Local ring at a non-singular point of a plane algebraic curve 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?
 A: $R=K[X,Y]_{(X,Y)}$ 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 $R$.
A: 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.
A: Your ring is an integrally closed noetherian local ring with Krull dimension one, and such a thing is a DVR.
A: 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.
