What schemes correspond to varieties in the sense of Weil? Out of (perhaps morbid) curiosity I am trying to learn the basics of Weil's foundations of algebraic geometry.
I tried to ask a question earlier but it turned out I had misunderstood some more basic points, so I want to confirm them before re-asking my original question.
Let $\mathbb{K}$ be a universal domain in the sense of Weil, i.e. an algebraically closed field of infinite transcendence degree over its prime field.
For convenience, by quantity we will mean an element of $\mathbb{K}$.
A point in $n$-space is an $n$-tuple of quantities.
A variety in $n$-space in the sense of Weil is a pair $(k, P)$ where:

*

*$k$ is a subfield of $\mathbb{K}$ such that $\mathbb{K}$ has infinite transcendence degree over $k$.

*$P$ is a point in $n$-space.

*The subfield $k (P) \subset \mathbb{K}$ generated by $k$ and the components of $P$ is separably generated over $k$, and $k$ is algebraically closed in $k (P)$.

If $V$ is a variety given by data $(k, P)$ as above, Weil says:

*

*$k$ is a field of definition of $V$.

*$V$ is defined over $k$.

*$V$ is the locus of $P$ over $k$.

*$P$ is a generic point of $V$ over $k$.

A point $Q$ in $n$-space is in $V$ if:

*

*For every polynomial $F$ in $n$ variables with coefficients in the field of definition $k$, $F (P) = 0$ implies $F (Q) = 0$.

Weil says two varieties in $n$-space are equivalent if they have the same points.
Note that equivalent varieties need not have the same field of definition.
Question 1. Is a variety in $n$-space in the sense of Weil, up to equivalence, the same information as a closed $\mathbb{K}$-subscheme of $\mathbb{A}^n_\mathbb{K}$ that is integral and of finite type over $\mathbb{K}$?
More precisely, suppose we are given a variety $V$ in $n$-space, with a field of definition $k$ and a generic point $P$ over $k$.
Let $I$ be the set of polynomials $F$ in $n$ variables with coefficients in $k$ such that $F (P) = 0$.
Then $I$ defines a closed $\mathbb{K}$-subscheme $\tilde{V}_\mathbb{K}$ of $\mathbb{A}^n_\mathbb{K}$.
I ask:

*

*Is this $\tilde{V}_\mathbb{K}$ integral and of finite type over $\mathbb{K}$?
(Yes: see this question.)

*Is the map $V \mapsto \tilde{V}_\mathbb{K}$ well defined and injective up to equivalence of varieties?

*Does every closed $\mathbb{K}$-subscheme of $\mathbb{A}^n_\mathbb{K}$ that is integral and of finite type over $\mathbb{K}$ arise as $\tilde{V}_\mathbb{K}$ for some $V$?

Furthermore, $I$ also defines a closed subscheme $\tilde{V}_k$ of $\mathbb{A}^n_k$, and $\tilde{V}_\mathbb{K} \cong \tilde{V}_k \times_{\operatorname{Spec} k} \operatorname{Spec} \mathbb{K}$.
So $\tilde{V}_k$ is a geometrically integral affine scheme of finite type over $k$.
Question 2. Is a variety in $n$-space defined over $k$, up to equivalence, the same information as a closed $k$-subscheme of $\mathbb{A}^n_k$ that is a geometrically integral affine scheme of finite type over $k$?
 A: The questions turn out to be fairly straightforward to answer once we know that the conditions on $k (P)$ are equivalent to geometric integrality.
This immediately answers question 2: the closed $k$-subschemes of $\mathbb{A}^n_k$ that correspond to varieties in $n$-space defined over $k$ are precisely the ones that are geometrically integral over $k$.
Next, to answer question 1.
Since affine varieties in $n$-space in the sense of Weil are equivalent iff they have the same points, the map $V \mapsto \tilde{V}_\mathbb{K}$ is well defined and injective on equivalence classes $V$.
This is basically the Nullstellensatz: $\mathbb{K}$ is algebraically closed, so the closed subspaces of $\mathbb{A}^n_\mathbb{K}$ are uniquely determined by their closed points.
Now we need to show that every closed $\mathbb{K}$-subscheme of $\mathbb{A}^n_\mathbb{K}$ that is integral and of finite type over $\mathbb{K}$ corresponds to some affine variety in $n$-space.
Well, such a subscheme corresponds to a finitely generated prime ideal $J$ of the algebra of polynomials in $n$ variables over $\mathbb{K}$.
Choose some generators of $J$, and let $k$ be a subfield $\mathbb{K}$ containing the coefficients of the chosen generators.
Since there are only finitely many coefficients, we can always choose $k$ small enough that $\mathbb{K}$ has infinite transcendence degree over $k$.
For simplicity we can also assume $k$ is algebraically closed in $\mathbb{K}$.
This will be our field of definition.
Let $I$ be the contraction of $J$, i.e. the subset of $J$ consisting of polynomials with coefficients in $k$.
Let $A$ be the algebra of polynomials over $k$ modulo $I$.
Then $A$ is an integral domain of finite type over $k$.
In particular, $\operatorname{Frac} (A)$ is a finitely generated field extension of $k$, hence there is a $k$-algebra embedding $\operatorname{Frac} (A) \to \mathbb{K}$.
Let $P$ be the $n$-tuple of images in $\mathbb{K}$ of the $n$ generators of $A$.
Since $A$ is an integral domain and $k$ is algebraically closed, $A$ is geometrically integral over $k$, and it follows that $\operatorname{Frac} (A) \cong k (P)$ is separable over $k$.
Hence $(k, P)$ is a variety in $n$-space in the sense of Weil, and by construction the ideal of polynomials over $k$ vanishing at $P$ is precisely $I$, so the locus is exactly the set of closed points of the subscheme we started with.
This completes the proof.
