The degree of a morphism of schemes

Question

Let $X$ and $Y$ be schemes over a field $k$, and let $\phi :X \to Y$ be a morphism of schemes.

1. What is the most general situation where one can define the degree of $f$? Is $X$ and $Y$ being geometrically integral enough?
2. How is the degree of $f$ affected by base change?

Answer 1

The most general definition I can think of is a finite locally free morphism, that is, a morphism of schemes $f\colon X \to Y$ such that:

• $f$ is an affine morphism, that is, if $U \subseteq Y$ is any affine open subset, then $f^{-1}(U) \subseteq X$ is also affine; and
• $f_* \mathcal{O}_X$ is a finite locally free $\mathcal{O}_Y$-module, meaning that for each $y \in Y$, there is an open neighborhood $U \subseteq Y$ of $y$ such that $f_* \mathcal{O}_X \rvert_U \cong \bigoplus_{i=1}^r \mathcal{O}_Y \rvert_U$ as an $\mathcal{O}_Y \rvert_U$-module for some finite $r$, called the rank or degree of $f$ at $y$.

In this case, $Y$ can be written as a disjoint union of open and closed subschemes on which the rank/degree is constant. So in particular, if $Y$ is connected, then the degree of $f$ is a constant.

To understand how this behaves under base change, we can pass to an affine open patch as above, so the situation is that $A \subseteq B$ is an extension of rings such that $B \cong \bigoplus_{i=1}^r A$ as an $A$-module, and $C$ is an $A$-algebra. Then $$B \otimes_A C \cong \Bigl( \bigoplus_{i=1}^r A \Bigr) \otimes_A C \cong \bigoplus_{i=1}^r (A \otimes_A C) \cong \bigoplus_{i=1}^r C.$$ Thus, $B \otimes_A C$ is a free $C$-module of rank $r$. (Note also that the base change of an affine morphism is affine.) This shows degree is invariant under base change.

A couple caveats:

1. For the notion of the rank of a free module over a ring to be well-defined, the ring must have the invariant basis number property. All nonzero commutative rings have this property, so this is mostly just of interest in noncommutative ring theory, not algebraic geometry. However, the zero ring is an edge case here, corresponding to the fact that if you base change to the empty scheme, you're left with the empty morphism, which does not have a well-defined degree.
2. Some sources (e.g. Qing Liu's Algebraic Geometry and Arithmetic Curves, or this MathOverflow answer) talk about the degree of a dominant finite morphism of integral, locally Noetherian schemes without requiring flatness. For example, the normalization morphism of a nodal cubic is finite of generic degree one, but fails to be flat at the node, as witnessed by the fact that the preimage of the node has two elements, not one. As discussed at the linked answer, the degree in such cases is preserved by flat base change but not by arbitrary base change. I would prefer to describe this notion as the "generic degree" rather than simply the "degree".