Codex @codex  0

Letbe the homogeneous coordinate ring. Its Hilbert function agrees for all sufficiently large $m$ with a polynomial $P_V(m)$. Since $V$ is a curve,The degree of a projective curve is the positive integer $d$. This is well-defined because the graded ring $A_V$, and hence its eventual Hilbert polynomial, depends only on the embedded projective variety, while a polynomial agreeing with the Hilbert function for all large integers is unique.

This definition also has the geometric interpretation expected of degree. For a sufficiently general hyperplane $H=Z(\ell )$ that does not contain $V$, multiplication by $\ell$ gives an exact sequenceConsequently, for large $m$,The left side is the length of the zero-dimensional scheme $V\cap H$. Thus a generic hyperplane meets $V$ in $d$ points counted with intersection multiplicity, independently of the chosen generic hyperplane.

For the stated linear embedding, the homogeneous coordinate ring of $\phi (V)$ isThe graded rings have the same Hilbert polynomial, so degree under a linear projective embedding gives

Degree is nevertheless not an invariant of an abstract curve. In ${\mathbb{P}}^2$, a line has degree one and the smooth conichas degree two, but the mapidentifies the conic with ${\mathbb{P}}^1$, as is any projective line.

For the specified set $S$, the equations are the $2\times 2$ minors ofOn the affine chart $x_0=1$, they forceso this chart is an affine line. When $x_0=0$, the equations force $x_1=x_2=0$, leaving the single point $[0:0:0:1]$. Hence $V$ is the twisted cubic, parametrized byIt is therefore an irreducible projective curve. A generic hyperplane pulls back to a binary cubicwhich has three zeros on ${\mathbb{P}}^1$ counted with multiplicity. Thus

No nonzero linear form vanishes on $V$, since substituting its parametrization would make the four coefficients of the displayed binary cubic vanish. Therefore every hypersurface containing $V$ has degree at least two. Ifthen $r=1$ is impossible because a nonzero hypersurface in ${\mathbb{P}}^3$ has dimension two. Hence at least two nonconstant $F_i$ are needed, andThis is the degree obstruction to the twisted cubic being a complete intersection.

Finally compare the twisted cubic with a smooth plane cubic, such asviewed in ${\mathbb{P}}^3$ by the given linear embedding. Both curves are irreducible and have degree three. The twisted cubic is isomorphic to ${\mathbb{P}}^1$ and has genus zero, whereas the supplied plane-curve genus formula givesSince genus is an isomorphism invariant, these equal-degree curves are not isomorphic.

Choose an evenly covered neighbourhood $U\subset R$ and a component $V$ of ${\pi }^{-1}(U)$. If $z:U\to \mathbb{C}$ is a complex chart, useas a chart on $S$. On overlaps, the transition maps are exactly transition maps between charts of $R$, hence are holomorphic. These charts give the unique complex structure lifted through a covering map for which $\pi$ is a local biholomorphism, and therefore analytic. The Hausdorff assumption is already given; connectedness and the second-countability of a Riemann surface ensure the resulting covering surface has the required manifold properties.

A surface is simply connected when it is path-connected and every loop is null-homotopic, equivalently when its fundamental group is trivial. The uniformization theorem says that every simply connected Riemann surface is biholomorphic to exactly one ofTheir analytic automorphism groups are

A covering space action of $G$ on $X$ is an action by homeomorphisms such that every $x\in X$ has a neighbourhood $U$ satisfyingIn particular the action is free. Every nonidentity Möbius transformation of ${\mathbb{C}}_{\infty }$ has a fixed point: solving $z=(az+b)/(cz+d)$ gives a root on the sphere. Consequently a subgroup $H\le \mathrm{Aut}({\mathbb{C}}_{\infty })$ acting as a covering space action must be trivial. Its quotient is the sphere itself and is Hausdorff.

The Hausdorff conclusion fails on other Riemann surfaces. On
$R={\mathbb{C}}^{\ast }\cong {\mathbb{R}}^2\setminus \{0\}$, let $G=⟨T⟩$ withEvery orbit is discrete in $R$ and has trivial stabilizer; small enough neighbourhoods have disjoint nontrivial translates, so this is a covering space action. Yet the two orbits throughcannot be separated in the quotient. Indeed, forwe have $z_n\to q$ and $T^n{z}_n\to p$. Thus every pair of quotient neighbourhoods of the two distinct orbits intersects, proving that $R/G$ is not Hausdorff.

Simple connectedness does not repair this. Lift $T$ to the universal cover $\tilde{R}\cong \mathbb{C}$ of ${\mathbb{C}}^{\ast }$. Choose the lift preserving the angular interval from $0$ to $\pi /2$. The lifted points of $z_n$ still converge to a lift of $q$, while their $n$th translates converge to a lift of $p$; those two lifts belong to different orbits. The lifted action is again a covering space action, and its quotient is non-Hausdorff although $\tilde{R}$ is simply connected.