Definition 33.3.1. Let $k$ be a field. A variety is a scheme $X$ over $k$ such that $X$ is integral and the structure morphism $X \to \mathop{\mathrm{Spec}}(k)$ is separated and of finite type.
33.3 Varieties
In the Stacks project we will use the following as our definition of a variety.
This definition has the following drawback. Suppose that $k'/k$ is an extension of fields. Suppose that $X$ is a variety over $k$. Then the base change $X_{k'} = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k')$ is not necessarily a variety over $k'$. This phenomenon (in greater generality) will be discussed in detail in the following sections. The product of two varieties need not be a variety (this is really the same phenomenon). Here is an example.
Example 33.3.2. Let $k = \mathbf{Q}$. Let $X = \mathop{\mathrm{Spec}}(\mathbf{Q}(i))$ and $Y = \mathop{\mathrm{Spec}}(\mathbf{Q}(i))$. Then the product $X \times _{\mathop{\mathrm{Spec}}(k)} Y$ of the varieties $X$ and $Y$ is not a variety, since it is reducible. (It is isomorphic to the disjoint union of two copies of $X$.)
If the ground field is algebraically closed however, then the product of varieties is a variety. This follows from the results in the algebra chapter, but there we treat much more general situations. There is also a simple direct proof of it which we present here.
Lemma 33.3.3.slogan Let $k$ be an algebraically closed field. Let $X$, $Y$ be varieties over $k$. Then $X \times _{\mathop{\mathrm{Spec}}(k)} Y$ is a variety over $k$.
Proof. The morphism $X \times _{\mathop{\mathrm{Spec}}(k)} Y \to \mathop{\mathrm{Spec}}(k)$ is of finite type and separated because it is the composition of the morphisms $X \times _{\mathop{\mathrm{Spec}}(k)} Y \to Y \to \mathop{\mathrm{Spec}}(k)$ which are separated and of finite type, see Morphisms, Lemmas 29.15.4 and 29.15.3 and Schemes, Lemma 26.21.12. To finish the proof it suffices to show that $X \times _{\mathop{\mathrm{Spec}}(k)} Y$ is integral. Let $X = \bigcup _{i = 1, \ldots , n} U_ i$, $Y = \bigcup _{j = 1, \ldots , m} V_ j$ be finite affine open coverings. If we can show that each $U_ i \times _{\mathop{\mathrm{Spec}}(k)} V_ j$ is integral, then we are done by Properties, Lemmas 28.3.2, 28.3.3, and 28.3.4. This reduces us to the affine case.
The affine case translates into the following algebra statement: Suppose that $A$, $B$ are integral domains and finitely generated $k$-algebras. Then $A \otimes _ k B$ is an integral domain. To get a contradiction suppose that
\[ (\sum \nolimits _{i = 1, \ldots , n} a_ i \otimes b_ i) (\sum \nolimits _{j = 1, \ldots , m} c_ j \otimes d_ j) = 0 \]
in $A \otimes _ k B$ with both factors nonzero in $A \otimes _ k B$. We may assume that $b_1, \ldots , b_ n$ are $k$-linearly independent in $B$, and that $d_1, \ldots , d_ m$ are $k$-linearly independent in $B$. Of course we may also assume that $a_1$ and $c_1$ are nonzero in $A$. Hence $D(a_1c_1) \subset \mathop{\mathrm{Spec}}(A)$ is nonempty. By the Hilbert Nullstellensatz (Algebra, Theorem 10.34.1) we can find a maximal ideal $\mathfrak m \subset A$ contained in $D(a_1c_1)$ and $A/\mathfrak m = k$ as $k$ is algebraically closed. Denote $\overline{a}_ i, \overline{c}_ j$ the residue classes of $a_ i, c_ j$ in $A/\mathfrak m = k$. The equation above becomes
\[ (\sum \nolimits _{i = 1, \ldots , n} \overline{a}_ i b_ i) (\sum \nolimits _{j = 1, \ldots , m} \overline{c}_ j d_ j) = 0 \]
which is a contradiction with $\mathfrak m \in D(a_1c_1)$, the linear independence of $b_1, \ldots , b_ n$ and $d_1, \ldots , d_ m$, and the fact that $B$ is a domain. $\square$
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