## 42.8 Cycles

Since we are not assuming our schemes are quasi-compact we have to be a little careful when defining cycles. We have to allow infinite sums because a rational function may have infinitely many poles for example. In any case, if $X$ is quasi-compact then a cycle is a finite sum as usual.

Definition 42.8.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $k \in \mathbf{Z}$.

A *cycle on $X$* is a formal sum

\[ \alpha = \sum n_ Z [Z] \]

where the sum is over integral closed subschemes $Z \subset X$, each $n_ Z \in \mathbf{Z}$, and the collection $\{ Z; n_ Z \not= 0\} $ is locally finite (Topology, Definition 5.28.4).

A *$k$-cycle* on $X$ is a cycle

\[ \alpha = \sum n_ Z [Z] \]

where $n_ Z \not= 0 \Rightarrow \dim _\delta (Z) = k$.

The abelian group of all $k$-cycles on $X$ is denoted $Z_ k(X)$.

In other words, a $k$-cycle on $X$ is a locally finite formal $\mathbf{Z}$-linear combination of integral closed subschemes of $\delta $-dimension $k$. Addition of $k$-cycles $\alpha = \sum n_ Z[Z]$ and $\beta = \sum m_ Z[Z]$ is given by

\[ \alpha + \beta = \sum (n_ Z + m_ Z)[Z], \]

i.e., by adding the coefficients.

Definition 42.8.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. The *support* of a cycle $\alpha = \sum n_ Z [Z]$ on $X$ is

\[ \text{Supp}(\alpha ) = \bigcup \nolimits _{n_ Z \not= 0} Z \subset X \]

Since the collection $\{ Z; n_ Z \not= 0\} $ is locally finite we see that $\text{Supp}(\alpha )$ is a closed subset of $X$. If $\alpha $ is a $k$-cycle, then every irreducible component $Z$ of $\text{Supp}(\alpha )$ has $\delta $-dimension $k$.

Definition 42.8.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. A cycle $\alpha $ on $X$ is *effective* if it can be written as $\alpha =\sum n_ Z [Z]$ with $n_ Z \geq 0$ for all $Z$.

The set of all effective cycles is a monoid because the sum of two effective cycles is effective, but it is not a group (unless $X = \emptyset $).

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