I am reading the book Algebraic Geometric Codes: Basic Notions by Tsfasman, Vladut and Nogin. The authors introduce some definitions about differential forms at the beginning of Section 2.2. These notions are as follows, but I have trouble in seeing how they coincide with the standard definition of Kähler differentials.

Let $X$ be a smooth projective curve over an algebraically closed field $k$. Let $P\in X$, and let $f\in\mathcal{O}_P$. Define the $\mathcal{k}$-linear map $d_P:\mathcal{O}_P\to\mathfrak{m}_P/\mathfrak{m}_P^2$ sending $f$ to the image of $f-f(P)$ in $\mathfrak{m}_P/\mathfrak{m}_P^2$. The element $d_P f$ is called the differential of $f$ at $P$. It can be verified that $d_P$ satisfies the Leibniz rule $d_P(fg)=f(P)d_P g+g(P)d_P f$.

For an open subset $U\subseteq X$, let $\Phi[U]$ be the set of maps $\varphi$ sending each point $P\in U$ to some element in $\mathfrak{m}_P/\mathfrak{m}_P^2$. Then $\Phi[U]$ is a $k[U]$-module. (They use $k[U]$ for the ring of regular functions over $U$, and I think the standard notion is $\mathcal{O}_X(U)$.) Any function $f\in k[U]$ defines an element $df\in \Phi[U]$ by $(df)(P)=d_P f$. An element $\varphi\in\Phi[U]$ is called a differential form regular on $U$ if for any $P\in U$ there exists a neighbourhood $V$ of $P$ such that $\varphi|_V$ lies in the $k[V]$-submodule of $\Phi[V]$ generated by elements $df$, $f\in k[V]$. Finally, differential forms regular on $U$ form a $k[U]$-module, denoted by $\Omega[U]$.

The authors avoid using the standard notions of sheaves and schemes. But their notions confuse me. I think $\Omega[U]$ here is the same as the global sections of Kähler differentials $\Omega_{X/k}$ over $U$. But why do they coincide? (The usual definition (I learned) of Kähler differentials $\Omega_{S/R}$ comes from the universal property, or is explicitly defined as $I/I^2$ where $I$ is the kernel of the multiplication map $S\otimes_R S\to S$.)

Also what is $\Phi[U]$? My guess is that $\Phi$ is presheaf and the sheaf $\Omega$ is constructed from it by taking the compatible germs.




You must log in to answer this question.

Browse other questions tagged .