This is a continuation of a previous question; I will use the same notation and terminology.
An abstract variety (or Variety, with a big V) in the sense of Weil consists of the following data:
- A finite number of varieties, $V_1, \ldots, V_h$, not necessarily all embedded in the same affine space.
- For each $\alpha$ (in $\{ 1, \ldots, h \}$), a non-empty open subset $U_\alpha \subseteq V_\alpha$.
- For each $\alpha, \beta$, a birational correspondence $T_{\beta, \alpha}$ between $V_\alpha$ and $V_\beta$ with the property that if $P_\alpha$ is a point in $U_\alpha$ and $P_\beta$ is a point in $U_\beta$ such that $P_\alpha$ and $P_\beta$ correspond under $T_{\beta, \alpha}$, then $P_\alpha$ is the unique point in $V_\alpha$ corresponding to $P_\beta$ and, vice versa, $P_\beta$ is the unique point in $V_\beta$ corresponding to $P_\alpha$.
- For each $\alpha, \beta, \gamma$, we have $T_{\gamma, \alpha} = T_{\gamma , \beta} \circ T_{\beta, \alpha}$.
The definition of "field of definition" is a bit subtle but also seems irrelevant to the question I have, so I omit it. The upshot seems to be that Weil's abstract varieties are same thing as separated integral schemes of finite type over $\mathbb{K}$. (Despite the presence of the affine $V_\alpha$ in the definition, it is the quasi-affine $U_\alpha$ that are being glued together, and the overlaps are given by $T_{\beta, \alpha} \cap (U_\beta \times U_\alpha)$.)
Let $\hat{\mathbb{K}} = \mathbb{K} \amalg \{ \infty \}$. A generalized quantity is an element of $\hat{\mathbb{K}}$; it is finite if it is a quantity, i.e. an element of $\mathbb{K}$. A pseudopoint in $n$-space is an element of $\hat{\mathbb{K}}^n$ that is not a point (i.e. at least one coordinate is not finite). Note that while $\hat{\mathbb{K}}$ may be identified with $\mathbb{P}^1 (\mathbb{K})$, $\hat{\mathbb{K}}^n$ corresponds to $(\mathbb{P}^1)^n (\mathbb{K})$, not $\mathbb{P}^n (\mathbb{K})$! A reciprocation is a map $\sigma : \hat{\mathbb{K}}^n \to \hat{\mathbb{K}}^n$ whose $i$-th component is either the projection $(x_1, \ldots, x_n) \mapsto x_i$ or its reciprocal $(x_1, \ldots, x_n) \mapsto 1 / x_i$, where as usual we define $1 / 0 = \infty$ and $1 / \infty = 0$. (We do not define any other arithmetic expressions involving $\infty$.)
Let $P$ and $P'$ be points or pseudopoints in $n$-space and let $k$ be a subfield of $\mathbb{K}$ such that $\mathbb{K}$ has infinite transcendence degree over $k$. We say $P'$ is a specialisation of $P$ over $k$ if there is a reciprocation $\sigma$ such that $\sigma (P)$ and $\sigma (P')$ are both points and, for every polynomial $F$ in $n$ variables with coefficients in $k$, $F (P) = 0$ implies $F (P') = 0$. It can be shown that the choice of reciprocation $\sigma$ is irrelevant. A specialisation is finite if it is a point (i.e. not a pseudopoint).
Now suppose we have an abstract variety as above. Let $k$ be a field of definition and choose points $M_1, \ldots, M_h$ with the following properties:
- For each $\alpha$, $M_\alpha$ is a generic point of $V_\alpha$ over $k$.
- For each $\alpha, \beta$, $M_\alpha$ and $M_\beta$ correspond under $T_{\beta, \alpha}$.
We can consider $(M_1, \ldots, M_h)$ as a point in some affine space (of high dimension). Weil says that the abstract variety is complete if it has the following property:
- For every specialisation $(M'_1, \ldots, M'_h)$ of $(M_1, \ldots, M_h)$, finite or otherwise, there is at least one $\alpha$ such that $M'_\alpha$ is a point in $U_\alpha$.
Question. Fix an abstract variety / separated integral scheme of finite type over $\mathbb{K}$. Does completeness in the sense of Weil correspond to properness in the sense of scheme theory? (Is this a one-way implication?)
A believe a direct translation of Weil's condition is something like this. Consider the closed subscheme $T \subseteq V_1 \times \cdots \times V_h$ defined by the following formula:
$$T = \bigcap_{\alpha, \beta} \{ (P_1, \ldots, P_h) : (P_\beta, P_\alpha) \in T_{\beta, \alpha} \}$$
Here, we are identifying the birational correspondence $T_{\beta, \alpha}$ with its graph, which is a closed subvariety of $V_\beta \times V_\alpha$. By definition we have a closed embedding $V_\alpha \hookrightarrow \mathbb{A}^{n_\alpha}_\mathbb{K}$, so these induce a closed embedding $T \hookrightarrow \mathbb{A}^n_\mathbb{K}$, where $n = n_1 + \cdots + n_h$. On the other hand, $\mathbb{A}^n_\mathbb{K}$ is a dense open subvariety of $(\mathbb{P}^1_\mathbb{K})^n$. Let $\bar{T}$ be the closure of $T$ in $(\mathbb{P}^1_\mathbb{K})^n$. I think Weil is asking that every (closed) point of $\bar{T}$ have the property that there is at least one $\alpha$ such that its projection lies in $U_\alpha$.
Notice that $\bar{T}$ is birationally equivalent to the abstract variety in question and is (by construction) a projective variety (since it is a closed subvariety of $(\mathbb{P}^1_\mathbb{K})^n$, which is a projective variety by the Segre embedding). Weil's condition would give us a surjective rational map from $\bar{T}$ to the abstract variety under discussion, I think.
