If $\mathcal O_P(C)$ is a DVR, then $P$ is non-singular Let $C$ be an irreducible curve over $\mathbb A^2$ and $P\in C$. I would like to prove if $$\mathcal O_P(C)=\{f\in k(C)\mid f=a/b, b(P)\neq 0\}$$ is a DVR, then $P$ is non-singular, i.e., the derivatives $F_X(P)\neq 0$ or $F_Y(P)\neq 0$.
MY ATTEMPT
I've already proved that $\dim_k(\mathfrak m/\mathfrak m^2)=1$, where $\mathfrak m$ is the maximal ideal of $\mathcal O_P(C)$ and $k=\mathcal O_P/\mathfrak m$. Someone could help me how to continue?
Thanks
 A: I thought about splitting this up into a bunch of algebra exercises but decided against it. Let's just plow through it.
Let's just assume that $P = (0, 0)$, corresponding to the maximal ideal $\mathfrak{m} = (x, y) \subseteq A = k[x, y]$. $C$ is cut out of $\mathbb{A}^2$ by some polynomial $f \in A$ and of course $f \in \mathfrak{m}$ if $P$ lies on $C$. The local ring $\mathcal{O}_{C, P}$ is obtained from $\mathcal{O}_{\mathbb{A}^2, P}$ by modding out by $f$. Now, $\mathfrak{m}_{C, P} = \mathfrak{m}_{\mathbb{A}^2, P}/(f)$ and $\mathfrak{m}_{C, P}^2 = (\mathfrak{m}_{\mathbb{A}^2, P}^2 + (f))/(f)$ and hence $\mathfrak{m}_{C, P}/\mathfrak{m}_{C, P}^2 \simeq \mathfrak{m}_{\mathbb{A}^2, P}/(\mathfrak{m}_{\mathbb{A}^2, P}^2 + (f))$.
This is a really important point: to get the cotangent space to $C$ at $P$ you take the cotangent space to $\mathbb{A}^2$ at $P$ and quotient out by (the residue class of) the defining equation for $C$. All that (introductory, granted) algebra to make an intuitive point. One more bit of algebra: I can identify $\mathfrak{m}_{\mathbb{A}^2,P}/\mathfrak{m}_{\mathbb{A}^2,P}^2$ with $\mathfrak{m}/\mathfrak{m}^2$. The former is the latter localized at $\mathfrak{m}$, but $A/\mathfrak{m}^2$ is already a local ring, so there's no need to localize.
Now, you can write $f$ as
$$
f(x, y) = ax + by + (\text{higher order terms})
$$
and here, really, $a = (\partial f/\partial x)(P)$ and $b = (\partial f/\partial y)(P)$. The residue class of $f$ mod $\mathfrak{m}^2$ is thus $a\bar{x} + b\bar{y}$. Here $\bar{x}$ and $\bar{y}$ form a basis for $\mathfrak{m}/\mathfrak{m}^2$. If the quotient is going to be one-dimensional then one of $a, b$ has to be nonzero.
