Let $A$ be an integral domain and $R=A[X]$. Let $Q$ be a prime ideal of $A$ and let $P=Q[X]$.

If $R_P$ is integrally closed (in its own fractions field), then is $A_Q$ integrally closed (in its own fraction field)? If not true in general, then does some additional assumption on $A$ make it true (like Noetherian or some restriction on dimension) ?

Is the converse true, i.e. if $A_Q$ is integrally closed, then is $R_P$ integrally closed ?


1 Answer 1


Yes, the two conditions are equivalent. Let $K$ be the fraction field of $A$; then the fraction field of $R$ is $K(X)$. Let (1) be the condition that $R_{P}$ is integrally closed, and let (2) be the condition that $A_{Q}$ is integrally closed. For (1) implies (2), use that the inclusion $A_{Q} \subseteq K \cap R_{P}$ (intersection inside $K(X)$) is an equality. For (2) implies (1), let $S := A \setminus Q$; then $A_{Q}[X]$ is integrally closed by Tag 030A; here $R_{P}$ is a localization of $A_{Q}[X]$ so it is also integrally closed by Tag 00GY.

  • $\begingroup$ why is $A_Q=K \cap R_P$ ? $\endgroup$
    – user521337
    Sep 10, 2018 at 0:44
  • $\begingroup$ why is $A[X]_{Q[X]}$ a localisation of $A_Q[X]$ ... ? at which set is it a localisation ... ? $\endgroup$
    – user521337
    Sep 10, 2018 at 0:46

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