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Use the graded ring homomorphismwhere source degree $m$ maps to target degree $3m$. The homogeneous ideal of the twisted cubic is $\ker \phi$: a homogeneous polynomial vanishes on the parametrized curve precisely when its substitution vanishes identically. Here $k$ is infinite, being an algebraically closed field, so vanishing at all parameter points implies the polynomial identity.

Let $J$ be generated byTheir substitutions vanish, hence $J\subseteq \ker \phi$. To prove the reverse inclusion as an equality of ideals, we need more than a set-theoretic description of the zero locus.

Modulo $J$, replace $x_1^2$ by $x_0{x}_2$, $x_2^2$ by $x_1{x}_3$, and $x_1{x}_2$ by $x_0{x}_3$. Each replacement strictly decreases the sum of the exponents of $x_1$ and $x_2$, so the reduction terminates. It shows that the degree-$m$ quotient is spanned by the standard monomials of the twisted cubicFor $m\ge 1$ there are $(m+1)+m+m=3m+1$ of these. For $m=0$ the sole monomial is $1$. Their substituted images have $t$-exponents respectively $3b$, $3b+1$ and $3b+2$, so they are exactly the distinct monomialsThese are linearly independent. Consequently any degree-$m$ polynomial in $\ker \phi$, after reduction modulo $J$, has all its reduced coefficients zero. It belongs to $J$. Applying this to each homogeneous component givesThis also proves that the quotient embeds in the domain $k[s,t]$, so the displayed ideal is prime; no radicalization or saturation correction is concealed in the argument. Its image is the third Veronese subring ${⨁}_{m\ge 0}k[s,t{]}_{3m}$, and the calculation gives the Hilbert polynomial $3m+1$, consistent with degree three and genus zero.

Let $X$ be a separated Noetherian scheme over a field $k$ and $\mathcal{F}$ a coherent sheaf. Coherent cohomology studies the groups $H^q(X,\mathcal{F})$: global sections in degree zero and the obstructions to gluing local sections in higher degrees. Coherence means finite generation locally, and on a Noetherian scheme it is preserved by kernels, cokernels and extensions. This makes coherent sheaves suitable for exact-sequence calculations.

For an ordered open cover $\mathcal{U}=(U_i)$, the Čech cochain groups arewith differentialEach double omission occurs twice with opposite signs, so ${\delta }^2=0$. The Čech cohomology of the cover is ${\check{H}}^q=\ker \delta /\mathrm{im}\delta$, and taking the direct limit over refinements defines cover-independent Čech cohomology. In degree zero the sheaf gluing axiom identifies it with $\Gamma (X,\mathcal{F})$.

The comparison with derived-functor sheaf cohomology is particularly effective algebraically. Positive-degree cohomology of a quasi-coherent sheaf on an affine scheme vanishes. A separated scheme has affine intersections of affine opens, so a finite affine cover is acyclic for $\mathcal{F}$. Its Čech cochain complex consequently computes $H^q(X,\mathcal{F})$ directly. For a short exact sequence of sheaves, sections over each affine intersection are exact on quasi-coherent sheaves. The resulting short exact sequence of Čech complexes gives the long exact sequence in sheaf cohomology. In particular, the connecting map measures the failure of a global section of the quotient to lift globally.

Several central projective results make the theory finite and computable. On a projective variety over $k$, every coherent cohomology group is finite-dimensional. The Grothendieck vanishing theorem gives $H^q(X,\mathcal{F})=0$ for $q>\dim X$. If ${\mathcal{O}}_X(1)$ is ample, Serre vanishing gives $H^{q>0}(X,\mathcal{F}(m))=0$ for sufficiently large $m$, and these large twists are globally generated sheaves. On projective space a coherent sheaf has a finite twisting resolution of a coherent sheaf on projective space: finite sums of $\mathcal{O}(a)$ resolve it. The explicit line-bundle cohomology below and the long exact sequences prove finiteness and positive-twist vanishing. For a closed embedding $i:X\hookrightarrow {\mathbb{P}}^n$, $i_{\ast }\mathcal{F}$ is coherent and $H^q(X,\mathcal{F})=H^q({\mathbb{P}}^n,i_{\ast }\mathcal{F})$, reducing these results to projective space. A sufficiently positive twist of a finite presentation by sums of $\mathcal{O}(-a)$ also proves global generation.

The Euler characteristic of a coherent sheafis therefore defined, and the long exact sequence makes it additive. With a fixed projective embedding, $\chi (X,\mathcal{F}(m))$ is its Hilbert polynomial, and for sufficiently large $m$ it equals $h^0(X,\mathcal{F}(m))$. On a smooth projective variety of dimension $r$, Serre duality gives, for a vector bundle $E$,For an arbitrary coherent sheaf the correct counterpart is $H^q(X,\mathcal{F}{)}^{\ast }\cong {\mathrm{Ext}}_X^{r-q}(\mathcal{F},{\omega }_X)$; replacing this Ext group indiscriminately by the cohomology of the ordinary sheaf dual would lose local extension information.

We now calculate the basic examples. For $n\ge 1$, use the standard cover $U_i=D_{+}(x_i)$ of ${\mathbb{P}}^n$. Sections of $\mathcal{O}(m)$ on an intersection are degree-$m$ Laurent monomials in which negative exponents are allowed only for coordinates inverted on that intersection. Decompose the Čech cochain complex by exponent vectors $(a_0,\dots ,a_n)$ with sum $m$. If no exponent is negative, the corresponding simplex complex contributes only in degree zero. If all exponents are negative, the monomial occurs only on the full intersection and contributes only in degree $n$. When the negative-index set is nonempty and proper, inserting any index outside it gives a contracting homotopy, so the monomial complex is acyclic. Hence the cohomology of twisting sheaves on projective space isIn the top group a basis consists of the Laurent monomials with every exponent at most $-1$. Writing $a_i=-1-b_i$ reduces the count to nonnegative solutions of $\sum b_i=-m-n-1$, giving the displayed binomial coefficient. In particular, all intermediate cohomology vanishes, and ${\omega }_{{\mathbb{P}}^n}=\mathcal{O}(-n-1)$ agrees with Serre duality. For ${\mathbb{P}}^0$ separately, every twist is trivial and $H^0=k$.

On the projective line this givesThus positive and negative degrees produce quite different groups, while their alternating difference is a single polynomial. The twisted cubic is isomorphic to ${\mathbb{P}}^1$ and its hyperplane bundle corresponds to ${\mathcal{O}}_{{\mathbb{P}}^1}(3)$. Consequently $h^0(X,{\mathcal{O}}_X(m))=3m+1$ for $m\ge 0$, agreeing with Q2, and $h^1(X,{\mathcal{O}}_X(m))=\max (-3m-1,0)$.

For a smooth plane curve $C$ of degree $d\ge 1$, the ideal-sheaf sequence of a projective hypersurface isThe long exact sequence in sheaf cohomology and the preceding formula give $H^0(C,{\mathcal{O}}_C)=k$ and $H^1(C,{\mathcal{O}}_C)\cong H^2({\mathbb{P}}^2,\mathcal{O}(-d))$. ThereforeFor $d=1,2$ this is zero; for a plane cubic it is one. Twisting the same exact sequence gives $\chi (C,{\mathcal{O}}_C(m))=dm+1-g$. These examples illustrate how Čech cohomology, vanishing, Serre duality and exact sequences convert local equations into global numerical invariants.

Use the polynomial graded ring $R[x_0,x_1,x_2,x_3]$, as in the PDF; the doubled brackets in the converted TeX are an OCR error. Let $I$ denote the homogeneous defining ideal. The four cubic coordinates in the proposed map never vanish simultaneously on ${\mathbb{P}}_F^1$, since $y_0^3$ and $y_1^3$ already have that property. Substitution makes each generator of $I$ zero, so they define a morphism into the generic fibre.

The two standard opens $D_{+}(x_0)$ and $D_{+}(x_3)$ cover $X$. Indeed, a homogeneous prime containing $x_0,x_3$ contains $x_2$ by $x_0{x}_3^2=x_2^3$, and then contains $x_1$ by $x_1^2={\pi }^2{x}_0{x}_2$. Such a prime contains the irrelevant ideal and gives no point of Proj.

On $D_{+}(x_0)$ put $a=x_1/x_0$, $b=x_2/x_0$, $c=x_3/x_0$. Over $F$, where $\pi$ is invertible, the equations giveThus the generic-fibre chart is exactly $\mathrm{Spec}F[t]$. The proposed map restricts to this isomorphism with $t=y_1/y_0$. On $D_{+}(x_3)$ put $s=x_2/x_3$. The equations giveso this chart is $\mathrm{Spec}F[s]$, and the map has $s=y_0/y_1$. On the overlap $s=t^{-1}$. These are the standard two charts of the projective line, proving the scheme-level isomorphismEquivalently, rescaling $x_1$ by ${\pi }^{-1}$ gives the usual twisted cubic. The chart proof also shows that the extra cubic relation introduces no additional generic-fibre scheme structure.

Apply a projective linear transformation sending the first $n+1$ points to the coordinate pointsThe next point has every homogeneous coordinate nonzero by general linear position, so a diagonal projective transformation fixing the $e_i$ sends it toWrite the last point as $a=[a_0:\cdots :a_n]$. Again general linear position implies that every $a_i$ is nonzero. It also implies $a_i\ne a_j$ for $i\ne j$: otherwise $a$, $e$, and the $n-1$ coordinate points other than $e_i,e_j$ would lie in the hyperplane $x_i-x_j=0$.

Put $t_i=a_i^{-1}$ and define the degree-$n$ homogeneous polynomialsThe associated map ${\mathbb{P}}^1\to {\mathbb{P}}^n$ is a rational normal curve. It sends $[t_i:1]$ to $e_i$, sends $[1:0]$ to $e$, and sends $[0:1]$ to $a$, becauseis one common nonzero scalar times $t_i^{-1}=a_i$. This proves existence.

For uniqueness, parametrize any rational normal curve through these points and choose a coordinate $[S:T]$ on its source ${\mathbb{P}}^1$ so that the preimages of $e$ and $a$ are respectively $[1:0]$ and $[0:1]$. If the preimage of $e_i$ is $[s_i:1]$, then its $i$th coordinate polynomial must have all the other $s_j$ as roots. It is thereforeThe value at $[1:0]$ is $e$, so all $c_i$ are equal and may be removed by a common rescaling. The value at $[0:1]$ is $a$, so $s_i^{-1}$ is proportional to $a_i$. Thus all $s_i$ are obtained from the $t_i$ by one common rescaling of the coordinate on ${\mathbb{P}}^1$. This merely reparametrizes the curve, proving uniqueness.

For $n=3$, choose the monomial basis $S^3,S^{2}T,S{T}^2,T^3$. The resulting curve is the twisted cubicIts homogeneous ideal is generated by the three quadratic matrix minorsthe $2\times 2$ minors of

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.

Modulo the three quadric relations of the twisted cubic, replace $x_1^2$ by $x_0{x}_2$, $x_2^2$ by $x_1{x}_3$, and $x_1{x}_2$ by $x_0{x}_3$. Each replacement reduces the total exponent of $x_1,x_2$. The resulting monomials are $x_0^a{x}_3^b$, $x_0^a{x}_1{x}_3^b$, and $x_0^a{x}_2{x}_3^b$. In degree $m$ their images under $x_i\mapsto s^{3-i}{t}^i$ are all $3m+1$ distinct monomials of degree $3m$. They are linearly independent, so the three quadrics generate the full homogeneous ideal, not merely its radical. The quotient is the third Veronese subring of $k[s,t]$.