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The arithmetic genus of a proper curve $C$ is $p_a(C)=1-\chi (C,{\mathcal{O}}_C)$. For an integral projective curve over an algebraically closed field, $H^0(C,{\mathcal{O}}_C)=k$, so $p_a=h^1\ge 0$. Its finite normalization has $p_a(C)=g(C^{\nu })+\mathrm{length}({\nu }_{\ast }{\mathcal{O}}_{C^{\nu }}/{\mathcal{O}}_C)$. Hence arithmetic genus zero forces the curve to be smooth and rational.

For a projective complete intersection curve cut out by a homogeneous regular sequence of positive degrees $a,b$ in ${\mathbb{P}}_k^3$, the Koszul resolution gives Hilbert polynomial $abm+ab(4-a-b)/2$. Hence its arithmetic genus is $1+ab(a+b-4)/2$. The same resolution shows $H^0(C,{\mathcal{O}}_C)=k$, and the dimension of a scheme is one, so this genus equals ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Smoothness and irreducibility of the curve are not required. For degrees five and seven, the value is $141$.

For a smooth projective curve with $H^0(C,{\mathcal{O}}_C)=k$, its genus is ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$, equivalently ${\dim }_k{H}^0(C,{\Omega }_C^1)$ by Serre duality. It equals the arithmetic genus $1-\chi ({\mathcal{O}}_C)$. Without the condition on constants, the general Riemann-Roch theorem instead uses $\chi ({\mathcal{O}}_C)=h^0({\mathcal{O}}_C)-h^1({\mathcal{O}}_C)$.

An algebraic curve which is projective, reduced and irreducible. Over an algebraically closed field, its only global regular functions are constants and its arithmetic genus is $h^1({\mathcal{O}}_C)\ge 0$.

Blow up the distinct singular points of the integral projective curve, once at each original point, and let $C^{\prime }$ be its strict transform of an algebraic subvariety. Write $H$ for the pullback of a projective line and $E_i$ for the algebraic exceptional divisors. The multiplicity of a plane curve at a point and the intersection formula for blowing up a surface giveThe canonical divisor formula for a surface blowup gives $K=-3H+{\sum }_i{E}_i$. Applying the arithmetic adjunction formula on a smooth surface, rather than a formula requiring a smooth strict transform, yieldsand henceThe strict transform of an algebraic subvariety is still an integral projective curve. Its arithmetic genus is $h^1(C^{\prime },{\mathcal{O}}_{C^{\prime }})\ge 0$, even if some singularities remain. Therefore the plane curve singularity multiplicity bound isThe argument uses only one blowup of a smooth algebraic surface at each original singular point; it does not assume these blowups resolve every singularity.

The genus-degree formula gives arithmetic genus three for a plane quartic. Each of its three double points has multiplicity of a plane curve at a point two, so the arithmetic genus drop under a point blowup is one at each point. The resulting integral projective curve $C^{\prime }$ hasFor its finite normalization of an algebraic curve $\nu :\tilde{C}\to C^{\prime }$, the arithmetic genus identity isBoth terms are nonnegative, so $g(\tilde{C})=0$. Over the algebraically closed field, the rationality of a smooth projective genus-zero curve identifies $\tilde{C}$ with the projective line: a point $P$ exists, and Riemann-Roch theorem gives a degree-one map from the two-dimensional space $L(P)$. Consequently $C$ is birational to ${\mathbb{P}}^1$. This proves the rationality of a plane quartic with three double points even when a double point is not an ordinary node.

Use the finite map $f:C\to {\mathbb{P}}^1$ from the preceding solution. The inverse images of the two standard affine charts and of their intersection are affine, because $f$ is finite. The vanishing of quasi-coherent cohomology on an affine scheme therefore lets their two-term Čech cochain complex compute $H^i(C,E)$. It has terms only in degrees zero and one, provingThis actually applies to every quasi-coherent sheaf on $C$.

For a smooth projective curve with $H^0(C,{\mathcal{O}}_C)=k$, define its genus of a smooth projective curve byThe preceding Serre duality also gives $g={\dim }_k{H}^0(C,{\Omega }_C^1)$. The condition on constants holds in particular for a geometrically connected smooth projective curve. It makes $g$ equal to the arithmetic genus $1-\chi ({\mathcal{O}}_C)$. We state the theorem with this usual convention and then give the formula valid without that condition.

For a divisor on an algebraic curve $D$ and a canonical divisor $K$ with $\mathcal{O}(K)\cong {\Omega }_C^1$, the Riemann-Roch theorem isThe degree over an arbitrary field is weighted by residue degrees: ${\mathrm{deg}}_{k}D={\sum }_p{n}_p[k(p):k]$ when $D={\sum }_p{n}_{p}p$. Equivalently, for a line bundle $L$,We now prove it via Riemann-Roch via elementary modifications.

For any closed point $p$ and any line bundle $L$, the exact sequencehas a skyscraper sheaf as its quotient, with $k$-dimension $[k(p):k]$. This follows locally from a uniformizer of the discrete valuation ring at $p$: a local line-bundle generator modulo that uniformizer spans one copy of $k(p)$. The quotient has zero higher cohomology, as it is the direct image from a finite affine scheme. The long exact sequence in sheaf cohomology and the proved higher vanishing giveEvery line bundle has a nonzero rational section of a line bundle, whose local orders give a divisor on an algebraic curve $D$ with $L\cong \mathcal{O}(D)$. Iterating the displayed identity, adding or subtracting closed points with their multiplicities, givesFor the constants convention above, $\chi ({\mathcal{O}}_C)=1-g$. The dualizing sheaf on a smooth projective curve gives $h^1(\mathcal{O}(D))=h^0(\mathcal{O}(K-D))$, which converts this Euler-characteristic identity into the stated Riemann-Roch theorem. This proves the formula over arbitrary $k$, without assuming the closed points are $k$-rational.

For vector bundles, the corresponding formula isTo see it, saturate a rational rank-one subspace to a line subbundle and induct on rank. Both Euler characteristic of a coherent sheaf and the degree of the determinant are additive in the resulting short exact sequence. The line-bundle formula just proved starts the induction.

Taking $L={\Omega }_C^1$ and using duality gives $h^0({\Omega }_C^1)=g$, $h^1({\Omega }_C^1)=1$, and hence ${\mathrm{deg}}_k{\Omega }_C^1=2g-2$. In particular a line bundle of degree greater than $2g-2$ has zero first cohomology: its dual twisted by ${\Omega }_C^1$ has negative degree, and a negative-degree line bundle has no nonzero section, since any section would have an effective divisor of zeros.

If the term “curve over $k$” is used without requiring $H^0({\mathcal{O}}_C)=k$, the unconditional formula we proved isWriting $g=h^1({\mathcal{O}}_C)$ then replaces $1-g$ by $h^0({\mathcal{O}}_C)-g$. For example, ${\mathbb{P}}_{k^{\prime }}^1$ viewed as a $k$-curve for a finite separable extension $k^{\prime }/k$ has $h^0({\mathcal{O}}_C)=[k^{\prime }:k]$, so a literal constant term $1$ would be incorrect. Alternatively the arithmetic genus $p_a=1-\chi ({\mathcal{O}}_C)$ gives the formula with constant $1-p_a$. The constants hypothesis for the usual genus is recorded in Stacks Project, genus of a curve.

Put $P={\mathbb{P}}_k^2$ and $S=k[t_0,t_1,t_2]$. A homogeneous polynomial $F$ of positive degree $d$ is nonzero, and $S$ is an integral domain. Multiplication by $F$ therefore gives an exact sequence of coherent sheaveswhere $i:X\hookrightarrow P$ is the closed immersion.

We compute the necessary cohomology of twisting sheaves on projective space. The standard three affine open subsets $D_{+}(t_i)$ and all their intersections are affine schemes. For a quasi-coherent sheaf, vanishing of quasi-coherent cohomology on an affine scheme and the acyclic cover theorem identify its sheaf cohomology with the Čech cohomology of this cover. In degree $m$, the Čech cochain complex isDecompose it into subcomplexes indexed by Laurent monomials $t_0^{a_0}{t}_1^{a_1}{t}_2^{a_2}$ with $a_0+a_1+a_2=m$. Let $N=\{i:a_i<0\}$. Such a monomial occurs in the summand for a nonempty index set $J$ precisely when $N\subseteq J$, since only the variables in $J$ are inverted.

If $N$ is empty, the monomial contributes the cochain complex of a full two-simplex: its degree-zero cohomology is $k$ and its higher cohomology vanishes. If $|N|=1$, the dimensions in degrees $0,1,2$ are $1,2,1$; the first map has rank one and the second map has rank one with the same kernel as the first map's image. If $|N|=2$, there is one copy of $k$ in each of degrees $1$ and $2$ and the map between them is an isomorphism. These two cases are acyclic over every field, including characteristic two. If $N=\{0,1,2\}$, only the degree-two term occurs, contributing one copy of $k$. It follows thatand that $H^0(P,{\mathcal{O}}_P)=k$, $H^0(P,{\mathcal{O}}_P(-d))=0$, and $H^2(P,{\mathcal{O}}_P)=0$. Moreover the Laurent-monomial description of top cohomology on projective space gives a basis of $H^2(P,{\mathcal{O}}_P(-d))$ consisting ofFor $d\ge 3$, subtracting one from each $b_i$ counts these as $\left(\genfrac{}{}{0ex}{}{d-1}{2}\right)$. For $d=1,2$ there are no such triples, and the formula $(d-1)(d-2)/2$ is also zero.

Use the long exact sequence in sheaf cohomology of the initial sequence, and cohomology under a closed immersion to identify the terms for $i_{\ast }{\mathcal{O}}_X$. The vanishings just proved giveThereforeThis proves the cohomology of a projective plane hypersurface without imposing smoothness, irreducibility or reducedness. In particular, the second number is an arithmetic genus, not a claim about the genus of a normalization.

For the Sheaf of relative Kähler differentials on $X={\mathbb{P}}_{\mathbb{C}}^3$, the cotangent form of the Euler sequence isThe last arrow is onto because at least one homogeneous coordinate is invertible on each standard affine chart. Locally its kernel is the rank-three module generated by the differentials of the three affine coordinates, yielding the displayed Sheaf of relative Kähler differentials. The cohomology of twisting sheaves on projective space gives $H^q(X,{\mathcal{O}}_X(-1))=0$ for every $q$, while $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$ and its positive-degree groups vanish. The long exact sequence in sheaf cohomology therefore gives the cotangent-sheaf cohomology of projective space in this dimension:The isomorphism for $q=1$ is the connecting map from the constant global sections of ${\mathcal{O}}_X$. The Euler characteristic of a coherent sheaf is the alternating sum of dimensions, explaining the minus sign.

For the curve, put $S=\mathbb{C}[t_0,t_1,t_2,t_3]$. It is a unique factorization domain, so irreducibility of $F$ makes $(F)$ a prime ideal. Also $F$ does not divide $G$: otherwise irreducibility of $G$ would make them associates, contrary to their distinct degrees. Thus the image of $G$ is a nonzero element of the integral domain $S/(F)$. It follows that $(F,G)$ is a regular sequence. Its Koszul complex is an exact sequence, and graded sheafification giveswhere $i:Z\hookrightarrow X$ is the closed immersion. The signs make the composite $F(-G)+GF=0$. The shifts come respectively from $5+7$, $5$, and $7$.

Let ${\mathcal{I}}_Z$ be the ideal sheaf of a closed subscheme. Split this Koszul resolution intoFor all integers $d$, $H^1(X,{\mathcal{O}}_X(d))=H^2(X,{\mathcal{O}}_X(d))=0$. Also $H^1(X,{\mathcal{O}}_X)=H^2(X,{\mathcal{O}}_X)=0$. The two long exact sequences in sheaf cohomology, together with sheaf cohomology under a closed inclusion, consequently identifyBy Serre duality, $H^3(X,{\mathcal{O}}_X(-m))\cong S_{m-4}^{\ast }$ for $m\ge 4$. Counting degree-$d$ monomials in four variables gives $\dim S_d=\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$. Thus the source has dimension $\left(\genfrac{}{}{0ex}{}{11}{3}\right)=165$, and the two target spaces have dimensions $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. The rank-nullity theorem proves the requested bound:

In fact equality holds. Under the Serre duality pairings, the dual of this map isIf $Ga=Fb$, primeness of $F$ and $F\nmid G$ imply $F\mid a$. But $a$ has degree one and $F$ has degree five, forcing $a=0$, and then $b=0$. The dual map is injective, so the original map is surjective and its kernel has dimension $141$. Thus the lower bound is attained for every pair allowed in the question. No smoothness assumption is needed. The regular sequence cuts out a projective complete intersection of dimension one. Also $H^0(X,{\mathcal{I}}_Z)=H^1(X,{\mathcal{I}}_Z)=0$, by the negative twists and the intermediate cohomology vanishing in the first short exact sequence, so the second gives $H^0(Z,{\mathcal{O}}_Z)=\mathbb{C}$. This is the genus of a complete-intersection space curve: the first cohomology has dimension $141$, equal to its arithmetic genus.

Eliminate $t_3$ using the linear equation. The homogeneous coordinate ring becomesThe cubic is a nonzero element of a polynomial integral domain, hence a non-zero-divisor. It gives the exact sequence of graded modulesThus $\dim B_m=\left(\genfrac{}{}{0ex}{}{m+2}{2}\right)-\left(\genfrac{}{}{0ex}{}{m-1}{2}\right)$ for $m\ge 3$. Equivalently, use the Hilbert series $(1-z^3)/(1-z{)}^3$, or the sheafified exact sequence on ${\mathbb{P}}^2$. The Hilbert polynomial of the plane cubic isIts degree is three and its arithmetic genus is one, although the curve is singular: its affine equation near $[1:0:0]$ is the cusp $t_2^2+t_1^3=0$. The Hilbert polynomial records the arithmetic genus, rather than the genus of the normalization.

Let $C\in \mathrm{Neg}(K_X)$. The arithmetic adjunction formula on a smooth surface givesWe first exclude $C^2\ge 0$, following the PDF's hint and the finiteness proved in (a).

If $C^2=0$, adjunction and $K_X\cdot C<0$ force $p_a(C)=0$ and $K_X\cdot C=-2$. An integral projective curve of arithmetic genus zero is a smooth rational curve: normalization and the nonnegative singularity-length correction show both normalization genus and singularity correction vanish. Part (iii) makes $C$ semiample. Choose a basepoint-free $|mC|$. Over the infinite algebraically closed field, a general section avoids containing any of the finitely many curves of $\mathrm{Neg}(K_X)$ as a component. Its effective divisor $G$ then has $K_X\cdot G\ge 0$, but $K_X\cdot G=m{K}_X\cdot C<0$, a contradiction.

If $C^2>0$, Riemann–Roch theorem for algebraic surfaces and Serre duality show $h^0(X,mC)$ is unbounded. Indeed $h^2(X,mC)=h^0(X,K_X-mC)=0$ for $m\gg 0$, because the latter divisor has negative intersection with a fixed ample divisor. HenceThere is therefore some $m$ with $h^0(X,mC)>h^0(X,(m-1)C)$, giving a section whose divisor does not contain $C$. The canonical section of $mC$ does not vanish identically along any other curve. A general linear combination of these two sections consequently contains none of the finite set $\mathrm{Neg}(K_X)$, again contradicting its negative canonical intersection. Thus $C^2<0$.

Now put $a=-C^2\ge 1$ and $b=-K_X\cdot C\ge 1$. Adjunction givesIt follows that $a=b=1$ and $p_a(C)=0$. HenceThe general-section argument only avoids finitely many proper linear subspaces, so it works over any algebraically closed field, including positive characteristic.

A plane cubic is a projective plane curve defined by a nonzero homogeneous polynomial of degree three. Its arithmetic genus is one by the genus-degree formula, and its Hilbert polynomial is $3m$. A singular cubic can have normalization of smaller genus; the Hilbert polynomial measures the arithmetic genus.

Blow up each distinct singular point of an integral projective curve in the projective plane once. The arithmetic genus drop under a point blowup and the genus-degree formula give arithmetic genus $(d-1)(d-2)/2-{\sum }_P{m}_P(m_P-1)/2$ for the strict transform. Its arithmetic genus is nonnegative, proving the bound in every characteristic. Infinitely near singularities need not have been resolved.

A degree-four projective plane curve. Its arithmetic genus is three by the genus-degree formula; singularities can lower the genus of its normalization of an algebraic curve.

Three distinct points of multiplicity two use the entire arithmetic genus three of an integral projective curve of degree four. One point blowup at each gives arithmetic genus zero. The normalization of an algebraic curve then has genus zero and is the projective line over an algebraically closed field, by rationality of a smooth projective genus-zero curve. This allows double points with one branch as well as ordinary nodes.

Over an algebraically closed field, a smooth projective rational curve is isomorphic to ${\mathbb{P}}^1$. Its line bundles are determined by degree, its degree-zero line bundle is trivial, and $H^1({\mathbb{P}}^1,\mathcal{O})=0$. An integral projective curve of arithmetic genus zero is automatically smooth and rational by the normalization genus formula.