Hartshorne III 10.4

Question

I'm reading Hartshorne III 10.4 which goes as follows:

Why the following two statements hold?

1. $X\rightarrow \text{Spec } k$ finite type,$k$ is algebraically closed,then $X\rightarrow \text{Spec}\ k$ is smooth of relative dimension $n$ iff $X$ is regular of dimension $n$.
2. $A$ is a localiztion of a finitely generated $k$-algebra,then $A\rightarrow A\otimes_k \bar{k}$ is faithfully flat.

statement 1 is used in paragraph 1.statement 2 is used in paragraph 2.statement 1 is also example 10.0.3.Here are example 10.0.3 and the definition of smooth morphism in Hartshorne:

What I have done:

statement 1: the (II,8.8) mentioned in example 10.0.3 is:

Regularity implies smoothness is obvious.When proving smoothness implies regularity,to use the above (II,8.8),it suffices to prove "over local ring $O_{p}$ at $p\in X$,$\Omega_{O_{p}/k}$ is always a free $O_p\text{-module}$".Now the only information I have is $\text{dim}_{k(x)}\Omega_{X/k}\otimes k(x)=n$.Here is lemma 8.9: So if $X$ is integral, it's done. But how to deal with the situation that $X$ is not integral? Or is there any other criterion for free-ness of a module over the local ring by just counting the dimension of the fibers like lemma 8.9?

statement 2:$\text{Spec }A\otimes_k \bar{k}\rightarrow \text{Spec }A$ being flat is obvious.So to prove it being faithfully flat,it suffices to prove it's surjective as map of sets.Note that all these properties are preserved when taking base change from $A$ to localization of $S^{-1}A$ or quotient of $A/I$,So I can set $A=k[x_1,...,x_n]$.Now because $\text{Spec }\bar{k}[x_1,...,x_n]\rightarrow \text{Spec} k[x_1,...,x_n]$ is surjective(I'm not sure...but this looks natural...),it's done.Did I prove statement 2 correctly?

Any help or reference is appreciated, thanks.

Answer 1

I address the question regarding Example III.10.0.3.

When $k$ is algebraically closed, Theorem 16.19 and Corollary 16.21 in Eisenbud (Commutative Algebra with a View Toward Algebraic Geometry) establish an equivalence between the local ring being regular of dimension $n$, the Jacobian matrix having co-rank $n$ and the module of derivations being locally free of rank $n$. Also, the cokernel of the Jacobian matrix computes the module of differentials (Exercise 21.2.E. in Vakil's November 2017 notes).

@KReiser: The local ring having a unique minimal prime does not alone make it into an integral domain.