Let $R$ be a commutative ring, and let $S\subseteq R$ be a subring. Consider a prime ideal $\mathfrak p\subseteq R$ and let $\mathfrak q=\mathfrak p\cap S$ be the restricted prime ideal in $S$. When does the embedding $$ S\hookrightarrow R$$ induce an embedding $$S_\mathfrak q\hookrightarrow R_\mathfrak p?$$ If this is not true in general, under what assumptions does it hold?

By the exactness of localization, we have an embedding $S_\mathfrak q\hookrightarrow S_\mathfrak q\cdot R$, but I don't if this can be extended to $S_\mathfrak qR\to R_\mathfrak p$.


1 Answer 1


About your first question, with $R=k[x,y]/(xy), p=(x),S=k[x],q=(x)$,

Since $y$ becomes a unit in $R_p$ then $x$ is in the kernel of $R\to R_p$ ie. $R_p=Frac(k[x,y]/(x))$,

Whereas $S_q = k[x]_{(x)}$.

  • $\begingroup$ What is the scheme theoretic word saying there is no such problem in a morphism $Spec(R)\to Spec(S)$ ? $\endgroup$
    – reuns
    Jan 20, 2020 at 19:30
  • $\begingroup$ I don't think I know the term your aiming at. Something like an "epimorphism" of schemes? $\endgroup$ Jan 21, 2020 at 15:14
  • $\begingroup$ And also: could this be true if we require $\mathfrak q$ and $\mathfrak p$ to be maximal? $\endgroup$ Feb 19, 2020 at 16:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .