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 $k \subset k'$? 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

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

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

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

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

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

for some $e_ i \in \mathbf{Z}$ and $c \in k_{12}$. In particular, the ring map

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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