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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?

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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 $R$.

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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.

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How do you prove that $A_{\mathfrak{m}}$ is an integrally closed domain? –  Makoto Kato Nov 13 '12 at 19:39
    
What have you tried? –  Mariano Suárez-Alvarez Nov 13 '12 at 19:58
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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
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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
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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.

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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.

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