MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a projective smooth variety $X$ of dimension $n$. Pick a very ample divisor $D \in Div(X)$, which induces an inclusion $\phi_D : X \rightarrow \mathbb{P}^N $. Then we can define the degree of any $m$ dimensional irreducible subvariety $Y \subset X $ by $$\text{deg}_D(Y):=\text{deg}({\phi_D}_*(\langle Y\rangle ) \cdot H^m )$$ where $H$ is any hyperplane in $\mathbb{P}^n$ and $\langle Y\rangle$ is the $m$-cycle class of $Y$. We can extend this linearly to the Chow's group of $X$. My question is : when the $n-1$ dimension Chow's group $C_{n-1}(X) = Pic(X) \cong \mathbb{Z}/n\mathbb{Z} $, then the degree map $\text{deg}_D : \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}$ is the zero map, which means all $n-1$ dimensional subvariety of $X$ has degree zero. This is weird and counter intuitive, can someone shed some light? (When can $Pic(X) \cong \mathbb{Z}/n\mathbb{Z} $ and what does it mean geometrically) Thanks.

share|cite|improve this question
$\mathrm{Pic}(X)$ of a projective variety cannot be $\mathbb{Z/nZ}$ exactly for the reason you find it weird. – Sándor Kovács Jun 11 '13 at 18:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.