# Question about localization of a local ring.

Given a noetherian local ring $$A$$, $$p$$ a prime ideal of $$A$$, and a subring $$R\subseteq A$$. Put $$p_0=p\cap R$$, $$S=R_{p_0}$$ and $$B'=A_{p_0}$$. Then it's said that $$B=A_p$$ is a localization of the local ring $$B'$$ by a maximal ideal.

I am not quite understand this statement. Usually, a localized ring is bigger than original ring. But $$p_0$$ is smaller than $$p$$, thus $$A_{p_0}$$ is bigger than $$A_p$$. Then why could we say $$B$$ is a localization of $$B'$$? Hope someone could help. Thanks!

First of all, we need to recall the definition of $$A_{p_0}$$. Let $$W = R \setminus p_0$$. Then $$A_{p_0} := W^{-1}A$$.
Since $$p_0 = R \cap p$$ and $$R \subset A$$, $$R \setminus p_0 \subset A \setminus p$$. Thus, $$A_p$$ is a further localization of $$A_{p_0}$$.
The last part of your question has an incorrect part. Let $$k$$ be a field, and let $$R = k, A = k[x,y]$$, and $$p = xA$$. Then $$p_0 = 0$$ and $$A_{p_0} = A$$. But $$p = xA$$ is not a maximal ideal. So, the word "maximal" needs to be replaced by the word "prime".
In general, a localization of a ring $$R$$ need not contain $$R$$, e.g., $$\mathbb{Z}/(6)$$.