Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $k$ ba field. Let $F(X, Y)$ be a non-constant polynomial in $k[X, Y]$. Suppose $F(0, 0) = 0$. Then $F(X, Y)$ is of the form $aX + bY +$ higher degree terms. Suppose $aX + bY \neq 0$. Let $A = k[X, Y]/(F)$. Let $x, y$ be the images of $X, Y$ in $A$ respectively. Let $\mathfrak{m} = (x, y)$. Is the localization $A_{\mathfrak{m}}$ of $A$ at $\mathfrak{m}$ a discrete valuation ring?

share|cite|improve this question
Yes. This is Theorem 1 in Chapter 3, Section 2 of Fulton's Algebraic Curves. It also follows (a little less directly) from Proposition 9.2 , implication $iv\to i$ in Atiyah- Macdonald's Commutative Algebra. –  Georges Elencwajg Nov 13 '12 at 19:13
@GeorgesElencwajg Dear Georges, but the theorem of the Fulton's book assumes $F$ is irreducible. –  Makoto Kato Nov 13 '12 at 19:32
Dear Makoto, yes you are right. However if $F$ factorizes into irreducibles as $F=F_1\cdot F_2\cdot \ldots$ and if $F_1(0,0)=0$ then we'll have $F_i(0,0)\neq 0$ for $i=2,3,...$ (because of your hypothesis that $F(X,Y)=aX+bY+...$) and these $F_i$'s will be invertible in $k[X,Y]_\frak m$ and won't change $A_\frak m$, so that you can reduce to the case where $F=F_1$ is irreducible. –  Georges Elencwajg Nov 13 '12 at 20:50
@GeorgesElencwajg Thanks. –  Makoto Kato Nov 13 '12 at 21:20
You are welcome, Makoto. –  Georges Elencwajg Nov 13 '12 at 21:23

4 Answers 4

up vote 3 down vote accepted

$R=K[X,Y]_{(X,Y)}$ is a local regular ring of dimension $2$ with maximal ideal $M=(X,Y)R$. Then $F\in M-M^2$, so $R/(F)$ is local regular of dimension $1$, hence DVR.

Edit. At Makoto Kato request I'll sketch a proof of the following assertion: $(R,M,k)$ local regular and $F\in M-M^2$, then $R^*=R/(F)$ is regular.

We have that $\dim R^*\ge\dim R-1$. On the other side, $\text{edim}(R^*)=\text{edim}(R)-1$, where $\text{edim}(R)$ is the minimal number of generators of $M$, i.e. $\dim_k M/M^2$. This can be proven easily by taking $F_1^*,\dots,F_n^*\in R^*$ a minimal system of generators for $M/(F)$ and showing that $F,F_1,\dots,F_n$ is a minimal system of generators for $M$. Now use the following inequality: $\dim R^*\le \text{edim}(R^*)$. We get $\dim R-1\le \dim R^*\le \text{edim}(R^*)=\text{edim}(R)-1$ and use the regularity of $R$.

share|cite|improve this answer
Could you explain why $R/(F)$ is regular? –  Makoto Kato Nov 13 '12 at 20:28
@MakotoKato Yes, but it's easier to give you a reference: Kaplansky, CR, Theorem 161. The idea is that the embedding dimension of $R/(F)$ is exactly the embedding dimension of $R$ minus $1$. –  user26857 Nov 13 '12 at 20:32
Unfortunately I don't have the Kaplansky's book and I don't have an easy access to a university library. –  Makoto Kato Nov 13 '12 at 20:37
+1 Now I understand. Thanks. Please let me wait for a few days. If there will be no better answer, I will accept this. –  Makoto Kato Nov 13 '12 at 21:33

Your ring is an integrally closed noetherian local ring with Krull dimension one, and such a thing is a DVR.

share|cite|improve this answer
How do you prove that $A_{\mathfrak{m}}$ is an integrally closed domain? –  Makoto Kato Nov 13 '12 at 19:39
Is that question relevant to the question how you prove that $A_{\mathfrak{m}}$ is integrally closed? –  Makoto Kato Nov 13 '12 at 20:09
If you personally have a problem proving it, then yes, it is relevant; if you know how to prove it, then you know how to prove it and I'll stop wasting my time; books, lots of books exist where you can read this proof. In any case, I should have known better. I'll go back to ignoring you. –  Mariano Suárez-Alvarez Nov 13 '12 at 20:13
Even if I know an answer, you know that it is perfectly legitimate to answer one's own question. Please note that there are usually several different answers to a question each of which has its own merit. Knowing them can be useful. And I don't claim I know all of them. –  Makoto Kato Nov 13 '12 at 21:53
My intention of asking questions is that: I would like to know various proofs of an interesting problem and share those proofs with users of this site. –  Makoto Kato Nov 15 '12 at 13:52

Yes. You are asking whether the origin is a nonsingular point of $C=\textrm{Spec}\,A\subset \mathbb A^2_k$. Write the homogeneous decomposition $F=\sum_{d\geq 1}f_d$, where $f_1=aX+bY\neq 0$. Let us show that $P$ is a regular point of $C$. If $P=(0,0)$ were singular, then (by definition) the two partial derivatives of $F$ would vanish at $P$. But then we would find \begin{equation} 0=\frac{\partial F}{\partial X}(P)=a+(\textrm{higher degree terms containing powers of}\, X \,\textrm{and}\, Y) \end{equation} \begin{equation} 0=\frac{\partial F}{\partial Y}(P)=b+(\textrm{higher degree terms containing powers of}\, X \,\textrm{and}\, Y). \end{equation} But this implies $a=0=b$, contradiction. Hence $P$ is regular.

Now I claim that saying $P$ is a regular point is equivalent to the assertion that $\mathcal O_{C,P}\,(\,=A_P)$ is regular as a local ring, that is, by definition: \begin{equation} \dim A_P=\dim T_{C,P}\,, \end{equation} where $T_{C,P}$ is the tangent space at $P$. If $P$ is regular then the tangent space at $P$ is a line, so $\dim T_{C,P}=1=\dim A=\dim A_P$. Conversely, if $\dim T_{C,P}=1$ then the partial derivatives of $F$, the generators of $T_{C,P}$, can't both vanish at $P$. Indeed, a point $(\alpha,\beta)\in \mathbb A^2$ is in $T_{C,P}$ if and only if \begin{equation} \frac{\partial F}{\partial X}(P)\cdot\alpha+\frac{\partial F}{\partial Y}(P)\cdot\beta=0. \end{equation} Hence $P$ is regular.

So far, we have established that $A_P$ is a regular local ring.

Finally, $\dim A_P=\dim A=\dim C=1$. Now, a DVR is a regular local ring of dimension one so your $\mathcal O_{C,P}$ is one such.

share|cite|improve this answer
How do you prove that the local ring at $P$ is a discrete valuation ring? –  Makoto Kato Nov 13 '12 at 19:27
The local ring at a nonsingular point of a curve is a DVR (regular local ring of dimension one). –  Brenin Nov 13 '12 at 20:13
If you know the proof of the statement you mentioned, please write it. –  Makoto Kato Nov 13 '12 at 20:26
Just edited my answer. –  Brenin Nov 13 '12 at 21:04
Could you explain why $A_P$ is regular? –  Makoto Kato Nov 13 '12 at 21:21

Lemma Let $A$ be a Noetherian local domain. Let $\mathfrak{m}$ be its unique maximal ideal. Suppose $\mathbb{m}$ is a non-zero principal ideal. Then $A$ is a discrete valuation ring.

Proof: Let $t$ be a generator of $\mathfrak{m}$. We claim that $\bigcap_n \mathfrak{m}^n = 0$. Let $x \in \bigcap_{n>0} \mathfrak{m}^n$. For every integer $n > 0$, there exists $y_n \in A$ such that $x = t^ny_n$. Since $t^ny_n = t^{n+1}y_{n+1}$, $y_n = ty_{n+1}$. Hence $(y_1) \subset (y_2) \subset \cdots$. Since $A$ is Noetherian, there exists $n$ such that $(y_n) = (y_{n+1})$. Hence there exists $a \in A$ such that $y_{n+1} = ay_n$. Hence $y_{n+1} = aty_{n+1}$. Hence $(1 - at)y_{n+1} = 0$. Since $1 - at$ is invertible, $y_{n+1} = 0$. Hence $x = 0$ as desired.

Let $x$ be a non-zero element of $\mathfrak{m}$. Since $\bigcap_n \mathfrak{m}^n = 0$. There exists integer $n > 0$ such that $x \in \mathfrak{m}^n - \mathfrak{m}^{n+1}$. Hence there exists $u \in A$ such that $x = t^nu$. Since $u$ is not contained in $\mathfrak{m}$, $u$ is invertible. Hence $A$ is a discrete valuation ring. QED

Let $R=K[X,Y]_{(X,Y)}$. As this question shows, there exists a canonical isomomorphism $A_{\mathfrak{m}} \cong R/(F)$. Let $F = F_1\cdots F_m$ be a factorization of $F$ into irreducible factors. Since $F(0, 0) = 0$, there exists $i$ such that $F_i(0, 0) = 0$. By the assumption that $F(X, Y) = aX + bY + \cdots$, $F_j(0, 0) \neq 0$ for $j \neq i$. Hence $F_j$ is invertible in $R$ for $j \neq i$. Hence $R/(F) = R(F_i)$. Hence $R/(F)$ is an integral domain. Therefore, by the lemma, it suffices to prove that $\mathfrak{m}$ is principal. By Nakayama's lemma, it suffices to prove that $dim_k \mathfrak{m}/\mathfrak{m}^2 = 1$.

Let $I = (X, Y)$ be the ideal generated by $X, Y$ in $k[X, Y]$. It is easy to see that $\mathfrak{m}/\mathfrak{m}^2$ is isomorphic to $I/((F) + I^2)$ as $k[X, Y]$-modules. In particular, it is isomorphic as $k$-vector spaces. Note that $dim_k I/I^2 = dim_k I/((F) + I^2) + dim_k ((F) + I^2)/I^2$. Let $x, y$ be the image of $X, Y$ by the canonical homomorphism $I \rightarrow I/I^2$ respectively. Clearly $x, y$ is a basis of the $k$-vector space $I/I^2$. Hence $dim_k I/I^2 = 2$. On the other hand, $((F) + I^2)/I^2$ is the vector subspace of $I/I^2$ generated by $ax + by$. By the assumption $ax + by \neq 0$. Hence $dim_k ((F) + I^2)/I^2 = 1$. Hence $dim_k \mathfrak{m}/\mathfrak{m}^2 = dim_k I/((F) + I^2) = 1$ as desired.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.