# Curves of arithmetic genus $1$

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?

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.

• Thank you! From $p_a(X)+h^0(F)=p_a(C)=1$, one already see $h^0(F)=0,1$. So the point is how to see ${\rm length}( F) =1$, implies it is either a node or cusp. My definition of node is it is analytically defined by $y^2=x^2$, and cusp $y^2=x^3$. So do you need something like classification of curve singularities to see that is the only possibilities? Commented Jan 1, 2018 at 3:17

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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 $\mathbb{C}[[x,y]/(y^2 - x^3)$.

• Thank you! This makes sense to me! Commented Jan 2, 2018 at 1:40
• I thought your argument again, how do you know the square is cartesian (i.e. $\hat{\mathcal{O}}_{C,x}$ is the maximal subring satisfies the commutative property)? For example, in the later case, why $\hat{\mathcal{O}}_{C,x}$ cannot just be $\mathbb C[[t^2]]$? Commented Jan 2, 2018 at 2:20
• This is proved here: mathoverflow.net/a/186650/12402. The point is that in both these cases $k(x)$ and $\mathcal{O}_{f^{-1}(x)}$ are $\widehat{\mathcal{O}}_{C}/I$ and $\widehat{\mathcal{O}}_X/I$ respectively where $I \subset \widehat{\mathcal{O}}_C \subset \widehat{\mathcal{O}}_X$ is the conductor ideal of the normalization. Commented Jan 2, 2018 at 4:23
• Sorry for keep asking. I feel the idea behind your argument is: one can use the residue fields of the points on $X$ and $C$, together with the completion on $X$ to get the completion of $C$. But my feeling is that this may not be enough as the residue fields do not carry much information. Hence what is crucial in your argument is the cartesian square, or in Karl Schwede's term, $\hat{\mathcal{O}_{C,x}}$ is a push-out of the diagram. But how do I know, a priori, it is a push-out? This is my question. Commented Jan 2, 2018 at 14:50

I include an answer here for the convenience of others:

Claim: Suppose that $$X$$ is a smooth projective surface with $$C \subseteq X$$ an irreducible divisor. If $$p_a(C)=1$$, then $$C$$ can be a smooth elliptic curve, a rational curve with a node and a rational curve with a cusp.

Proof: After a sequence of blowups of $$X$$, let $$\tilde C \to C$$ be a resolution of $$C$$ where $$\tilde C$$ is the strictly transform of $$C$$ on the blowup. By [Hartshorne $$\S$$ V Example 3.9.2], $$p_a(\tilde C) = p_a(C) - \sum_{p} \frac{1}{2}r_p(r_p-1),$$ where $$p$$ is a center of a blowup and $$r_p$$ is the multiplicity (of $$C$$) at $$p$$ (see Page 388 for definition). Because $$p_a(C)=1$$, if $$C$$ is not smooth, then $$p_a(\tilde C) <1$$. Hence $$p_a(\tilde C) =0$$, and there exists a unique singular point $$p \in C$$ with multiplicity $$r_p=2$$ which can be resolved by a single blowup.

Because $$X$$ is smooth, choose a local coordinate $$x,y$$, and by the definition of multiplicity, $$C$$ has local equation $$f=f_2(x,y)+g$$ with $$\deg f_2=2, \deg g>2$$. After an analytic change of variables, $$f_2(x,y)+g$$ becomes $$f=y^2-x^n$$ for $$n\geq 2$$. Now I follow the notation of [Hartshorne $$\S$$ V Proposition 3.6]. After a blowup $$\pi$$ of $$p=(0,0)$$, in an affine space, $$\pi^*f = x^2(u^2-x^{n-2}),$$ where $$\{x=0\}$$ defines the exceptional divisor and $$u^2-x^{n-2}=0$$ defines the strict transform of $$C$$. If $$n>3$$, then $$\{u^2-x^{n-2}=0\}$$ is a singular curve, this contradicts to the fact that only one blowup is needed to resolve $$C$$. Hence $$n=2,3$$ and $$p_a(\tilde C)=0$$.