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!
Best Answer
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., $\mathbb{Z}/(6)$.