The arithmetic genus of non-reduced curves Let $(X,h)$ be a smooth projective variety, and let $C\subset X$ be a smooth rational curve. Then $C$ has arithmetic genus $0$. (That $p_a(C)=0$ is not important, just to fix ideas).
But if I am interested in other curves in $X$, whose associated $1$-cycle is $dC$ for some $d>1$, how to compute their arithmetic genus? Actually, this problem splits into two questions:


*

*Do all curves $E\subset X$ with $1$-cycle is $dC$ share the same Hilbert polynomial (and therefore the same $p_a$), regardless of the scheme structure? 

*How to compute the arithmetic genus of a curve $E\subset X$ whose cycle is $dC$?


Thanks!
 A: Certainly the arithmetic genus depends on the scheme structure, and not just
the underlying cycle; see the example in Hartshorne of two skew lines in $\mathbb P^3$ coming together in a flat family (and acquiring an embedded point) if you
want an illustration of this.  (Or just compute the arithmetic genus
of the subscheme $XY = Y^2 = 0$ of $\mathbb P^2$, and compare it with arithmetic
genus of the underlying reduced subscheme.)
In particular, if $C$ is not a codimension one cycle (a divisor), then I'm not sure
that this question has that much of an answer, the reason being that
arithmetic genus depends on having an actual subscheme of $X$ (or if you like,
a point of the Hilbert scheme of the ambient variety $X$), whereas a cycle
is less data than that.  
If $X$ is a surface, so that $C$ is a Cartier divisor, then we can interpret
$dC$ as being the $d$th power of this Cartier divisor, and hence get a well-defined
subscheme, which varies in a flat family if $C$ does, and hence for which it
makes sense to discuss the arithmetic genus.  As Cantlog suggests in comments,
the arithmetic genus will then be given by the adjunction formula.  In particular,
the self intersection $C\cdot C$ will play a role.
You can see this by comparing the case of $X$ being $\mathbb P^2$ vs. the case
of $X$ being a smooth quadric surface (i.e. $\mathbb P^1 \times \mathbb P^1$).
In the first case, if we take $C$ to be a line, then $2 C$ is the linear equivalence
class of conics, whose arithmetic genus is again $0$.    On the other hand,
if we take $C$ to be one of the lines in one of the rulings of a quadric (i.e.
$C = \mathbb P^1 \times \text{ a point }$), then $2C$ is linear equivalent to
the disjoint union of two lines in the ruling, and has arithmetic genus equal to $-1$ (assuming I have the sign in the formula for arithmetic genus right).  
The point is that the two lines crossing (with is another member of the linear
equivalence class of $C$ in the first case) has one less point then two disjoint
lines (the crossing point appears just once altogether instead of once on 
each line), thus $\chi(\mathcal O_C)$ is one less in the crossing case than in
the disjoint case, thus $p_a := 1 - \chi(\mathcal O_C)$ is one more, i.e. is
$0$ in the first case rather than $-1$ in the second case.  
From a more formal point of view: the adjunction formula says that $2 p_a - 2  = K_X \cdot C + C \cdot C,$ and
this is not linear in $C$;  e.g. you find that the arithmetic genus $p_a(2 C)$
is equal to $2p_a(C) - 1 + C \cdot C$.
A: I think in general what you desire is not possible. An explanation is furnished by the example of double lines in $\mathbb P^3$ worked out in Eisenbud and Harris's The Geometry of Schemes, p. 136-8. (I think I've heard these referred to as "ribbons".) 
These double lines must all have cycle class $2[C]$, where $C$ is the underlying line $V_+(x-y)$. However, the double line defined by $X_d = \operatorname{Proj}\left(k[x,y,z,w]/ (x^2,xy,y^2,z^dx-w^dy)\right)$, $d\ge 0,$ has arithmetic genus $p_a(X_d) = -d$, showing that any genus $\le 0$ is possible in this case of curves $E$ satisfying $[E] = 2[C]$.
