Let $X$ be a projective scheme over a field $k$. Let $\mathcal{O}(1)$ be an ample line bundle on $X$, then the Hilbert polynomial $P(E)$ is given by $m\mapsto\chi(E ⊗ O(m))$. The explicit polynomial form is given by the following result

$\textbf{Lemma}$: Let $E$ be a coherent sheaf of dimension $d$ and let $H_1 , . . . , H_d \in |\mathcal{O}(1)|$ be an $E$-regular sequence. Then $P(E, m) = \chi(E ⊗ \mathcal{O}(m)) =\Sigma_{i=0}^{d}\chi(E|_{\cap_{j\leq i }H_j}){m+i-1\choose i}$.

So this polynomial can be uniquely written in the form $\Sigma_{i=0}^d\alpha_i(E)\frac{m^i}{i!}$.

We then define the rank as $rk(E)=\frac{\alpha_d(E)}{\alpha_d(\mathcal{O}_X)}$. But for an integral scheme, the rank is defined to be the rank at the generic point. How are these two notions the same?

Similarly, the degree of $E$ is defined to be $deg(E)=\alpha_{d-1}(E)-rk(E).\alpha_{d-1}(\mathcal{O}_X)$. Again, for a projective variety, the degree is defined to be $c_1(E).H^{d-1}$. How do we show that these two are the same. Any help will be greatly appreciated!

  • $\begingroup$ See also this answer for a more down-to-earth approach, at least showing that the two notions of rank agree. $\endgroup$
    – Remy
    Nov 5, 2016 at 13:52

1 Answer 1


Use the Hirzebruch-Riemann-Roch Theorem. $$ \chi(O(m)) = \int \exp(mH) \cdot td(X) = m^dH^d/d! + m^{d-1}H^{d-1}td_1/(d-1)! + \cdots $$ so we see that $a_d(O) = H^d$ and $a_{d-1}(O) = H^{d-1}td_1$.

Next \begin{align*} \chi(E(m)) &= \int ch(E) \cdot \exp(mH) \cdot td(X) \\ &= rm^dH^d/d! + c_1(E)H^{d-1}m^{d-1}/(d-1)! + rm^{d-1}H^{d-1}td_1/(d-1)! + \cdots \end{align*} so we see that $a_d(E) = rH^d$ and $a_{d-1} = c_1(E)H^{d-1} + r\cdot td_1 H^{d-1}$.

If you take for granted that the rank as you defined is equal to the zeroth chern character, everything you want follows.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .