Question of Hartshorne II8.15

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.

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.

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 $$X$$.

• @6666 Hartshorne calls this "generization", and the concept is introduced in exercise II.3.17 (e), which is stated on pages 93-94. I have added more details at the end about the reason every point is a generalization of a closed point. (Why did you delete your comment?) Apr 24 '20 at 1:35
• Sorry about that, I deleted because I found generalization later on Hartshorne's book.
– 6666
Apr 24 '20 at 14:23
• Isn't a closed subscheme of a Noetherian scheme Noetherian? Do we have to work on the setting of quasi-compact scheme? Can we work via Noetherian scheme?
– 6666
Apr 24 '20 at 14:33
• @6666 a Noetherian scheme is quasicompact by definition (or by a very direct argument: given an open cover, we can construct a descending chain of closed subsets, and it stabilizing is equivalent to the cover being finite). Closed subschemes of Noetherian schemes are again Noetherian, and more is true: any subspace of a Noetherian topological space is Noetherian. Apr 24 '20 at 16:53
• Thanks, I am asking because I proved the statement last night by showing a Notherian scheme has a closed point
– 6666
Apr 24 '20 at 16:56