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    Curves of genus $0$

    PR-VGA2L

    A genus-zero curve with a rational point is $\PP^1$; without one, the anticanonical system presents it as a smooth conic.Suggest a slogan

    Let \(C\) be a smooth projective curve of genus \(0\) over a field \(k\).

    • If \(C\) has a \(k\)-rational point, then \(C \cong \PP^1_k\).

    • In general \(C\) embeds as a smooth conic in \(\PP^2_k\) by the anticanonical system \(\abs{-K}\), which has degree \(2\) and \(\ell(-K) = 3\).

    • Over \(k = \bar{k}\), and over any finite field, the first case always holds.

    Riemann–Roch does the work. With \(g = 0\) and \(\deg D = 1\), \(\ell(D) = 2\) and \(\ell(K - D) = 0\) since \(\deg(K-D) = -3 < 0\), so a degree-one divisor gives a degree-one map to \(\PP^1\), which is an isomorphism. A \(k\)-point is exactly such a divisor, and that is where the rationality hypothesis enters.

    Without a rational point the anticanonical divisor still has degree \(2\) and \(\ell = 3\), giving the conic. The conic \(x^2 + y^2 + z^2 = 0\) over \(\RR\) is the standard curve of genus zero that is not \(\PP^1\).

    Over a finite field the second case cannot occur: a smooth conic over \(\FF_q\) always has a rational point by the Chevalley–Warning theorem, so every genus-zero curve there is \(\PP^1\). Thus over a finite field every smooth projective genus-zero curve has a rational point and is isomorphic to \(\PP^1\).

    What can you say about curves of genus $0$?
    Is every genus-zero curve isomorphic to $\PP^1$?

    Dependencies

    • T-MWDVLRiemann--Roch and Serre duality on a curveuses

    Backlinks

    • D-VARSEVBRAUERSeveri--Brauer varieties, and the Brauer class of a twisted projective spacerelated-to

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