# 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 $$k$$?

## 1 Answer

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.