Question of Hartshorne II8.15

Question

The theorem $8.15$ (p.177) from the Hartshorne's book "Algebraic Geometry" says:

" Let $X$ be an irreducible separated scheme of finite type over an algebraically closed field $k$. Then $\Omega_{X/k}$ is a locally free sheaf of rank $n= \dim \ X$ iff $X$ is nonsingular variety over $k$."

Proof: If $x\in X$ is a closed point, then the local ring $B =\mathcal{O}_{x,X}$ has dimension $n$, residue field $k$, and is a localization of a $k$-algebra of finite type. Furthermore, the module $\Omega_{B/k}$ of differentials of $B$ over $k$ is equal to the stalk $(\Omega_{X/k})_x$ of the sheaf $\Omega_{X/k}$· Thus we can apply (8.8) and we see that $(\Omega_{X/k})_x$ is free of rank $n$ if and only if $B$ is a regular local ring. Now the theorem follows in view of (8.14A), which says any localization of a regular local ring at a prime ideal is again a regular local ring. and (Ex. 5.7), which says that $\Omega_{X/k}$ is locally free iff $(\Omega_{X/k})_y$ is free for all $y\in X$.

My confusion lies in the last sentence, I can see if $\Omega_{X/k}$ is a locally free sheaf of rank $n= \dim \ X$, then by (Ex. 5.7), $(\Omega_{X/k})_x$ is free of rank $n$ for any closed point, thus $\mathcal{O}_{X,x}$ is a regular local ring at each closed point. Now apply (8.14A), we know $\mathcal{O}_{X,x}$ is a regular local ring at non-closed points, thus $X$ is nonsingular. However, I feel confused with the other direction: Suppose $X$ is smooth, then by (8.8), we know $(\Omega_{X/k})_x$ is free of rank $n$ for any closed point $x$. But how do we show $(\Omega_{X/k})_x$ is free of rank $n$ for any nonclosed point? Here to apply (8.8), it requires $k(x)=k$, but which is true for closed points. I don't think we can use (8.8) for non-closed points.

Answer 1

This is a direct application of exercise II.5.7 (a):

Let $X$ be a noetherian scheme and let $\mathcal{F}$ be a coherent sheaf. If the stalk $\mathcal{F}_x$ is a free $\mathcal{O}_{X,x}$ module for some $x\in X$, then there is a neighborhood of $U$ such that $\mathcal{F}|_U$ is free.

For any closed point $x\in X$, such a neighborhood $U$ contains all nonclosed points which are generalizations of $x$ (that is, if $x'$ is a point with $x\in \overline{\{x'\}}$, then $x'\in U$). Why? If not, then $U^c$ is a closed subset containing $x'$ but not $x$, so $x\notin \overline{\{x'\}}$ by the definition of the closure as the smallest closed subset containing $x'$. So the stalk of $\Omega_{X/k}$ is free of rank $n$ at every point which is a generalization of a closed point, and since we are working over a Noetherian scheme, this is every point.

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Edit: As requested (in a now-deleted comment), here are some more details about why every point is a generalization of a closed point in a noetherian scheme $X$. It suffices to prove that every point has a closed point in it's closure, or equivalently every closed subscheme has a closed point. But then this is implied by the standard statement that a quasi-compact scheme has a closed point, as every subscheme of a Noetherian scheme is quasi-compact.

To find a closed point in an arbitrary quasi-compact scheme $X$, let $X=\bigcup_{i=1}^n U_i$ be a finite decomposition of $X$ as a union of open affines so that no $U_i$ is contained inside $\bigcup_{j\neq i} U_j$. Now $(U_2\cup \cdots \cup U_n)^c\cap U_1\subset U_1$ is a closed subset of an affine scheme and thus affine. Since every affine scheme has a closed point (this is the geometry version of every ring having a maximal ideal), we can find a closed point $p\in (U_2\cup \cdots \cup U_n)^c\cap U_1 \subset U_1$, which is then closed in every $U_i$ and thus defines a closed point of $X$.