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For an embedded projective variety $X\subseteq {\mathbb{P}}_k^n$, let $I(X)$ be its homogeneous vanishing ideal. Its homogeneous coordinate ring is the graded ringThe grading records polynomial degree, and $\mathrm{Proj}S(X)$ recovers the embedded projective variety. The affine scheme $\mathrm{Spec}S(X)$ is its affine cone.

The chosen homogeneous ideal matters: different graded quotients can have the same Proj construction. For the usual degree-one polynomial grading, saturating an ideal by the irrelevant ideal recovers the ideal determined by the associated closed closed subscheme. Thus an arbitrary presentation $X=V_{+}(I)$ should not silently identify $I$ with $I(X)$.^[ref]

The ring also depends on the projective embedding. A projective line in its standard embedding has homogeneous coordinate ring $k[s,t]$, while its degree-two Veronese map image has homogeneous coordinate ring $k[X,Y,Z]/(XZ-Y^2)$.

Euler number and self-intersection. Orient the rank-two normal bundle ${\nu }_S$ by the orientations of $S$ and $M$: the ordered tangent and normal spaces have the ambient orientation. Choose a smooth normal section $s$ transverse to the zero section. Its zeros are isolated, and their signed number is the evaluation $⟨e({\nu }_S),[S]⟩$ of the Euler class on the fundamental class. Scale the section sufficiently small to lie in a tubular neighborhood. Its graph is a push-off $S^{\prime }$ isotopic to $S$.

The intersections of $S$ with $S^{\prime }$ occur exactly at the zeros of $s$. In a positively oriented local splitting $TS\oplus {\nu }_S$, the basis formed from the tangent spaces to $S$ and to the graph has block matrixIts determinant is $\det Ds$, so the local intersection sign is precisely the local zero index of the section. Summing proves the Euler number equals self-intersection identity

The conic. The given map is the degree-two Veronese map $[u:v]\mapsto [u^2:uv:v^2]$ followed by the projective linear transformationThe Veronese map is injective with nonvanishing differential: on the chart $u\ne 0$ its coordinates are $[1:t:t^2]$, and on $v\ne 0$ they are $[s^2:s:1]$. It is therefore a smooth embedding, and $S$ is a smooth plane conic diffeomorphic to ${\mathbb{CP}}^1\cong S^2$. Its equation in the given coordinates is $z_0^2-z_2^2=4z_1^2$.

Let $L$ be a projective line, with its complex orientation. The cohomology ring of complex projective space gives $[L]\cdot [L]=1$ in ${\mathbb{CP}}^2$. The hyperplane $z_1=0$ meets $S$ at $[u:v]=[1:0]$ and $[0:1]$. The zeros of $uv$ are simple in the respective local coordinates, and complex intersections have positive signs. Hence $[S]\cdot [L]=2$, so $[S]=2[L]$. The self-intersection number and the normal Euler class arewhere $u\in H^2(S^2;\mathbb{Z})$ has $⟨u,[S^2]⟩=1$.

The boundary of the tubular neighborhood. Write $N=\nu (S)$ for the closed disk tubular neighborhood and $E=\partial N$ for its boundary. This distinguishes the disk neighborhood $N$ from the vector normal bundle ${\nu }_S$. The space $E$ is the oriented circle bundle of ${\nu }_S$, with Euler class $4u$. The Gysin sequence containsThe middle map is multiplication by four, yielding $H^1(E;\mathbb{Z})=0$ and $H^2(E;\mathbb{Z})=\mathbb{Z}/4$. The same Gysin sequence gives $H^0(E;\mathbb{Z})=H^3(E;\mathbb{Z})=\mathbb{Z}$. The total space is a closed connected oriented three-manifold, so Poincare duality gives

The exterior. Work first with $W={\mathbb{CP}}^2\setminus \mathrm{int}N$. A collar neighborhood of $\partial W$ shows that its interior, the requested ${\mathbb{CP}}^2\setminus N$, has the same homotopy equivalence type: push the boundary a small positive distance into the collar. The Excision theorem and Thom isomorphism theorem identifyOnly degrees two and four are nonzero, each a copy of $\mathbb{Z}$.

In degree four the map $H_4({\mathbb{CP}}^2;\mathbb{Z})\to H_4(N,E;\mathbb{Z})$ sends the ambient fundamental class to the relative fundamental class of $N$. Under the Thom isomorphism theorem this becomes $[S]$. Thus this map is multiplication by one with compatible orientations. The long exact sequence in homology givesso $H_4(W)=H_3(W)=0$.

In degree two, the Thom isomorphism theorem identifies the map $H_2({\mathbb{CP}}^2)\to H_0(S^2)$ with intersection against $[S]$. A projective line intersects the conic twice, so this map is multiplication by two. The long exact sequence in homology isIt yields $H_2(W)=0$ and $H_1(W)=\mathbb{Z}/2$. In degree zero the relative groups vanish, so $H_0(W)\cong H_0({\mathbb{CP}}^2)\cong \mathbb{Z}$. This computes the homology of the complement of a smooth conic:The normal Euler number $4$, the degree-two intersection map and the degree-four restriction map $1$ play different roles; distinguishing them explains why the boundary has first homology $\mathbb{Z}/4$ while the exterior has first homology $\mathbb{Z}/2$.

Use coefficient coordinates $[a:b:c]$ on $\mathbb{P}(V)$. A nonzero binary quadratic has a repeated linear factor precisely when its discriminant vanishes, soIts gradient is $(-4c,2b,-4a)$, which cannot vanish at a projective point. Therefore $C$ is a nonsingular projective conic. The parametrization $[\alpha :\beta ]\mapsto [{\alpha }^2:2\alpha \beta :{\beta }^2]$ identifies it as a degree-two Veronese map.

A projective conic is the zero locus in a projective plane of a nonzero homogeneous degree-two polynomial. The equation is considered up to multiplication by a nonzero scalar. Over the complex numbers, a symmetric matrix for the quadratic form has rank three, two, or one: these give a nonsingular conic, two distinct projective lines, or one line counted twice. Diagonalization of a quadratic form and rescaling its nonzero diagonal entries prove this classification. The smooth model $XZ-Y^2=0$ is parametrized by the degree-two Veronese map $[s:t]\mapsto [s^2:st:t^2]$.

A rational normal curve in ${\mathbb{P}}^n$ is the image of the degree-$n$ Veronese map from ${\mathbb{P}}^1$. After projective changes of coordinates it has parametrization

A smooth complex degree-two projective plane curve is a projective linear image of the degree-two Veronese map and is diffeomorphic to a sphere. Its class in ${\mathbb{CP}}^2$ is twice the class of a projective line. Its self-intersection number is four, so its oriented normal bundle has Euler class evaluating to four.

For an integer $d\ge 1$, the degree-$d$ Veronese map sends a projective point to all degree-$d$ monomials in its homogeneous coordinates:Multiplying the original coordinates by $\lambda \ne 0$ multiplies every output by ${\lambda }^d$, so the map is well-defined on projective points. It is a closed immersion, given by the complete linear system of $\mathcal{O}(d)$.

For $n=1,d=2$, the formula is $[s:t]\mapsto [s^2:st:t^2]$. Its image in the projective plane satisfies $XZ=Y^2$, a smooth plane conic. On the chart $X\ne 0$, the ratio $Y/X$ recovers $t/s$; the chart $Z\ne 0$ recovers $s/t$. The degree-$d$ construction similarly makes homogeneous degree-$d$ expressions into linear coordinates in a larger projective space.