Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 23I (Algebraic Geometry)

For $Q=\phi (P)$, the local ring of a nonsingular curve at $Q$ is a discrete valuation ring. Choose a uniformizer $t$ there. The ramification index of a morphism of curves isEquivalently, ${\phi }^{\ast }t=u{z}^{e_P}$ for a local uniformizer $z$ at $P$ and a unit in a ring $u$. A different choice $t^{\prime }=vt$ changes its pullback by a unit in a ring, so the index is well-defined. This local definition applies also to inseparable morphisms; conclusions involving separability or tame ramification require those extra hypotheses.

The cubic equation is $x_0{x}_2^2-x_1^3+x_0^2{x}_1=0$. On $x_0\ne 0$, put $u=x_1/x_0$, $v=x_2/x_0$, giving $v^2=u^3-u$. The projection is $u$, so its function-field extension is generated by a square root of $u^3-u$. This is not a square in $k(u)$, since its zeros at $0,1,-1$ have odd order. In characteristic different from two the extension is separable of degree two.

The only apparent basepoint is $P_{\infty }=(0:0:1)$. In the chart $x_2=1$, put $s=x_0$, $t=x_1$. The equation is $s-t^3+s^{2}t=0$, so $t$ is a local parameter and $s=t^3/(1+st)$. The ratio $s/t=t^2/(1+st)$ is regular and the projection extends there as $(s/t:1)$, with value $(0:1)$. Thus the displayed rational projection does define a morphism after this unique extension; it is not literally evaluable by the raw pair $(x_0,x_1)$ at that point.

At a finite point with $v\ne 0$, the derivative with respect to $v$ is $2v\ne 0$, so $u-u(P)$ is a parameter and $e_P=1$. At $v=0$, the roots $u=0,\pm 1$ are simple (also in characteristic three), and $v$ is a parameter with $u-u(P)$ of order two. At infinity the target parameter $s/t$ likewise has order two. ThereforeThe affine gradient and the nonzero $\partial f/\partial x_0$ at infinity also verify nonsingularity for every characteristic other than two. The ramification description is on geometric points; it descends to these stated rational points over $k$.

For a nonsingular projective integral curve, apply the Riemann-Roch theorem to the divisor on an algebraic curve $mP$. It gives $\ell (mP)\ge m\mathrm{deg}P+1-g$, where $g$ is the geometric genus. For sufficiently large $m$, this exceeds the dimension of the space of constant functions; over the usual geometrically integral convention that dimension is one. Thus $L(mP)$ contains a nonconstant rational function.

By the definition of $L(mP)$, the principal divisor on an algebraic curve of that function plus $mP$ is effective, so it has no pole away from $P$. Therefore there is a nonconstant rational function regular on $X\setminus \{P\}$. If $P$ is not a rational point, using its closed-point degree in the same inequality handles that case. A positive pole order at $P$ is unavoidable, since regular functions on a projective integral curve are constant in its constant field.