# algebraic curves with negative (arithmetic) genus?

By an algebraic curve I mean a projective reduced connected scheme of pure dimension 1 over a field.

My question is: Is there a lower bound for the arithmetic genus of such curves? If the answer is no, is there a general method to construct such curves with arbitrary negative genus? Any reference is welcome, thank you!

Notice that, as pointed out by Andrea in the comment, when the curve is geometrically connected, the arithmetic genus is always non-negative.

• If $X$ is geometrically connected, then $p_a(X) = \dim_k H^1(X, \mathcal{O}_X) \geq 0$. – Andrea May 28 '12 at 16:30
• @Andrea: Yes, and this is why I didn't add "geometrically"(shouldn't it be "geometrically connected AND GEOMETRICALLY REDUCED"?). – Ch Zh May 28 '12 at 16:39

Let $K / k$ be finite extension of fields of degree $d$. Let $X$ be smooth geometrically connected projective curve over $K$ of genus $g$. The morphism $\mathrm{Spec} K \to \mathrm{Spec} k$ is projective (Exercise 3.3.22 of Liu's Algebraic geometry and arithmetic curves), therefore $X$ is projective over $k$. Then $$p_{a,k}(X) = 1- \chi_k(X) = 1 - d \cdot \chi_K(X) = 1 - d (1-g).$$ You can take $g = 0$.
EDIT. More concretely, let $k$ be a field, let $f \in k[t]$ be a monic irreducible polynomial of degree $d$, let $\alpha \in \bar{k}$ be a root of $f$ and let $K = k(\alpha)$. If $f(t) = \sum_{i=0}^d a_i t^i$, then $\mathrm{Spec} K$ is isomorphic to $$\mathrm{Proj} k[x_0,x_1] / (\sum_{i=0}^d a_i x_0^i x_1^{d-i} ).$$ Then $$X = \mathbb{P}^1_K = \mathbb{P}^1_k \times_k \mathrm{Spec} K = \mathbb{P}^1_k \times_k \mathrm{Proj} k[x_0,x_1] / (\sum_{i=0}^d a_i x_0^i x_1^{d-i} )$$ is a closed subscheme of $\mathbb{P}^1_k \times_k \mathbb{P}^1_k$, hence it is a closed subscheme of $\mathbb{P}^3_k$ by Segre embedding, i.e. $$X = \mathrm{Proj} k[z_0, z_1, z_2, z_3] / (z_0 z_3 - z_1 z_2, \sum_{i=0}^d a_i z_0^i z_1^{d-i}, \sum_{i=0}^d a_i z_2^i z_3^{d-i}).$$
• Many thanks for this nice answer! But I am a bit confused: Let $X_k$ be the $X$ over $k$. Then, as $X$ is smooth and geometrically connected (i.e. $X_{\bar{k}}$ is regular and connected), so is $X_k$, so we should have $P_{a,k}(X_k)\geq0$, where is wrong? – Ch Zh May 28 '12 at 17:59
• $X$ is geometrically connected over $K$, but is not geometrically connected over $k$. Moreover $X \neq X \times_k K$. – Andrea May 28 '12 at 18:01