Exercise 5.11

Question

Let $f: A 	o B$ be a flat homomorphism of rings. Then $f$ has the going-down property.

Amit Awasthi · Accepted Answer

\begin{proof}
  By Exercise $3.18$, we know that if $\mathfrak{q} \in \operatorname{Spec}(B)$ and $\mathfrak{p} = \mathfrak{q}^c$, then $f^* : \operatorname{Spec}(B_{\mathfrak{q}} 	o \operatorname{Spec}(A_{\mathfrak{p}}))$ is surjective. By AM Exercise $5.10ii)$ above, this implies $f$ has the going-down property.
\end{proof}

Exercise 5.11

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Exercise 5.11

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