Localization of a one-dimensional noetherian integral domain

Question

Let $o$ be a one-dimensional noetherian integral domain, and $a$ be a non-zero ideal of $o$. Suppose $p$ be a prime ideal containing $a$. Show that $o/o \cap ao_p= o_p/ao_p$. (Here $o_p$ is the localization of $o$ at $p$.)

The following is my idea: Let $f$ be the canonical mapping $o$ to $o_p/ao_p$, then the kernel of $f$ is exactly $o \cap ao_p$, but why is it surjective? I know this is equivalent to $o+ao_p=o_p$ but I don't know how to prove it.

Thanks in advance!

Answer 1

In fact, by considering a primary decomposition of $\mathfrak a$, we may assume that $\mathfrak a$ is $\mathfrak p$-primary (since $\mathfrak a\mathcal O_{\mathfrak p}=\mathfrak q\mathcal O_{\mathfrak p}$ for some $\mathfrak p$-primary ideal $\mathfrak q$), and now all is clear: $\mathcal O\cap\mathfrak q\mathcal O_{\mathfrak p}=\mathfrak q$ and $\mathcal O/\mathfrak q$ is an artinian local ring with the maximal ideal $\mathfrak p/\mathfrak q$, so $\mathcal O/\mathfrak q\simeq(\mathcal O/\mathfrak q)_{\mathfrak p/\mathfrak q}\simeq\mathcal O_{\mathfrak p}/\mathfrak q\mathcal O_{\mathfrak p}$.

Answer 2

First note that $o_p$ and $a o_p$ and in general $o_q$ for a prime $q$ of $o$ are subrings and $o$--submodules of $k=Q(o)$, the quotient field of $o$. So $o \cap a o_p$ and similar expressions are well defined.

Now $o/(o \cap a o_p) = (o + a o_p)/(a o_p)$ by Noether's isomorphism theorem. So indeed $o/(o \cap a o_p) = o_p/(a o_p)$ is equivalent to

$$o + a o_p = o_p$$

Now call $x = o + a o_p$ and $y = o_p$. Note that for an arbitrary prime $q \neq 0$ we have $(o_p)_q = k$ when $q \neq p$ and $(o_p)_q = o_p$ when $q = p$. The first assertion follows from the fact that $()_q$ destroys the prime $p o_p$ and so $(o_p)_q$ has only the prime $(0)$.

So it is $x_q = y_q = k$ for all $q \neq p$ and $x_q = y_q = o_p$ for $q=p$. So $x=y$ by the local-global principle.