Product and coproduct of schemes

Question

Do product and coproduct exist in the category of schemes? I only saw fiber product but didn't see product and coproduct of schemes, so do they not exist in general?

Actually, I am curious, in the definition of separated morphism/schemes, why we replace the product by fiber product?

Answer 1

You've asked three questions, so here are three answers.

First, for every scheme $X$, there is one and only one morphism of schemes $X \rightarrow \operatorname{Spec}\mathbb Z$. It follows that for two schemes $X$ and $Y$, the fiber product $X \times_{\operatorname{Spec} \mathbb Z} Y$ is the product of $X$ and $Y$ in the category of schemes.

Second, the coproduct $Z$ of any two schemes $X$ and $Y$ does exist in the category of schemes. Topologically, $Z$ is the disjoint union of $X$ and $Y$ with the disjoint union topology. If $W$ is open in $Z$, then the ring $\mathcal O_Z(W)$ is the product ring $\mathcal O_X(W \cap X) \times \mathcal O_Y(W \cap Y)$.

For the final question about why fibre product is used instead of product, I'll give an answer when your schemes are taken over a field $k$. One reason is that the fibre product over $k$ is a much nicer object than the product.

For example, suppose we consider the affine $k$-scheme $X = \operatorname{Spec} k[t]$. The fibre product is $X \times_{\operatorname{Spec} k} X$ is nothing more than $\operatorname{Spec}$ of the polynomial ring $k[x,y]$ in two variables. On the other hand, the product $X \times X$ is the spectrum of the horrible object $k[t] \otimes_{\mathbb Z} k[t]$.

Answer 2

In categories with a terminal object $Z$, the product $X\times Y$ is the fiber product $X \times_{\mathbb Z} Y$. So in the category of schemes, since $\text{Spec }\mathbb{Z}$ is the final object (which follows from the fact that $\mathbb{Z}$ is initial in the category of rings), the product of two schemes $X$ and $Y$ is the fiber product $X \times_{\text{Spec }\mathbb{Z}} Y$.

The reason for this is because the data of maps $W \to X$ and $W\to Y$ is the same as the data of a commutative square as we see in fibered products. This happens because the data of maps $X \to \text{Spec }\mathbb{Z}$ and $Y \to \text{Spec }\mathbb{Z}$ making this diagram commute is "automatic", with both existence of these maps and the commutativity of the resulting diagram coming from the existence and uniqueness of maps to the final object.

In the category of $A$-schemes, where $A$ is a ring, we can easily adapt this, $\text{Spec }A$ is now the final object and so the product in this category will be given by fiber products over $\text{Spec }A.$