# Applications of valuation rings

Some background: I am in the process of writing a research paper for an undergraduate abstract algebra course. I've chosen to write my paper on valuation rings and discrete valuation rings. The goal of the paper is to broaden my own and my classmates' understanding of abstract algebra by independently researching a topic not covered in the course. So far, my paper consists of surveying the properties of valuation rings and giving examples of valuation rings.

What I would like to know: I would like to be able to comment on how valuation rings are utilized in various fields of mathematics. Thus far I've had a hard time finding examples that are both explicit and accessible to me, as all of the literature I've found on valuation rings have been graduate texts. It seems like valuation rings are often used in number theory and algebraic geometry, but how are they applied in those fields? What other fields find valuation rings of significant usefulness, and how are they applied? I would also find so-called real world applications useful for my understanding of the topic, but personally I'm more interested in how mathematicians make use of them.

Assume I know linear and abstract algebra at an undergraduate level. Do not assume I know very much about geometry, number theory, or analysis. I'm currently taking a differential geometry course, but what I'm learning seems entirely separate from anything I've seen relating valuation rings to geometry.

My apologies for the broad question. Please let me know if I can clarify or specify in any way.

• You can check for the valuative criterion for properness. It is a bit technical but it states properness (a topological property of an algebraic variety) in term of extending map of valuation rings. There are also probably exercises in Atiyah-Macdonald which are worth working. – user171326 Apr 12 '17 at 5:26
• @N.H. IMO even in a graduate course the valuative criteria are not easy to digest, so this doesn't seem to satisfy the requirements of the OP. – user347489 Apr 12 '17 at 6:03

If you're familiar with complex analysis, the collection of meromorphic functions on an open subset $U \subseteq \mathbb{C}$ can be endowed with many discrete valuations, one for each point of $U$. Given a meromorphic function $f$, for each $a \in U$ we can write $f(z) = (z - a)^v g(z)$ for some $v \in \mathbb{Z}$, where $g$ is a function holomorphic at $a$ with $g(a) \neq 0$. We define the order (of vanishing) of $f$ at $a$, denoted $\operatorname{ord}_a(f)$, to be this $v$. One can show that $\operatorname{ord}_a$ is a discrete valuation.

Similarly, in algebraic geometry DVRs can be used to measure the order of vanishing of a function at a point. Given a curve $C$ and a point $P \in C$, then the local ring at the point $P$ is a DVR iff $C$ is nonsingular at $P$. Basically, a function $f$ regular at $P$ has order of vanishing $v$ if $f \in \mathfrak{m}^v$ and $v$ is the smallest such positive integer, where $\mathfrak{m}$ is the maximal ideal corresponding to $P$.

This also allows us to detect the singularities of a curve algebraically. For instance, consider the cuspidal cubic $C: y^2 = x^3$. The origin is a singular point of $C$, as is clear from looking at a plot of the curve, or by computing partial derivatives, and this is reflected by the fact that the local ring $\left(\frac{k[x,y]}{(y^2 - x^3)}\right)_{(x,y)}$ is not a DVR.

• great answer. I don't think I know of any simpler use of DVRs in algebraic geometry than this. The keyword here is Weil divisor. – user347489 Apr 12 '17 at 6:02
• I am not familiar with complex analysis (I edited my question to reflect that), but this is still helpful information. I'll have to dig around and find the definitions of some of the terms you used (e.g. meromorphic function, order of vanishing), but I really appreciate your response. Thank you. – rosterherik Apr 12 '17 at 18:01

A very natural set of examples come from number theory(which is apparently where the name divisor comes from!), namely number fields (i.e. finite extensions of the field of rational numbers $\mathbf{Q}$). The simplest case is $\mathbf{Q}$ itself considered as such an extension. One fixes a prime number $p$, and given any rational number $\alpha = \frac{a}{b}$ write $\alpha$ as $p^n\frac{a'}{b'}$ where the integers $a'$ and $b'$ are relatively prime and are not divisible by the fixed prime number $p$. It is easy to see (but needs proof) that such an $n \in \mathbf{Z}$ is uniquely determined(this is simply because $\mathbf{Z}$ is a unique factorization domain). Then, one gets a function on $\mathbf{Q}$ by defining $\nu_p(\frac{a}{b}):=n$, as found above. It is then an exercise to prove that $\nu_p$ is a valuation on $\mathbf{Q}$. It turns out that (Theorem of Ostrowski) apart from the usual absolute value, these are all valuations on $\mathbf{Q}$.

A parallel line of construction can be carried out by replacing $\mathbf{Q}$ with the field of rational functions over an algebraically closed field $k$, namely $k(X)$. In this case for any element $\alpha \in k$, there is a unique integer $n$ so that: $$f(X) = \frac{p(X)}{q(X)} = (X-\alpha)^n\frac{p'(X)}{q'(X)};$$ where $p'$ and $q'$ are relatively prime polynomials in $k[X]$. As was the case before, $k[X]$ is a UFD, hence $n$ is uniquely determined. Once again, we define $\nu_\alpha(f(X)) := n$. One must check that this defines a valuation on $k(X)$. An analogue for Ostrowski's theorem is still valid here with the only remaining function on $k(X)$ defining a valuation being the $\mathrm{deg}$ function: which is defined as the difference of the degrees of $p$ and $q$.

It must be stressed that the construction for $\mathbf{Q}$ can be carried out for any number field, and the construction for $k(X)$ can be carried out for any algebraic curve.

There is, of course, the tropical geometry world where valuations play a central role. this tool is used frequently to solve, at least, certain enumerative problems in algebraic geometry.

• For your example regarding the degree of a rational function, where we define $\nu:k(X)\rightarrow\mathbb{Z}$ by $\nu(\frac{f}{g})=\deg g - \deg f$, is there any restriction on the domain? I've found this example mentioned briefly elsewhere, and the author stated that we restrict the domain to rational functions $\frac{f}{g}$ over $k$ where $\deg f \leq \deg g$. They did not say why however and I'm not sure if this is restriction is necessary. – rosterherik Apr 12 '17 at 18:29
• This discussion is discussion of a valuation defined on a field. To each such valuation one may attach its valuation ring, defined as $$\mathcal{O}_\nu :=\{\alpha \, : \, \nu(\alpha) \geq 0\}.$$ Conversely, from an integral domain with a valuation one obtains a valuation on the field of fractions of this integral domain. These two theories are equivalent. In the example you are referring to, the author takes the second approach, and the set he is describing, namely the set of rational functions $f/g$ with $\mathrm{deg}(f) \leq \mathrm{deg}(g)$ is exactly the valuation ring. – ayberk Apr 13 '17 at 8:28
• Ah! This makes a lot of sense. In my research I have yet to encounter the second approach stated explicitly. Thank you for this! – rosterherik Apr 14 '17 at 0:49