# Does the concept of localizing at an extension of a prime ideal make sense?

If $A,B$ are commutative rings with $1$, $p$ is a prime ideal in $A$ and $f:A\rightarrow B$ makes $B$ an $A$-algebra, I want to know if it is possible to define the localization $B_p$ of $B$ at the extension of $p$.

I suspect it is possible, because if we view $B$ as an $A$-module, we obtain the $A_p$-module $B_p$. This is isomorphic to $B\otimes_A A_p$, which carries the structure of an $A$-algebra. Thus, $B_p$ is an $A$-algebra.

However, it is not clear to me that $B\setminus p$ is a multiplicative subset of $B$, so I'm stuck.

-

## 2 Answers

That's good, because generally $B \setminus f(p)B$ isn't a multiplicative subset.

But it's clear that $B \otimes_A A_p$ is a localization of $B$: it's formed by inverting the multiplicative subset $S = \{ f(a) \mid a \in A \setminus p \}$.

In some nice situations, this is a semi-localization of $B$ at an ideal. e.g. if $f : A \to B$ is an extension of rings of integers in number fields (e.g. the inclusion $\mathbb{Z} \to \mathbb{Z}[i]$), then $pB$ factors as a product of prime ideals $q_i$. Then $B_{pB}$ is defined to be the localization by the multiplicative subset

$$\bigcap_i B \setminus q_i$$

and you have $B_{pB} \cong B \otimes_A A_p$ which is a semi-local ring whose maximal ideas are precisely the (extensions of the) ideals $q_i$.

-
My description of nice situations is rather narrow -- I wanted to make sure I was saying something true, but I'm sure something similar can be said in great generality. – Hurkyl Feb 11 '13 at 6:19
Well, I feel silly for not considering $f(A\setminus p)$. Thanks for your answer. – dsm4 Feb 11 '13 at 6:35

Let $S = f(A \setminus \mathfrak p)$. Then the localization $B_{\mathfrak p}$ of $B$ as an $A$ module is isomorphic to the localization $S^{-1}B$ of $B$ as a ring. This is why $B_{\mathfrak p}$ is an $A$-algebra.

But it's not true in general that $B \setminus f(\mathfrak p)$ is multiplicatively closed. For an example consider the inclusion $k \to k[x]/x^2$ and let $\mathfrak p = (0)$. Then $B \setminus f(\mathfrak p)$ is the set of nonzero elements in $k[x]/x^2$, which is not multiplicatively closed because $x^2 = 0$.

-