Third Chern class of a sheaf on a threefold

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Let $X$ be a smooth projective threefold. Let $D$ be a smooth irreducible divisor on $X$ and $Z$ be an irreducible smooth divisor on $D$. We have inclusions: $$ j:Z\hookrightarrow D\text{ and } i :D\hookrightarrow X.$$

Consider the sheaf $M=i_*j_*O_Z$ on $X$. I am trying to compute the Chern classes of $M$. I have obtained the following:

$c_1(M)=0$ since $Z$ is of codimension 2 in $X$.

$c_2(M)=-[Z]$.

Is this correct? What about $c_3$. Is it zero? I am not able to justify that.

Also if $L$ is a line bundle on $D$, what are the Chern classes of $N=i_*L$. I know that $c_1(N)=[D]$. What are $c_2$ and $c_3$?

Thanks in advance!

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If $Z=E\cap D$, wuth $E,D$ divisors on $X$, we have an exact sequence, $0\to O_X(-D-E)\to O_X(-D)\oplus O_X(-E)\to O_X\to O_Z\to 0$. So, Whitney sum formula says, $c(O_Z)c(O_X(-D)\oplus c(O_X(-E))=c(O_X)c(O_X(-D-E))$, where $c$ is the total chern class. If we denote by $a=c_1(O_X(D))$, $a_2=c_1(O_X(E))$, we get, $c(O_Z)(1-a)(1-b)=1-(a+b)$. From this, you can calculate all the chern classes of $O_Z$ easily by elementary algebra. We have $c(O_Z)=(1-(a+b))(1-a)^{-1}(1-b)^{-1}= (1-(a+b))(1+a+a^2+a^3)(1+b+b+b^2+b^3)$. Thus $c_1(O_Z)=0$ as you said. $c_2=a^2+ab+b^2-(a+b)^2=-ab$, again, this is the class of $-[Z]$ as you said. Finally, $c_3=a^3+a^2b+ab^2+b^3-(a+b)(a^2+ab+b^2)=-ba^2-ab^2=-(a+b)ab$.

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One can use Grothendieck--Riemann--Roch for such computations. Let us, for instance, compute the Chern classes of $i_*L$, where $L$ is a line bundle on the divisor $D \subset X$. GRR says: $$ ch(i_*(L)) = i_*(ch(L)td(X/D)^{-1}), $$ where $ch$ is the Chern character and $td$ is the Todd class. Denote $C := c_1(L)$. Then $$ ch(L) = 1 + C + C^2/2, $$ $$ td(X/D) = (1 + D/2 + D^2/12)^{-1} = 1 - D/2 + D^2/6. $$ Altogether, $$ ch(i_*L) = i_*\Big((1 + C + C^2/2)(1 - D/2 + D^2/6)\Big) = i_*(1 + (C - D/2) + (C^2/2 - CD/2 + D^2/6)). $$ Thus $$ ch_1(i_*L) = D, \qquad ch_2(i_*L) = i_*(C) - D^2/2, \qquad ch_3(i_*L) = i_*(C^2/2 - CD/2) + D^3/6. $$ Computing the Chern classes, one gets $$ c_1(i_*L) = D, \qquad c_2(i_*L) = D^2 - i_*(C), \qquad c_3(i_*L) = D^3 + i_*(C^2 - 2CD). $$ Hope that helps.