Nonsingular curve and normality condition I tried to prove the classical result that in $\mathbb{C}[x,y]$ ,an irreducible polynomial $f$ defines a nonsingular variety iff its coordinate ring is integrally closed. To do so, i first localized in a point . Let it be $(0,0)$ wlog. I was hinted to demonstrate that said $A=\mathbb{C}[x,y]/(f)$ if $A_{(x,y)}$ is normal then the tangent space has dimension $1$. I wrote $f= f_1+f_2+\cdots f_d$ with $f_i$ homogenous of degree $d$ but i didn't really understand what to do. Any help would be appreciated. Ty
 A: There are several things going between everything here, so I'll break it down.
Traditionally, nonsingular would mean that "there exists a tangent space", thus we will assume then that $X=Z(f)$ is nonsingular at $P \in X$ if $rank(\nabla(f)(P))=1$ (i.e. in this case $\nabla(f)(P) \neq 0$).
Now let the coordinate ring of $X$ be $A(X) = \mathbb{C}[x,y]/(f)$ and $P \in X$ correspond to the max ideal $m_P = (x-a,y-b)$ in $\mathbb{C}[x,y]$.
Now, we know via commutative algebra that $A(X)$ is integrally closed if and only if $A(X)_{m_P}$ are integrally closed for all $P\in X$. Thus we reduce to showing $X$ is nonsingular at $P$ if and only if $A(X)_{m_P}$ is integrally closed. Further, as $A(X)_{m_P}$ is dimension 1 (dim($X$)=1), this is equivalent to it being a DVR and also to it being a regular local ring.
Thus, how do you show dim($m_P/m_P^2$) = $rank(\nabla(f)(P))$? (Hint: $\nabla: \mathbb{C}[x,y] \to \mathbb{C}^2$ induces an isomorphism $m_P/m_P^2 \to \mathbb{C}^2$, $\nabla((f))(P)$ is a $rank(\nabla(f)(P))$ subspace that you can pull back to $m_P/m_P^2$, and relate this to the max ideal of $A(X)_{m_P}$.)
More generally, this is in R. Hartshorne's Algebraic Geometry Theorem I.5.1.
