$\newcommand{\oh}{\mathcal{O}} \newcommand{\ch}{\mathrm{ch}} \newcommand{\Hom}{\mathrm{Hom}} \newcommand{\Ext}{\mathrm{Ext}}$Let $i : C \hookrightarrow X$ be a non-smooth conic in a smooth projective threefold $X$. I'm trying to find the Chern character $\ch(I_C)$ where $I_C$ is the ideal of $C$.
I know that for a smooth conic, we have $\ch(I_C)=(1,0,-2L,0)$ where $L$ is the class of a line, essentially by Hirzebruch-Riemann-Roch and being able to compute the cohomology groups $H^i(X, I_C)$ because $C \cong \mathbb{P}^1$.
For a non-smooth conic (either reducible, i.e. union of two transversal lines; or non-reduced, i.e. $C_{\mathrm{red}} = L$ a line) however, I'm not entirely sure what the cohomology groups $H^i(X, I_C)$ are. I've taken cohomology of the short exact sequence $0 \to I_C \to \oh_X \to i_* \oh_C \to 0$ which gives the long exact sequence
$$ \begin{align} 0 & \to H^0(X, I_C) \to \mathbb{C} \to H^0(X, i_* \oh_C) \\
&\to H^1(X, I_C) \to 0 \to H^1(X, i_* \oh_C) \\
&\to H^2(X, I_C) \to 0 \to 0 \to \cdots \end{align} $$
Here I've used $H^i(X, i_* \oh_C) \cong H^i(C, \oh_C)$ and that this cohomology vanishes for $i > \dim(C)=1$ (Grothendieck). I've also used that $H^i(X, \oh_X)$ is $\mathbb{C}$ for $i=0$ and $0$ else.
So my question essentially becomes: how does one compute the cohomologies $H^i(X, i_* \oh_C)$ for $i=0,1$ in the non-smooth case, and how to interpret the resulting sequence $ 0 \to H^0(X, I_C) \to \mathbb{C} \to H^0(X, i_* \oh_C) \to H^1(X, I_C) \to 0$?
Thanks.
Edit: I think I recall reading somewhere that for any conic, $H^1(C, \oh_C)=0$ (but I still don't know how to prove this?). If this is the case, then regardless of what $H^i(X, I_C)$ are for $i=0,1$ (I'd still be interested to know though), they'd have to be of equal dimension (by dimension counting the long exact sequence in cohomology). Then if $\ch(I_C)=(1,0,-2L, w)$, we have $\chi(I_C) = w = h^0(X,I_C)-h^1(X,I_C) = 0$ which would mean that for any conic $C$, $\ch(I_C)=(1,0,-2L,0)$. Is this correct?
Best Answer
To compute the cohomology of $\mathcal{O}_C$ one can use the exact sequence $$ 0 \to \mathcal{O}_C \to \mathcal{O}_{L_1} \oplus \mathcal{O}_{L_2} \to \mathcal{O}_P \to 0 $$ for $C = L_1 \cup L_2$, where $P$ is the intersection point of the lines and the maps are induced by the restriction of functions, and the exact sequence $$ 0 \to \mathcal{O}_L(-1) \to \mathcal{O}_C \to \mathcal{O}_L \to 0 $$ for $C$ being the double line $L$.
Anyway, the Chern character of $C$ does not depend on the type of $C$.