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Suppose $K$ is a field and $R=K[x,y]/(y^2-x^3)$. The question requires to show the localization at $(x,y)$ is not a discrete valuation ring.

I can find the unit element in the localization is the polynomial a with a non trivial constant term. The the valuation should be determined by the value at $x$ and $y$. Since $x^3=y^2$, there is $3v(x)=2v(y)$. I do not know how to proceed from here.

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Is the ring a principal ideal domain? –  Mariano Suárez-Alvarez Sep 24 '12 at 20:28
If you know what are the units, look at the complement, and try to show that is not a maximal ideal. –  M Turgeon Sep 24 '12 at 20:38
Also, if you look at this question from a geometric point of view, we know that the curve associated to $R$ has a cusp at $(0,0)$, which means that if you localise at the ideal $(x,y)$, you don't get an integrally closed domain. –  M Turgeon Sep 24 '12 at 20:40
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2 Answers

up vote 5 down vote accepted

Remark that the equation $(y/x)^2 = y^2/x^2 = x^3/x^2 = x$ holds in the quotient field of $R$. This shows that $y/x$ is integral over $R$. But $y/x$ is not in the localization of $R$ at $(x,y)$, hence this localization is not integrally closed in its field of fractions, and therefore cannot be a D.V.R.

As M Turgeon points out, this is due to the fact that the curve $C: y^2=x^3$ is singular at the origin. The function $y/x$, despite not belonging to the local ring of $C$ at the origin, does not have a singularity there when restricted to $C$ - rather, it behaves just like $\sqrt x$. This illustrates the fact that near a singular point, there are "more" functions than there should be.

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Instead of solving the OP's problem directly, we will prove a general proposition and apply it to the problem. In the following proof, we will borrow an idea of Hartshorne's algebraic geometry.

Let $A = K[x, y]$. Let $f(x, y) \in K[x, y]$ be an irreducible polynomial such that $f(0, 0) = 0$. Let $\mathfrak{a} = (x, y)$, $\mathfrak{b} = (f(x, y))$. Since $f(0, 0) = 0$, $\mathfrak{b} \subset \mathfrak{a}$. Let $R = A/\mathfrak{b}$. Let $R_{\mathfrak{p}}$ be the localization of $R$ at $\mathfrak{p} = \mathfrak{a}/\mathfrak{b}$. Let $\mathfrak{m}$ be the maximal ideal of $R_{\mathfrak{p}}$.

Let $J_0(f) = (\frac{\partial f}{\partial x}(0),\frac{\partial f}{\partial y}(0))$. We regard $J_0(f)$ as a $(1, 2)$-matrix. We will prove that $R_{\mathfrak{p}}$ is a discrete valuation ring if and only if rank $J_0(f) = 1$. The OP's problem can be solved immediately by this result.

We define a linear map $\theta\colon A \rightarrow K^2$ by $\theta(F) = (\frac{\partial F}{\partial x}(0),\frac{\partial F}{\partial y}(0))$. Since $\theta(x) = (1, 0)$, $\theta(y) = (0, 1)$, $\theta$ is surjective. Since $\theta(\mathfrak{a}^2) = 0$, $\theta$ induces a surjective linear map $\bar\theta: \mathfrak{a}/\mathfrak{a}^2 \rightarrow K^2$. Since dim$_K \mathfrak{a}/\mathfrak{a}^2 = 2$, this map is an isomorphism. Then $\theta(\mathfrak{b})$ is isomorphic to $(\mathfrak{b} + \mathfrak{a}^2)/\mathfrak{a}^2$. Since dim$_K \theta(\mathfrak{b}) =$ rank $J_0(f)$, rank $J_0(f) =$ dim$_K (\mathfrak{b} + \mathfrak{a}^2)/\mathfrak{a}^2$

It is easy to see that $$\mathfrak{m}/\mathfrak{m}^2 \approx \mathfrak{a}/(\mathfrak{b} + \mathfrak{a}^2)$$

Hence dim $\mathfrak{m}/\mathfrak{m}^2 +$ rank $J_0(f) =$ dim$_K \mathfrak{a}/\mathfrak{a}^2 = 2$.

Hence dim $\mathfrak{m}/\mathfrak{m}^2 = 1$ if and only if rank $J_0(f) = 1$. By Nakayama's lemma, this is equivalent to that $\mathfrak{m}$ is prinicipal. Hence this is equivalent to that $R_{\mathfrak{p}}$ is a discrete valuation ring(e.g. Atiyah-MacDonald).

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