Segre classes of subvarieties I am following 13.2 of "3264 and All That".
I read the definition of "Segre classes of subvarieties" but I couldn't calculate with describing specific examples.
The definition is as following.
Let $C \subset X$ be a subvariety cut out by ideal sheaf $\cal I$,
$\cal S$ be the sheaf of graded ring $\oplus {\cal I}^n/{\cal I}^{n+1}$ and $\pi: E = \mathbb{P}{\cal S} \to C$ be the projectivization,
and $\zeta$ be the line bundle on $E$ associated to ${\cal S}(1)$.
Then the Segre class of $C$ in $X$ is defined as $s_k(C,X) = \pi_{*}(c_1(\zeta^k))$.
I want to calculate this in two cases with $X=\mathbb{P}^3 = {\rm Proj}\ k[x,y,z,w]$.
One is $C \simeq {\rm Proj}\ k[z,w] $ cut out by ${\cal I}=(x,y)$,
another is $C'$ cut out by ${\cal I} '=(x^2,y^2)$.
I considered affines $U_1 = {\rm Spec}\ k[x/w,y/w,z/w]$ and $U_2 = {\rm Spec}\ k[x/z,y/z,w/z]$ in $X$.
Then $C$ is covered by $V_1 = C \cap U_1 \simeq {\rm Spec}\ k[z/w]$ and $V_2 = C \cap U_2 \simeq {\rm Spec}\ k[w/z]$.
${\cal I}(V_1)$ is a module generated by $(x/w,y/w)$ over $k[x/w,y/w,z/w]$.
${\cal S}(V_1)$ is a graded ring generated by $a_1 = x/w \bmod {\cal I}(V_1)^2, b_1=y/w \bmod {\cal I}(V_1)^2$ over $k[z/w]$.
Similarly,
${\cal I}(V_2)$ is a module generated by $(x/z,y/z)$ over $k[x/z,y/z,w/z]$,
${\cal S}(V_2)$ is a graded ring generated by $a_2 = x/w \bmod {\cal I}(V_2)^2, b_2=y/w \bmod {\cal I}(V_2)^2$ over $k[w/z]$.
So I described E as covered by 4 affines:
$V_{1a} ={\rm Spec}\ [z/w,a_1/b_1], V_{1b} = {\rm Spec}\ [z/w,b_1/a_1]$,
$V_{2a} ={\rm Spec}\ [w/z,a_2/b_2], V_{2b} = {\rm Spec}\ [w/z,b_2/a_2]$
Take rational section of $\zeta$ is informally of the form $Aa_1+Bb_1$, that is,
$Aa_1/b_1+B$ on $V_{1a}$,
$A+Bb_1/a_1$ on $V_{1b}$,
$Aa_2/b_2+B$ on $V_{2a}$,
$A+Bb_2/a_2$ on $V_{2b}$,
that vanishes in $a_1/b_1 = -B/A$, a line.
So $c_1(\zeta)$ is the class in $E$ corresponding this line, but I don't see how to pushforward this class to $C$.

In the another case with $C'$, doing similarly, $E'$ is covered by 
$V'_{1a} = {\rm Spec}\ k[x/w,y/w,z/w,a_1/b_1]/(x^2/w^2,y^2/w^2)$,
$V'_{1b} = {\rm Spec}\ k[x/w,y/w,z/w,b_1/a_1]/(x^2/w^2,y^2/w^2)$,
$V'_{2a} = {\rm Spec}\ k[x/z,y/z,w/z,a_2/b_2]/(x^2/z^2,y^2/z^2)$,
$V'_{2b} = {\rm Spec}\ k[x/z,y/z,w/z,b_2/a_2]/(x^2/z^2,y^2/z^2)$,
I thought $c_1(\zeta)$ is again the class in $E'$ corresponding a line $a_1/b_1=const.$ in $V'_{1a}$, but I don't see how to pushforward it.
(From the book I know I can use conormal sheaf in the first case and the answer is $1-2\alpha$ where $\alpha$ is the class of a point in $C$, but it is not available in the second case.)

EDIT: I found an argument with non-reduced case in 13.3.5, but I could not follow the argument. There, $S$ is 2 dimensional variety parametrized by $[aa:bb:cc:ab:bc:ca]$ in $\mathbb{P}^5$
, $\cal I$ is the ideal sheaf that cut out $S$,
and T is non-reduced subvariety cut out by ${\cal I}^2$.
Then there says: Blow-up along $T$ is the same as the blow-up along $S$, but with the exceptional divisor doubled, so there is the relation $s_k(T,\mathbb{P}^5) = 2^{k+3}s_k(S,\mathbb{P}^5)$, but I could not find the reason.
Why isn't it simply $s_k(T,\mathbb{P}^5) = 2s_k(S,\mathbb{P}^5)$?
 A: I reviewed the definition of pushforward and I noticed I misunderstood the definition.
While $s(C,X) = \sum_{k\geq 0} π_*(\zeta^k)$, $\pi_*$ preserves dimension but not codimension, so in this case with $\dim(C)=1, \dim(E)=2$, the relation is actually $s_k(C,X) = π_*(\zeta^{k+1})$.
Thus, $\pi_* (ζ) = 1$, $\pi_*( ζ^2) = -2\alpha$.
The latter is not trivial. One way is using the blow-up calculated as following:
Let $j:E\to B$ be blow up of $i:C \to X$.
Let $[E], [H_C], [H]$ be respectively the cycle of the exceptional divisor, proper transform of plane including C, proper transform not including C. For example considering rational function $x/z$, I see rational relation $[E]+[H_C]=[H]$.
Then for example using, Proposition 13.12, 
$j_*(1) = [E],$ 
$j_*(ζ) = -j_*(1)\cdot j_*(1) = -[E]\cdot[E],$
$j_*(ζ^2) = -j_*(1)\cdot j_*(\zeta) = [E]\cdot[E]\cdot[E] $
Using the relation $[H]\cdot[H]\cdot[H] = [H]\cdot[H]\cdot[H_C] = 1$ point, $[H_C]\cdot [H_C]=0$, I got $j_*(ζ^2) = -2$ points.

For the case in last EDIT, since $\dim S=2, \dim\mathbb{P}^5=5$, the relation should be $s_k(C,X) = π_*(\zeta^{k+3})$, that is where additional factor $2^3$ comes from.
EDIT: I still had mistake. In this case, the dimension of corresponding exceptional divisor is $4$, so the relation is $s_k(C,X) = π_*(\zeta^{k+2})$. Another factor $2$ seems to come from the "doubleness" itself of $T$.
