How do I prove the differentials of a smooth curve is 1 dimensional? Let $C$ be a smooth curve over a field $k$, with function field ${k(C)}$. Let ${\Omega_C}$ be the module of differentials of ${k(C)}$ over $k$. I am trying to show that ${\Omega_C}$ is a $1$ dimensional vector space with basis given by ${dt}$ where ${t}$ is any local parameter of $C$.
My idea has been to take an ${f \in k(C)}$, and let ${P}$ be a point in the open subset of $C$ for which ${f}$ is regular. Then ${f-f(P)}$ has a $0$ at $P$, and thus lies in the maximal ideal of the Discrete Valuation Ring ${\mathcal{O}_{C,P}}$ generated by any local parameter $t$ at $P$. So ${f - f(P) = ut^n}$ for ${u}$ a unit in ${\mathcal{O}_{C,P}}$ and ${n\geq 0}$. Thus ${d(f-f(P)) = df}$ (since ${f(P) \in k}$) ${ = d(ut^n) = t^ndu + nut^{n-1}dt}$. My issue is now that I don't know what to do about this ${du}$ term. Any tips?
 A: Question: "My idea has been to take an ${f \in k(C)}$, and let ${P}$ be a point in the open subset of $C$ for which ${f}$ is regular. Then ${f-f(P)}$ has a $0$ at $P$, and thus lies in the maximal ideal of the Discrete Valuation Ring ${\mathcal{O}_{C,P}}$ generated by any local parameter $t$ at $P$. So ${f - f(P) = ut^n}$ for ${u}$ a unit in ${\mathcal{O}_{C,P}}$ and ${n\geq 0}$. Thus ${d(f-f(P)) = df}$ (since ${f(P) \in k}$) ${ = d(ut^n) = t^ndu + nut^{n-1}dt}$. My issue is now that I don't know what to do about this ${du}$ term. Any tips?"
Answer: Assume your curve $C$ (assume it is projective) is embedded as a closed subscheme in some projective space. It follows $C$ is birational to a plane projective curve $C'$ and $C'$ contains an open affine curve $C_1$ birational to $C$. hence $K(C) \cong K(C_1)$ have the same function field. If we assume $C$ is irreducible it follows $C_1$ is irreducible and we may write $C_1=Spec(A)$ where $A:=k[x,y]/(f(x,y))$ and where $f(x,y)$ is an irreducible polynomial. It follows the module of Kahler differentials $\Omega:=\Omega^1_{A/k}$ is the free rank 2 module on $dx,dy$ modulo the submodule $U$
$$\Omega \cong A\{dx,dy\}/U$$
with $U:=Adf$ where $df:=f_xdx+f_ydy$ and where $f_x,f_y$ are the partial derivatives wrto the $x$ and $y$ variables. If $df=0$ it follows $\Omega$ is a free rank 2 module over $A$. Hence
$$\Omega^1_{K(C_1)/k} \cong K(C_1)\{dx,dy\}$$
is a 2-dimensional vector space over $k(C_1)$. If $df\neq 0$ you may choose a basis for $\Omega^1_{K(C_1)/k}$ depending on wether $f_x=0$ or $f_y=0$. In this case
$$dim_{K(C_1)}(\Omega^1_{K(C_1)/k})=1.$$
Hence if $df=0$ it follows $dim=2$ and if $df \neq 0$ it follows $dim=1$.
Ex: If char $k=p>0$ any polynomial $f\in k[x^p,y^p]$ will have $df=0$ but such a curve will not be smooth.
