Let $X$ be a smooth projective surface. Let $Z$ be a dimensional $0$ subscheme of length $l$. Suppose $I_Z$ is the ideal sheaf of $Z$.

Then it claimed that $c_1(I_Z) = 0$ and $c_2(I_Z) = l$.

(1)Why this is true?

(Here the Chern class is about a sheaf rather than vector bundles. Hence, we use a resolution of $I_Z$ by vector bundles to define its Chern classes. I want to use $0 \to I_Z \to \mathcal{O}_X \to i_*\mathcal{O}_Z \to 0$ to compute Chern classes. This means that at least I have to show that $c_2(i_*\mathcal{O}_Z) = -l$, this seems very wired to me because $i_*\mathcal{O}_Z$ supported in dimensional $0$.)

(2) Besides, I want to know how to define the length of zero dimensional scheme (may be reduced)?

Edit I found that the post https://mathoverflow.net/questions/106148/chern-classes-of-ideal-sheaf-of-an-analytic-subset related to this problem, but I am still looking for an elementary proof (just use HRR theorem).


1 Answer 1


I just realized that this is a consequence of Grothendieck–Riemann–Roch.

Let $i: Z \to X$ be the closed embedding, by GRR, we have $$ch(i_! \mathcal{O}_Z) td(X) = i_*(ch(\mathcal{O}_Z) td(Z)),$$ where $i_i\mathcal{O}_Z = i_*\mathcal{O}_Z$ because $i$ is an affine morphism and $R^ji_*\mathcal{O}_Z = 0$, for $j >0$.

The left hand side of the formula is $$[0 + c_1(i_*\mathcal{O}_Z) + \frac 1 2 (c_1^2(i_*\mathcal{O}_Z) - 2c_2(i_*\mathcal{O}_Z))][1 + \frac 1 2 c_1(X) + \frac 1 {12} (c_1^2(X)+ c_2(X))],$$ while the right hand side is just $l \cdot 1$.

By comparing the coefficients, we have $c_1(i_*\mathcal{O}_Z) = 1$ and $c_2(i_*\mathcal{O}_Z) = -l$.

Finally, by using $0 \to I_Z \to \mathcal{O}_X \to i_*\mathcal{O}_Z \to 0$, and $$1 = c_t(\mathcal{O}_X) = c_t(I_Z)\cdot c_t(i_*\mathcal{O}_Z),$$ one can find that $c_1(I_Z)=0, c_2(I_Z) = l$.

  • $\begingroup$ Shouldn't it be $c_1(i_*\mathcal{O}_Z)=0$? Also the 4th line should be $i_!\mathcal{O}_Z=i_*\mathcal{O}_Z$. $\endgroup$
    – AG learner
    Feb 28, 2021 at 4:27

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