Diagonal morphism (algebraic geometry)

As an example, consider an algebraic variety over an algebraically closed field k and ${\displaystyle p:X\to \operatorname {Spec} (k)}$  the structure map. Then, identifying X with the set of its k-rational points, ${\displaystyle X\times _{k}X=\{(x,y)\in X\times X\}}$  and ${\displaystyle \delta :X\to X\times _{k}X}$  is given as ${\displaystyle x\mapsto (x,x)}$ ; whence the name diagonal morphism.

A separated morphism is a morphism ${\displaystyle f}$  such that the fiber product of ${\displaystyle f}$  with itself along ${\displaystyle f}$  has its diagonal as a closed subscheme — in other words, the diagonal morphism is a closed immersion.

As a consequence, a scheme ${\displaystyle X}$  is separated when the diagonal of ${\displaystyle X}$  within the scheme product of ${\displaystyle X}$  with itself is a closed immersion. Emphasizing the relative point of view, one might equivalently define a scheme to be separated if the unique morphism ${\displaystyle X\rightarrow {\textrm {Spec}}(\mathbb {Z} )}$  is separated.

Notice that a topological space Y is Hausdorff iff the diagonal embedding

${\displaystyle Y{\stackrel {\Delta }{\longrightarrow }}Y\times Y,\,y\mapsto (y,y)}$ 

is closed. In algebraic geometry, the above formulation is used because a scheme which is a Hausdorff space is necessarily empty or zero-dimensional. The difference between the topological and algebro-geometric context comes from the topological structure of the fiber product (in the category of schemes) ${\displaystyle X\times _{{\textrm {Spec}}(\mathbb {Z} )}X}$ , which is different from the product of topological spaces.

Any affine scheme Spec A is separated, because the diagonal corresponds to the surjective map of rings (hence is a closed immersion of schemes):

${\displaystyle A\otimes _{\mathbb {Z} }A\rightarrow A,a\otimes a'\mapsto a\cdot a'}$ .

Let ${\displaystyle S}$  be a scheme obtained by identifying two affine lines through the identity map except at the origins (see gluing scheme#Examples). It is not separated.^[1] Indeed, the image of the diagonal morphism ${\displaystyle S\to S\times S}$  image has two origins, while its closure contains four origins.

A classic way to define the intersection product of algebraic cycles ${\displaystyle A,B}$  on a smooth variety X is by intersecting (restricting) their cartesian product with (to) the diagonal: precisely,

${\displaystyle A\cdot B=\delta ^{*}(A\times B)}$ 

where ${\displaystyle \delta ^{*}}$  is the pullback along the diagonal embedding ${\displaystyle \delta :X\to X\times X}$ .

• regular embedding
• Diagonal morphism

• Hartshorne, Robin (1977), Algebraic Geometry, Graduate Texts in Mathematics, vol. 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9, MR 0463157