A Geometric Reason to Care About Discrete Valuation Rings

Perhaps those that have taken an introductory commutative algebra class have come across the profoundly opaque definition of a discrete valuation ring (DVR), thinking to themselves “why have the math gods condemned me?” From a purely algebraic perspective, it is a remarkable that a ring A is a valuation ring if and only if it is a domain so that for any element in the fraction field for A, x or x^{-1} is contained in  A. Even further, if a valuation ring is Noetherian (and not a field) it is a DVR. Again, algebraically, there are a lot of reasons to care about these rings, but today isn’t about algebra. While there are a huge number of equivalences we can speak of, but instead we will make use of the following definition:


Definition:  A ring A is said to be a DVR if it is a local Noetherian ring with principal maximal ideal.

Geometrically, it is very easy to come up with many examples of Noetherian local rings. For example, consider some curve Z(f) \subseteq \mathbb A^2 (let’s just assume that k is algebraically closed.) We can then form its co ordinate ring k[X,Y]/(f) and localize at the maximal ideal corresponding to some point in  p=(p_1, p_2) \in Z(f), as in (x-p_1,y-p_2). Since f vanishes at p, it is contained in the maximal ideal, so it survives the localization and localization is exact , we could have just as easily localized first and taken the co-ordinate ring either. Either way, now we have this associated ring of functions around some point, denoted \mathcal{O}_{f,p}, and in particular, we should notice that it is local Noetherian, almost by construction.

Now, this point of view is interesting, but what would happen if the maximal ideal were principal, making \mathcal{O}_{p,f} a DVR? Well in this case, note   that \mathfrak{m}/\mathfrak{m}^2 has exactly dimension 1 as a \mathcal{O}_{p,f}/\mathfrak{m}-vector space.  The converse is true as well,  since Nakayama’s lemma implies that in a local Noetherian ring, a basis for \mathfrak{m}/\mathfrak{m}^2 lifts to a generating set for all of \mathfrak{m}, implying that it is principal as an ideal. [Proposition 6.6 here should do it.]

Geometrically, \mathfrak{m}/\mathfrak{m}^2 is known as the Zariski cotangent space which is heuristically all of the linear functions (degree one) defined around our point. The main point is that by taking the dual of the cotangent space, we get another vector space, which is precisely is actually the tangent space, denoted \Theta_{p,f}. Another way to see it, is that given a polynomial on a variety vanishing at our point, we should ask what function these polynomials restrict to on our tangent space, which are the linear ones. Here is a potentially an unnecessary formalism: restricting our attention to curves, we can define

    \[d_p(f):=\frac{\partial f}{\partial x}(x-p_1)+\frac{\partial f}{\partial y} (y-p_2).\]

 From this, we solve for where the differentials of each defining polynomial are simultaneously zero. By restricting to each polynomial onto the tangent space, d_p defines a homomorphism from polynomials onto linear forms, but since d_x(\alpha)=0 for all constant functions \alpha, we really can just consider d_p:m_{x} \to \Theta_{p,f}^{*} and do great algebraic battle to conclude that this induces an isomorphism \mathfrak{m}/\mathrak{m}^2 \to \Theta_{p,f}^*.

With this in mind, we will give a few basic definitions and then state what all of this is really about:

Definition: A point of p \in C, with C an irreducible curve,  is called simple if f has nonvanishing derivative at p, so that \frac{\partial f}{\partial x}(p) \neq 0 or \frac{\partial f}{\partial y}(p) \neq 0.

We should note that by a linear change in co-ordinate shift p to the origin if after decomposing f into homogenous components, f has a nonzero linear term.

These considerations give rise to a theorem that makes me smile:


Theorem: If C \subseteq \mathbb A^2 is an irreducible curve given by V(f), then p \in C is simple if and only if \mathcal{O}_{f,p} is a DVR.

proof: Without a loss of generality, suppose that p=0 is simple. Furthermore, we can define the tangent line T, but by some rotation we can assume that T=y. We will show that \mathfrak{m}_p=(x). Letting f be the polynomial associated to C, we see that in our nice co-ordinates, f=y+\{\mathrm{higher \, order \, terms}\}. We collect all of those terms that have a power of y:

    \[f=y \cdot g(x,y)-x^2h(x),\]

where g(x,y) has a constant term. In the local ring, \mathcal{O}_{f,p}, the polynomial g becomes invertible, so we can factor y=x^2 h(x) \cdot g^{-1}(x,y), and see that y \in (x), so m=(x,y) is really the same as (x).


Conversely, suppose that A:=\mathcal{O}_{f,p} is a DVR. Again supposing that p=0, we can show that it is simple by showing that f \notin (x,y)^2 (so it has linear terms, and thus nonvanishing derivative.) Assume for contradiction, that f \in (x,y)^2. Well, then (x,y) would survive in A since f \in (x,y)^2, which is a contradiction. In particular, we can consider \mathfrak{m}^{\prime}:= (x,y) \subseteq k[x,y]_{\mathfrak{m}}, and note that (f) \to \frac{\mathfrak{m}^{\prime}}{\mathfrak{m}^{\prime}^2} \twoheadrightarrow \frac{\mathfrak{m}}{\mathfrak{m}^2} is  a composition with kernel (f), and the middle term is two dimensional as a vector space. If f \in (x,y)^2, this would contradict the claim that \mathfrak{m}/\mathfrak{m}^2 is one dimensional as a vector space.

As an aside, there is a more detailed and difficult proof in many texts, that define simple (nonsingular points) by way of dimension: \mathrm{dim}_k(\mathfrak{m}/\mathfrak{m}^2)=dim(X) for an affine variety X.


There is some more work to be done, regarding the condition that \mathcal{O}_{f,p} be a DVR and the dimension of the Zariski cotangent space, but there is a sense in which this proof belongs to a greater generality, instead really being about regular local rings.