# How to embed the localization of a subring

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 $$S_\mathfrak qR\to R_\mathfrak p$$.

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 $$S_q = k[x]_{(x)}$$.
• What is the scheme theoretic word saying there is no such problem in a morphism $Spec(R)\to Spec(S)$ ? Jan 20, 2020 at 19:30
• And also: could this be true if we require $\mathfrak q$ and $\mathfrak p$ to be maximal? Feb 19, 2020 at 16:37