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. 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. $\square$
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)