Let $R\subset S$ be two local PIDs that have the same fields of fractions. How to prove that they are equal?
-
$\begingroup$ Arguably R=k[[T]] and S=Frac(R) gives you a counterexample. Perhaps you have different definition of local and/or PID than I do though. $\endgroup$– Kevin BuzzardMar 28, 2010 at 12:03
-
1$\begingroup$ Probably means to assume that the inclusion of R into S is a local homomorphism. In that case it is valid more generally for $R$ a valuation ring and $S$ merely a local domain, and as such it is at the level of a homework exercise. It should not be posted here as a question. $\endgroup$– BCnrdMar 28, 2010 at 12:35
-
$\begingroup$ Why hasn't this question been closed already? $\endgroup$– user4324Mar 28, 2010 at 13:30
-
$\begingroup$ Let me also leave a hint for a direct proof: show that every ring in between a PID and its fraction field is a localization. Then think about what the possibilities are for localizing a PID with only one prime element. $\endgroup$– Pete L. ClarkMar 28, 2010 at 17:52
1 Answer
I believe that the OP meant to include the condition that each of the local PIDs is not a field. In this case the result is true, and as several people have said, is a rather standard exercise.
At this moment it seems to me that if we get asked a rather standard question that has not been asked on MO before, it would be nice to use it as an opportunity to explain something a little deeper / slightly less standard related to the question. In this regard, let me mention a generalization:
A local PID $R$ (which is not a field!) with fraction field $K$ is precisely a discrete valuation ring, i.e., is the valuation ring $R = \{x \in K \ | \ |x| \leq 1 \}$ of a norm $| \ |: K \rightarrow \mathbb{R}^{\geq 0}$ such that $|K^{\times}|$ is a discrete subgroup of $\mathbb{R}^{\times}$. Now for nontrivial norms $| \ |_1$, $| \ |_2$ (Archimedean or not) on a field $K$, there is the following result:
Theorem: The following are equivalent:
(i) There exists $\alpha > 0$ such that $| \ |_2 = | \ |_1^{\alpha}$.
(ii) For all $x \in K$, $|x|_1 < 1 \implies |x|_2 < 1$.
(iii) For all $x \in K$, $|x|_1 \leq 1 \implies |x|_2 \leq 1$.
(See e.g. http://alpha.math.uga.edu/~pete/8410Chapter1.pdf, p. 4, for a proof.)
Now the implication (iii) $\implies$ (i) shows that there can be no proper containments among DVRs with the same fraction field. The same holds for all rank one valuation ring because, by definition, a rank one valuation ring is one whose value group is a subgroup of $\mathbb{R}$; therefore the data of a rank one valuation is equivalent to that of a non-Archimedean norm. (Note that if $| \ |$ is a non-Archimedean norm, then $v = - \log | \ |$ is a valuation, and conversely if $v$ is a rank one valuation, then $| \ | = e^{-v}$ is a non-Archimedean norm.) It is not true for valuation rings of higher rank.
-
$\begingroup$ I like the answer, and I especially like your second paragraph. Given the choice between an explanation of why a question is interesting and explanation of why it isn't, in situations like this when both are available, I think we all get more out of the former. $\endgroup$ Mar 28, 2010 at 17:45
-
$\begingroup$ typical for mathoverflow. the answers are voted up which deal with a lot of known and fancy material, no matter if it actually answers the question ... $\endgroup$ Mar 29, 2010 at 8:44
-
$\begingroup$ Martin, ultimately votes are absolutely irrelevant... $\endgroup$ Mar 29, 2010 at 19:35
-