Curves of arithmetic genus $1$

Question

If $C$ is a projective integral curve over $\mathbb C$, suppose its arithmetic genus $p_a(C)=1$, then what kind of curve $C$ could be?

According to Hartshorn Ex. 1.8 page 298, if $\tilde C \to C$ is the normalization, then $p_a(\tilde C)+\delta = p_a(C)=1$, hence either $C$ is an elliptic curve or $\tilde C = \mathbb P^1$ and $\delta =1$. Ex. 1.8 (c) claims that node or cusp will have $\delta=1$.

It seems that these are the only possibilities, but why?

Answer 1

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Expanding on Mohan's answer, let us suppose that that we are in the $\operatorname{length}(F) = 1$ case. Then $F = \mathcal{O}_{C,x}/\mathfrak{m}_x = k(x)$. If $f^{-1}(x) = P + Q$ with $P \neq Q$ then after localizing and completing we have an inclusion $0 \to \widehat{\mathcal{O}}_{C,x} \to \widehat{\mathcal{O}}_{X,P} \oplus \widehat{\mathcal{O}}_{X,Q} \cong \mathbb{C}[[ t_1]]\oplus \mathbb{C} [[t_2 ]]$ and a commutative diagram $$ \begin{CD} \widehat{\mathcal{O}}_{C,x} @>>> \mathbb{C}[[ t_1]]\oplus \mathbb{C} [[t_2 ]]\\ @VVV @VVV\\ k(x) @>>> k(P) \oplus k(Q) \end{CD} $$ where the bottom horizontal map is the diagonal map $\mathbb{C} \to \mathbb{C}^2$. This is a fiber product of $\mathbb{C}$-algebras from which one can see that $\widehat{\mathcal{O}}_{C,x} = \mathbb{C} \oplus t_1 \mathbb{C}[[t_1]] \oplus t_2 \mathbb{C}[[t_2]] \subset \mathbb{C}[[t_1]] \oplus \mathbb{C}[[t_2]]$ which is isomorphic to $\mathbb{C}[[x,y]]/(x^2 - y^2)$.

On the other hand, if $P = Q$ so that $f^{-1}(x) = 2P$ then we have $0 \to \widehat{\mathcal{O}}_{C,x} \to \widehat{\mathcal{O}}_{X,P} \cong \mathbb{C}[[t]]$, $\mathcal{O}_{f^{-1}(x)} \cong \mathbb{C}[[t]]/(t^2)$ and there is a cartesian square $$ \begin{CD} \widehat{\mathcal{O}}_{C,x} @>>> \mathbb{C}[[ t]]\\ @VVV @VVV\\ k(x) @>>> \mathcal{O}_{f^{-1}(x)} \end{CD} $$ where the bottom map is the inclusion $\mathbb{C} \to \mathbb{C}[[t]]/(t^2)$. Then $\mathcal{O}_{C,x} = \mathbb{C}[[t^2, t^3]]$ as a subring of $\mathbb{C}[[t]]$ which is isomorphic to $\mathbb{C}[[x,y]/(y^2 - x^3)$.

Answer 2

Let $f:X\to C$ be the normalization (easier to write $X$ than $\tilde{C}$). Then we have an exact sequence, $0\to \mathcal{O}_C\to f_*\mathcal{O}_X\to F\to 0$, where $F$ is a skyscraper sheaf. Taking cohomologies, we have $0\to H^0(F)\to H^1(\mathcal{O}_C)\to H^1(\mathcal{O}_X)\to 0$. Since $p_a(C)=1$, we have only two possibilities for $H^1(\mathcal{O}_X)$, namely it has dimension one or zero. If it is one, we get $H^0(F)=0$ and then $F=0$, showing that $f$ is an isomorphism. If it is zero, then $X=\mathbb{P}^1$ and $F=k(x)$ for some point $x\in C$. This says, $f$ is an isomorphism outside $x$ and $f^{-1}(x)$ is a length 2 effective divisor. If it is $P+Q$, $P\neq Q$, we get a node and if $P=Q$, we get a cusp.