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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)