a closed subscheme and this makes it completely different from the situation you are considering.