# Equality of chern classes and isomorphism

Given two torsion free coherent sheaves $M$ and $N$ wit $rk(M)=rk(N)=r$ on an smooth projective surface $S$, by definition $det(M):=\Lambda^r(M)^{\*\*}$.

Is the following criterion correct?

$M\cong N$ $\Leftrightarrow$ $M \hookrightarrow N$ and $c_i(M)=c_i(N)$ for $i=0,1,2$

One only has to look at "$\Leftarrow$":

So we have $0\rightarrow M\rightarrow N\rightarrow Q \rightarrow0$. Because of $c_i(M)=c_i(N)$ we see that $c_1(Q)=0$ and $c_2(Q)=0$. Since $M$ and $N$ have the same rank $codim(supp(Q))\geq 1$. We also have an induced map $det(M)\hookrightarrow det(N)$ of line bundles, i.e. $det(N)\cong det(M)\otimes O_S(D)$ for some effective divisor $D$. Now $det(Q)\cong det(N)\otimes det(M)^{-1}\cong O_S(D)$.

So by definition $c_1(Q)=c_1(det(Q))=D$, but $c_1(Q)=0$, so $D$ is effective and 0, i.e. $Q$ has no support in codimension 1, so $codim(supp(Q))\geq 2$.

So $Q$ is an Artinian sheaf, and for those one has $c_2(Q)=-dim(H^0(S,Q))$. Since $c_2(Q)=0$ we have $H^0(S,Q)=0$, but $H^0(S,Q)=\bigoplus\limits_{s\in supp(Q)} Q_s$. So $Q_s=0$ for all $s\in supp(Q)$, i.e. $Q=0$. So we have $M\cong N$.

-

That's correct. A slightly shorter argument is: if $\mathcal{Q}$ has support in codimension $d$, then its Chern character $\mathrm{ch}_d(\mathcal{Q})$ is non-zero and effective. So a sheaf is trivial if and onlfy if $\mathrm{ch} = 0$, which is true if and only if the rank and the Chern classes vanish. In particular, the result holds for any dimension.
Thanks, that's interesting and more general. What does effective mean in this case, e.g. in the Chow ring? $ch_d(Q)$ is a sum of codimension d subvarieties, and all coefficient are non negative? And how can i see that $ch_d(Q)$ has to be effective in this case? – –  TonyS Oct 29 '10 at 15:21