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I'm reading Matsumura's proof on independence of valuation (p.87, "Commutative Ring Theory"), and I understand most of the proof, except the last bit about principal ideal ring, and if anyone could help me I would greatly appreciate it:

$K$ a field, $R_1$,...,$R_n\subset K$ DVRs of $K$ not containing one another, $A=\cap_{i}^{n}R_i$. Let $\mathfrak{m}_i\subset R_i$ be the maximal ideals and $\mathfrak{p}_i:=\mathfrak{m}_i\cap A$. I can see that $A$ is a semi-local ring with maximal ideals $\mathfrak{p}_1$,...,$\mathfrak{p}_n$, and that $A_{\mathfrak{p}_i}=R_i$. But I don't see why $A$ is a PID. Here is a passage from the book:

$\mathfrak{m}_i\neq\mathfrak{m}_i^2$, and hence $\mathfrak{p}_i\neq\mathfrak{p}_i^{(2)}$, where $\mathfrak{p}^{(2)}:=\mathfrak{p}^2A_{\mathfrak{p}}\cap A$. Thus there exists $x_i\in\mathfrak{p}_i$ such that $x_i\notin\mathfrak{p}_i^{(2)}$, and $x_i\notin\mathfrak{p}_j$ for $i\neq j$ (I understand up to here); then $\mathfrak{p}_i=x_iA$ (I don't understand this). If $I$ is any ideal of $A$ and $IR_i=x^{\nu_i}_iR_i$ for $i=1$,...,$n$ then it is easy to see that $I=x_{1}^{\nu_1}...x_{n}^{\nu_n}A$ (I don't see this).

I'm sorry for this rather pedantic question but I have no one to help me...

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Well, $x_iR_i = \mathfrak{m}_i$, since $x_i\in\mathfrak{m}_i$ and $x_i\notin\mathfrak{m}_i^2$. Does that suffice to show $\mathfrak{p}_i = x_iA$ ? – Arturo Magidin Mar 6 '11 at 4:26
I don't think $x_iR_i=\mathfrak{m}_i$ necessarily implies $\mathfrak{p}_i=x_iA$ in this situation. – ashpool Mar 6 '11 at 4:36
Possibly not; alas, my copy of Matsumura is behind chained doors at the moment... – Arturo Magidin Mar 6 '11 at 4:53
Could someone please explain the notation $A_{\mathfrak{p}}$? – joriki Mar 6 '11 at 5:10
@joriki: $A$ localized at $\mathfrak{p}$. Intuitively, make everything not in $\mathfrak{p}$ a unit. Formally, – Arturo Magidin Mar 6 '11 at 5:31
up vote 1 down vote accepted

If you have a map of $R$-modules $f: M \rightarrow N$, whether it is an injection / surjection / bijection can be checked locally: i.e., it is enough to check the corresponding property for $f_{\mathfrak{m}}: M_{\mathfrak{m}} \rightarrow N_{\mathfrak{m}}$ for all maximal ideals $\mathfrak{m}$ of $R$. (See for instance $\S 7.5$ of my commutative algebra notes. I'm sure it's in Matsumura as well...)

In this case you have an inclusion of $R$-modules $\iota: x_i A \rightarrow \mathfrak{p}_i$ (moral: don't forget that "ideal" = "$R$-submodule of $R$"!). I think you'll find the hypotheses are rigged so that upon extending to each of the finitely many maximal ideals, $\iota$ is a bijection. So $\iota$ is itself a bijection.

The second statement you ask about can be proved in a similar way.

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Let $v_i$ be the valuations corresponding to the DVR $R_i$. Note that according to your choice we have $v_i(x_j) = 0$ if $i \neq j$ but $v_i(x_j) =1$ if $i = j$. So $x_iA \subset \mathfrak{p}_i$. We have to show the opposite inclusion. Choose any any $y \in \mathfrak{p}_i$. Then $y = x_it$ for some $t \in R_i$ since $\mathfrak{m}_i = x_i R_i$. Now it is easy to see that values of $t = y/x_i$ under all valuations $v_i$ are non negative. This means that $t \in A$. So we are done. in fact you can also show that $\mathfrak{p}_i^{k} = \mathfrak{p}_i^{(k)}$ for any natural number $k$. Now note that the element $x_{1}^{\nu_1}...x_{n}^{\nu_n} \in I_{\mathfrak{p_i}}$ for all $\mathfrak{p_i}$. This means that $x_{1}^{\nu_1}...x_{n}^{\nu_n}A \in I$. Now the two ideals are equal locally. Therefore they are equal globally.

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