Proposition 33.31.1. Let X be a scheme. Let a : X \to \mathop{\mathrm{Spec}}(k_1) and b : X \to \mathop{\mathrm{Spec}}(k_2) be morphisms from X to spectra of fields. Assume a, b are locally of finite type, and X is reduced, and connected. Then we have k_1' = k_2', where k_ i' \subset \Gamma (X, \mathcal{O}_ X) is the integral closure of k_ i in \Gamma (X, \mathcal{O}_ X).
33.31 Uniqueness of base field
The phrase “let X be a scheme over k” means that X is a scheme which comes equipped with a morphism X \to \mathop{\mathrm{Spec}}(k). Now we can ask whether the field k is uniquely determined by the scheme X. Of course this is not the case, since for example \mathbf{A}^1_{\mathbf{C}} which we ordinarily consider as a scheme over the field \mathbf{C} of complex numbers, could also be considered as a scheme over \mathbf{Q}. But what if we ask that the morphism X \to \mathop{\mathrm{Spec}}(k) does not factor as X \to \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k) for any nontrivial field extension k'/k? In other words we ask that k is somehow maximal such that X lives over k.
An example to show that this still does not guarantee uniqueness of k is the scheme
X = \mathop{\mathrm{Spec}}\left( \mathbf{Q}(x)[y]\left[\frac{1}{P(y)}, P \in \mathbf{Q}[y], P \not= 0\right] \right)
At first sight this seems to be a scheme over \mathbf{Q}(x), but on a second look it is clear that it is also a scheme over \mathbf{Q}(y). Moreover, the fields \mathbf{Q}(x) and \mathbf{Q}(y) are subfields of R = \Gamma (X, \mathcal{O}_ X) which are maximal among the subfields of R (details omitted). In particular, both \mathbf{Q}(x) and \mathbf{Q}(y) are maximal in the sense above. Note that both morphisms X \to \mathop{\mathrm{Spec}}(\mathbf{Q}(x)) and X \to \mathop{\mathrm{Spec}}(\mathbf{Q}(y)) are “essentially of finite type” (i.e., the corresponding ring map is essentially of finite type). Hence X is a Noetherian scheme of finite dimension, i.e., it is not completely pathological.
Another issue that can prevent uniqueness is that the scheme X may be nonreduced. In that case there can be many different morphisms from X to the spectrum of a given field. As an explicit example consider the dual numbers D = \mathbf{C}[y]/(y^2) = \mathbf{C} \oplus \epsilon \mathbf{C}. Given any derivation \theta : \mathbf{C} \to \mathbf{C} over \mathbf{Q} we get a ring map
\mathbf{C} \longrightarrow D, \quad c \longmapsto c + \epsilon \theta (c).
The subfield of \mathbf{C} on which all of these maps are the same is the algebraic closure of \mathbf{Q}. This means that taking the intersection of all the fields that X can live over may end up being a very small field if X is nonreduced.
One observation in this regard is the following: given a field k and two subfields k_1, k_2 of k such that k is finite over k_1 and over k_2, then in general it is not the case that k is finite over k_1 \cap k_2. An example is the field k = \mathbf{Q}(t) and its subfields k_1 = \mathbf{Q}(t^2) and \mathbf{Q}((t + 1)^2). Namely we have k_1 \cap k_2 = \mathbf{Q} in this case. So in the following we have to be careful when taking intersections of fields.
Having said all of this we now show that if X is locally of finite type over a field, then some uniqueness holds. Here is the precise result.
Proof. First, assume the lemma holds in case X is quasi-compact (we will do the quasi-compact case below). As X is locally of finite type over a field, it is locally Noetherian, see Morphisms, Lemma 29.15.6. In particular this means that it is locally connected, connected components of open subsets are open, and intersections of quasi-compact opens are quasi-compact, see Properties, Lemma 28.5.5, Topology, Lemma 5.7.11, Topology, Section 5.9, and Topology, Lemma 5.16.1. Pick an open covering X = \bigcup _{i \in I} U_ i such that each U_ i is quasi-compact and connected. For each i let K_ i \subset \mathcal{O}_ X(U_ i) be the integral closure of k_1 and of k_2. For each pair i, j \in I we decompose
U_ i \cap U_ j = \coprod U_{i, j, l}
into its finitely many connected components. Write K_{i, j, l} \subset \mathcal{O}(U_{i, j, l}) for the integral closure of k_1 and of k_2. By Lemma 33.28.4 the rings K_ i and K_{i, j, l} are fields. Now we claim that k_1' and k_2' both equal the kernel of the map
\prod K_ i \longrightarrow \prod K_{i, j, l}, \quad (x_ i)_ i \longmapsto x_ i|_{U_{i, j, l}} - x_ j|_{U_{i, j, l}}
which proves what we want. Namely, it is clear that k_1' is contained in this kernel. On the other hand, suppose that (x_ i)_ i is in the kernel. By the sheaf condition (x_ i)_ i corresponds to f \in \mathcal{O}(X). Pick some i_0 \in I and let P(T) \in k_1[T] be a monic polynomial with P(x_{i_0}) = 0. Then we claim that P(f) = 0 which proves that f \in k_1. To prove this we have to show that P(x_ i) = 0 for all i. Pick i \in I. As X is connected there exists a sequence i_0, i_1, \ldots , i_ n = i \in I such that U_{i_ t} \cap U_{i_{t + 1}} \not= \emptyset . Now this means that for each t there exists an l_ t such that x_{i_ t} and x_{i_{t + 1}} map to the same element of the field K_{i, j, l}. Hence if P(x_{i_ t}) = 0, then P(x_{i_{t + 1}}) = 0. By induction, starting with P(x_{i_0}) = 0 we deduce that P(x_ i) = 0 as desired.
To finish the proof of the lemma we prove the lemma under the additional hypothesis that X is quasi-compact. By Lemma 33.28.4 after replacing k_ i by k_ i' we may assume that k_ i is integrally closed in \Gamma (X, \mathcal{O}_ X). This implies that \mathcal{O}(X)^*/k_ i^* is a finitely generated abelian group, see Proposition 33.28.5. Let k_{12} = k_1 \cap k_2 as a subring of \mathcal{O}(X). Note that k_{12} is a field. Since
k_1^*/k_{12}^* \longrightarrow \mathcal{O}(X)^*/k_2^*
we see that k_1^*/k_{12}^* is a finitely generated abelian group as well. Hence there exist \alpha _1, \ldots , \alpha _ n \in k_1^* such that every element \lambda \in k_1 has the form
\lambda = c \alpha _1^{e_1} \ldots \alpha _ n^{e_ n}
for some e_ i \in \mathbf{Z} and c \in k_{12}. In particular, the ring map
k_{12}[x_1, \ldots , x_ n, \frac{1}{x_1 \ldots x_ n}] \longrightarrow k_1, \quad x_ i \longmapsto \alpha _ i
is surjective. By the Hilbert Nullstellensatz, Algebra, Theorem 10.34.1 we conclude that k_1 is a finite extension of k_{12}. In the same way we conclude that k_2 is a finite extension of k_{12}. In particular both k_1 and k_2 are contained in the integral closure k_{12}' of k_{12} in \Gamma (X, \mathcal{O}_ X). But since k_{12}' is a field by Lemma 33.28.4 and since we chose k_ i to be integrally closed in \Gamma (X, \mathcal{O}_ X) we conclude that k_1 = k_{12} = k_2 as desired. \square
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