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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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    @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

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$R=K[X,Y]_{(X,Y)}$ 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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    +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
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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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    I'm just trying to make clear your proof. Please explain why dim $(m_P/m_P^2)^* = 1$.2012-11-15
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Your ring is an integrally closed noetherian local ring with Krull dimension one, and such a thing is a DVR.

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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.2012-11-15
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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.