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?
Local ring at a non-singular point of a plane algebraic curve
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1Yes. 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*. – 2012-11-13
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0@GeorgesElencwajg Dear Georges, but the theorem of the Fulton's book assumes $F$ is irreducible. – 2012-11-13
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0Dear 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. – 2012-11-13
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0@GeorgesElencwajg Thanks. – 2012-11-13
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0You are welcome, Makoto. – 2012-11-13
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0FYI, you could have replaced sentences 2 through 5 with "Let $F(X,Y) = aX + bY + \text{higher degree terms} \in k[X,Y]$ with $aX + bY \neq 0$." or something similar, eliminating some unnecessary exposition. – 2012-11-14
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0@Hurkyl That $F(X, Y)$ has the stated form is rather a result from the assumption that $F(0, 0) = 0$. – 2012-11-14
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0@Makoto: Right. The point is that making the derivation doesn't really add anything to the question in its current form. Since it's been a problem in the past, I'm trying to give examples of the sorts of things you can do to simplify your exposition in the future. – 2012-11-15
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0@Hurkyl It is not essential that $P = (0, 0)$. $P$ can be any point of the curve $F(X, Y) = 0$. If $P = (s, t)$, then $F(X, Y)$ is of the form $a(X - s) + b(X - t)$ + higher terms. And the question would be essentially the same. – 2012-11-15
4 Answers
$R=K[X,Y]_{(X,Y)}$ 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 $R$.
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0Could you explain why $R/(F)$ is regular? – 2012-11-13
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0@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 $1$. – 2012-11-13
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0Unfortunately I don't have the Kaplansky's book and I don't have an easy access to a university library. – 2012-11-13
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0+1 Now I understand. Thanks. Please let me wait for a few days. If there will be no better answer, I will accept this. – 2012-11-13
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.
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0How do you prove that the local ring at $P$ is a discrete valuation ring? – 2012-11-13
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0The local ring at a nonsingular point of a curve is a DVR (regular local ring of dimension one). – 2012-11-13
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0If you know the proof of the statement you mentioned, please write it. – 2012-11-13
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0Just edited my answer. – 2012-11-13
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0Could you explain why $A_P$ is regular? – 2012-11-13
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0It's the definition of regular point. – 2012-11-13
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0Why $P$ is a regular point? – 2012-11-13
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0I proved it in my answer – 2012-11-13
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0You just proved the obvious fact that the two partial derivatives of $F$ at $P$ don't vanish at the same time. – 2012-11-13
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0... which is the definition of regular point. – 2012-11-13
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0But you said that the definition of a regular point is that $A_P$ is a regular local ring. – 2012-11-13
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0let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6414/discussion-between-atricolf-and-makoto-kato) – 2012-11-14
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0Why not here? Please explain why $A_P$ is a regular local ring. – 2012-11-14
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0You are showing no effort in understanding your own problem. For the last time, the following are equivalent: at least one partial derivative of $F$ is nonzero at $P$ $\Leftrightarrow$ $P$ is a regular point $\Leftrightarrow$ $A_P$ is a regular local ring. These are all straightforward equivalences. In my answer I prove that $P$ is a regular point. And because its local ring is of dimension one, you conclude. I already wrote all this above. – 2012-11-14
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0You said that the definition of a regular point $P$ is that $A_P$ is regular. You also said that the definition of a regular point is that at lease one partial derivative of $F$ at $P$ is nonzero at $P$. Which is your definition? – 2012-11-14
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0What is your definition of a regular local ring? – 2012-11-14
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0Your definition of a regular local ring uses the tangent space of a local ring. Please explain the definition of it. – 2012-11-15
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0The tangent space of a local ring does not make sense. $T_{C,P}:=(m_P/m_P^2)^\vee$ denotes the tangent space of $C$ at $P$, in EVERY book you might wonder to look up at. Now let us stop with these comments, I am exhausted by the total absence of your efforts. If you don't like my answer, don't worry, I can survive. – 2012-11-15
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0I'm just trying to make clear your proof. Please explain why dim $(m_P/m_P^2)^* = 1$. – 2012-11-15
Your ring is an integrally closed noetherian local ring with Krull dimension one, and such a thing is a DVR.
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1How do you prove that $A_{\mathfrak{m}}$ is an integrally closed domain? – 2012-11-13
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1Is that question relevant to the question how you prove that $A_{\mathfrak{m}}$ is integrally closed? – 2012-11-13
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5If 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. – 2012-11-13
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0Even 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. – 2012-11-13
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0«Answering your own question» has absolutely nothing to do with this situation... – 2012-11-13
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0Why? You need to know an answer before you answer your own question. – 2012-11-13
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0I think you know this, but just in case. "if you have a question that you already know the answer to if you’d like to document it in public so others (including yourself) can find it later it is OK to ask, and answer, your own question on a relevant Stack Exchange site." http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/ – 2012-11-15
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0As you know, I find that to be a waste of time. I prefer not to participate in that, and as I have told you a few times, I think it would be best if you were explicit about your intentions **when you ask the question**. Since you apparently have some problem with being upfront with what your intention is in asking questions, I will try to keep ignoring you. – 2012-11-15
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1My 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. – 2012-11-15
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.