Here's a question I am working on but I'm unable to make sense of it because the notation is used very liberally. This question is based out of the chapter about Derivations in Matsumura, Commutative ring theory.
If $X$ is a variety over a field $k$ (let's assume $k$ is algebraically closed) and $P$ is a $k$-point on $X$ with maximal ideal $m_P$ then $\Omega^1_{X/k}$ modulo $m_P$ is a $k$-vector space naturally isomorphic to the dual of $m_P/m_P^2$. Now the rank of $\Omega^1_{X/k}$, as an $O_X$-module, equals the dimension of $X$, and so $\Omega^1_{X/k}$ is locally free in a neighbourhood of P if and only if X is smooth at P.
I believe $\Omega^1_{X/k}$ means $\Omega^1_{k[X]/k}$, where $k[X]$ is the coordinate ring. I believe the $\mathcal{O}_X$ also means the coordinate ring $k[X]$. The points of an affine variety are in one to one correspondence with maximal ideals in the variety's coordinate ring. So that's where maximal ideal associated with $P\in X$ comes from.
The terms I can't understand are "$\Omega^1_{X/k}$ modulo $m_P$", "dimension of X" and "locally free". I am not meant to use scheaves or schemes as in Hartsthorne to answer this.
Anyone got any ideas how to decode this?
Best Answer
You are correct that you may interpret $\Omega^1_{X/k}$ as $\Omega^1_{k[X]/k}$ and $\mathcal{O}_X$ as $k[X]$ if you want to avoid speaking of sheaves. Now, to answer your questions:
$\Omega^1_{X/k}$ modulo $m_P$: Let $A$ be a commutative ring, let $\mathfrak{m}$ be an ideal of $A$, and let $M$ be an $A$-module. Then $\mathfrak{m}M := \{am\mid a\in\mathfrak{m}, m\in M\}$ is an $A$-submodule of $M,$ and we may form the quotient module $M/\mathfrak{m}M,$ which is naturally a module over $A/\mathfrak{m}.$ As a side note, $M/\mathfrak{m}M$ is isomorphic to $M\otimes_A A/\mathfrak{m}$.
Dimension of $X$: The dimension of a variety or scheme is its dimension as a topological space. This means that the dimension of $X$ is the supremum of the lengths of chains of irreducible closed subsets of $X$: $$\dim X := \sup_{n}\{Z_0\subsetneq Z_1\subsetneq \cdots\subsetneq Z_n\subseteq X\mid Z_i\textrm{ closed irreducible subspace of }X\}.$$ In an affine scheme $X = \operatorname{Spec}A,$ such a chain corresponds to a chain of prime ideals in $A,$ so that the dimension of $X$ is the same as the Krull dimension of $A.$
Locally free: This one is a little harder to make sense of without sheaves. You should know that to any morphism of schemes $f : X\to S,$ we may associate a sheaf of $\mathcal{O}_X$-modules $\Omega^1_{X/S}$ which is obtained by gluing together the sheaves of Kähler differentials on affine pieces. A sheaf $\mathcal{F}$ on $X$ is called locally free if for every point $x\in X,$ there exists an open neighborhood $U\ni x$ such that $\left.\mathcal{F}\right|_U\cong\mathcal{O}_U^n$ for some $n.$
Translating this back into a condition on your module, to be locally free in a neighborhood of $P$ means that there exists some $f\in A\setminus m_P$ such that the localization $M_f = M[f^{-1}]$ is isomorphic to a free $A_f = A[f^{-1}]$-module of some rank.