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For a nonhyperelliptic smooth projective curve of geometric genus four, the canonical map is an embedding of degree six. Dimension counts $10>h^0(2K)=9$ and $20>h^0(3K)=15$ supply a quadric and a cubic not divisible by it. The quadric is irreducible because the curve is nondegenerate. Degree six and unmixedness of a complete intersection then identify the intersection with the curve as a scheme.

A genus one curve is a smooth projective curve of geometric genus one. Choosing a rational point makes it an elliptic curve.

For a smooth complex projective plane curve of degree $d$,Project generically from a point outside the curve to a projective line. Its degree is $d$, by intersection with a generic line through the center. Ramification occurs when such a line is tangent. In coordinates for projection $[x:y:z]\mapsto [x:z]$, these points satisfy $F=F_y=0$. The polar has degree $d-1$, and Bézout gives total ramification $d(d-1)$; generic projection makes the ramification simple, or the intersection multiplicities can be read as the corresponding ramification orders. The Riemann-Hurwitz formula yields $2g-2=-2d+d(d-1)$ and hence the formula. The smoothness qualification is essential for geometric genus: singular plane curves have the same arithmetic degree-genus expression, while the genus of their normalization can be smaller.

Let $X$ be a connected compact Riemann surface of geometric genus $g$, equivalently a smooth projective curve over $\mathbb{C}$, and put $V=H^0(X,K_X)$. Choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$ with $a_i\cdot b_j={\delta }_{ij}$. For a closed differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The underlying bilinear identity isIt holds for any two smooth closed one-forms. In particular, the Riemann bilinear relations for holomorphic differential forms arefor $\omega ,\eta \in V$ and $\omega \ne 0$ in the inequality. The inequality fixes the orientation and the sign convention.

To prove the identity, cut $X$ along the symplectic basis to a fundamental polygon $P$. On $P$, a closed one-form $\alpha$ has a primitive $F$. The Stokes theorem givesOpposite copies of each cut have opposite orientations, while their values of $F$ differ by the corresponding period of $\alpha$. Pairing those edges therefore leaves $A_i(\alpha ){\int }_{b_i}\beta -B_i(\alpha ){\int }_{a_i}\beta$ for the $i$th handle. Summing proves the identity. The same calculation can be made with a small disk about the polygon's vertices removed; its extra boundary terms tend to zero, so the vertices cause no additional term.

For holomorphic differential forms $\omega ,\eta$, their wedge product of differential forms vanishes: locally both are multiples of $dz$. This proves the first relation. If $\omega =f(z)dz$ and $z=x+iy$, thenA nonzero holomorphic differential form is nonzero on an open set, so the integral is strictly positive. Applying the identity to $\omega ,\overline{\omega }$ proves the second relation.

These relations first prove that the map $V\to {\mathbb{C}}^g$ taking $a$-periods is injective: if every $A_i(\omega )$ vanishes, the positive integral just computed would vanish. Since $\dim V=g$ by the Riemann-Roch theorem, it is an isomorphism. Consequently there is a unique basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$. Set ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. The first relation gives ${\tau }_{ij}={\tau }_{ji}$, and the second gives, for any nonzero column $c$,Thus the normalized period matrix of a complex torus is symmetric and has positive-definite imaginary part:

The integration map sends a cycle $\gamma$ to the functional $\omega \mapsto {\int }_{\gamma }\omega$ in $V^{\ast }$. In the normalized basis its image is the period latticeIndeed, the real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: its imaginary part is $Yy$, and $Y$ is invertible. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and has compact quotient. This proves thatis a well-defined $g$-dimensional complex torus, rather than merely a quotient by an arbitrary subgroup.

The relations also provide its canonical principal polarization of a complex torus. With the convention that a Hermitian form is conjugate-linear in its first argument, put $H(z,w)={\overline{z}}^{,t}{Y}^{-1}w$ and $E=\mathrm{Im}H$. Symmetry of $\tau$ givesIn particular $E$ is integral and unimodular on $\Lambda$, and it is compatible with the complex structure. These are exactly the positivity and integrality conditions for a Riemann form on a complex torus; they make the complex torus a Jacobian variety with a principal polarization of a complex torus. If instead one uses a first-argument-linear Hermitian form, the same alternating form is $-\mathrm{Im}H$. Changing the symplectic basis changes the coordinates but not this polarized quotient. For $g=0$, $V=0$ and the Jacobian variety is a point.

Assume $X$ is a connected compact Riemann surface of geometric genus $g\ge 1$. Work first in ${\mathrm{Pic}}^{g-1}(X)$, translating back to the Jacobian variety by tensoring with ${\mathcal{O}}_X(-(g-1)p_0)$ when desired. PutThis is the image of the Abelian sum map from $X^{(g-1)}$. The Riemann singularity theorem asserts that this image is a reduced Cartier divisor and thatHere multiplicity means the order of a local defining equation in the maximal ideal of the smooth ambient Jacobian variety. Consequently the singular points are precisely the classes with at least two independent sections. We will prove the multiplicity equality, including the nonvanishing of its proposed leading term.

Choose an effective divisor $E$ of degree $n\ge g$, and a local holomorphic family $\mathcal{L}$ of degree-$g-1$ line bundles about $L$ on $X\times U$, where $U$ is a small neighborhood in ${\mathrm{Pic}}^{g-1}(X)$. Such a family is obtained by varying the Čech cocycles of a line bundle. Since $\mathrm{deg}{L}^{\prime }(E)>2g-2$ for all $L^{\prime }\in U$, Serre duality gives $H^1(L^{\prime }(E))=0$; the Riemann-Roch theorem gives $h^0(L^{\prime }(E))=n$. The cohomology and base change for line bundles on a curve then identifiesas vector bundles of rank $n$. Evaluation defines a matrix $M:A\to B$. For each $L^{\prime }$, its exact sequence isThus $f=\det M$ vanishes exactly on $\Theta$.

The function $f$ is not identically zero. To exhibit a degree-$g-1$ line bundle with no section, choose $g$ distinct points $p_1,\dots ,p_g$ giving independent evaluation functionals on $H^0(K_X)$. Such points can be chosen successively because a nonzero holomorphic differential form cannot vanish everywhere. For $F=\sum p_i$, the Riemann-Roch theorem gives $h^0({\mathcal{O}}_X(F))=1$. Choose $q$ outside $F$. The unique section of ${\mathcal{O}}_X(F)$ does not vanish at $q$, so ${\mathcal{O}}_X(F-q)$ has no section. Therefore the evaluation determinant genuinely cuts out a hypersurface. Its support is irreducible, since it is the image of the irreducible symmetric product of a curve $X^{(g-1)}$.

Fix now $L\in \Theta$, and put $r=h^0(L)=h^1(L)$, the equality following from the Riemann-Roch theorem. The matrix $M(L)$ has an invertible $(n-r)$ by $(n-r)$ block. After invertible holomorphic row and column operations, its local form iswhere $N$ is $r$ by $r$. More explicitly, eliminate the invertible block by its Schur complement; the discarded determinant is a holomorphic unit. Hence $f$ is a unit times $\det N$. Every entry of $N$ lies in the maximal ideal, so its order is at least $r$.

The linear term of $N$ is the cup-product mapTo verify this, deform the transition functions as $g_{ij}(1+\epsilon {\xi }_{ij})$ and try to lift a section $s$ to $s+\epsilon s^{\prime }$. Its first-order gluing equation has right side ${\xi }_{ij}{s}_j$, whose Čech cohomology class is exactly the obstruction to the lift. The kernel-to-cokernel derivative of the evaluation matrix gives that same obstruction. Under Serre duality, if $s_1,\dots ,s_r$ and $t_1,\dots ,t_r$ are bases of $H^0(L)$ and $H^0(K_X\otimes L^{-1})$, thenWe must show that this matrix is invertible for some $\xi$; otherwise the lower bound on multiplicity would not be an equality.

For each of these two $r$-dimensional section spaces, there are $r$ distinct points with independent evaluations. Choose them successively, using a section in the remaining kernel to find the next point. The set of such $r$-tuples is a nonempty open subset of $X^r$. The two open sets intersect because $X^r$ is irreducible, and we may avoid their diagonals. Choose $q_1,\dots ,q_r$ in that intersection and trivialize $L$ and $K_X$ there. Both matricesare invertible. Let $\xi \in H^0(K_X{)}^{\ast }=H^1({\mathcal{O}}_X)$ be the sum of the evaluation functionals at these points, with any nonzero weights $c_k$. Thenis invertible. It follows that the degree-$r$ polynomial $\det N_1$ is nonzero, so ${\mathrm{ord}}_{L}f=r$. This proves the needed invertible cup-product direction for a line bundle on a curve by an explicit construction.

We still need to identify this determinant hypersurface with the reduced image, rather than a multiple of it. Choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$. For their sum $D$, the Riemann-Roch theorem gives $h^0(D)=h^0(K_X-D)=1$. At its class, the calculation above gives order one. The determinant hypersurface has irreducible support, so its sole divisor on an algebraic curve coefficient is one everywhere. A local hypersurface in a smooth space with these divisor on an algebraic curve coefficients is reduced. Thus it is exactly the reduced image $W_{g-1}$, and the multiplicity formula holds for that image itself.

The argument gives the useful additional descriptionFor $r=1$, its nonzero linear equation is $⟨\xi ,st⟩=0$. This agrees with the derivative of the Abelian sum map: if $\mathrm{div}(s)=D$, then $st$ is the unique holomorphic differential form vanishing along $D$. For $r\ge 2$, the local equation has no linear term and the point is singular. When $g=1$, $W_0$ is the single trivial line bundle in ${\mathrm{Pic}}^0(X)$, with multiplicity one; the argument includes this case using the empty effective divisor. In genus zero there is no $W_{g-1}$ of effective divisors on an algebraic curve, so that hypothesis is necessary.

Let $g$ be the geometric genus of $X$. Choose an integer $N\ge 2g+1$. For each prescribed point $x_i$, both $N{x}_i$ and $(N-1)x_i$ have degree greater than $2g-2$, so their complementary canonical divisors have negative degree and their complementary Riemann-Roch spaces vanish. The Riemann-Roch theorem therefore givesChoose $f_i\in L(N{x}_i)\setminus L((N-1)x_i)$. This is possible because the two dimensions differ by one. Such a rational function has its only pole at $x_i$, and its pole order there is exactly $N$. Set $f={\sum }_{i=1}^m{f}_i$.

At $x_i$ all terms except $f_i$ are regular. In a local parameter, the term of order $-N$ in $f_i$ has a nonzero coefficient, and none of the other terms has a negative-power term. Hence that coefficient cannot cancel. Away from the prescribed points every summand is regular. We obtain the stronger conclusionThus $f$ is regular on $U$ and has a pole at every $x_i$. This rational functions with prescribed poles construction works in every characteristic and does not require a generic-coefficient choice. If the prescribed list is empty, a constant function meets the vacuous requirements.

Let $E$ be the elliptic curve, and choose a point $O\in E(k)$ to serve as identity. Its geometric genus is one and its canonical divisor $K$ has degree zero. For any divisor $D$ of degree one, the Riemann-Roch theorem givesSince $K-D$ has degree $-1$, its Riemann-Roch space is zero: a nonzero section would produce an effective divisor of negative degree. Hence $\ell (D)=1$. Choose a nonzero $f\in L(D)$. The divisor $D+(f)$ is effective of degree one, so over the algebraically closed field it is a single point $P$. Thus every degree-one divisor class has a point representative.

This representative is unique. If $P$ and $Q$ are linearly equivalent, choose $h$ with $(h)=P-Q$. Then $h\in L(Q)$. That space has dimension one and already contains the constants, so $h$ is constant and $P=Q$. Consequentlyis injective. It is also surjective: if $D$ represents a degree-zero divisor class, apply the preceding result to $D+O$ to obtain its unique point representative $P$, giving $[D]=[P-O]$.

The degree-zero Picard group of a curve is a group because divisors add commutatively, principal divisors form a subgroup, and degree passes to the quotient. DefineEquivalently, $R=P\oplus Q$ is the unique point satisfying the linear equivalence of divisors $P+Q\sim R+O$, where the plus signs in this equivalence denote divisor addition. The zero class corresponds to $O$, so $O$ is the identity. The inverse of $P$ is the unique point $P^{\prime }$ with $P+P^{\prime }\sim 2O$. Associativity follows because both $(P\oplus Q)\oplus R$ and $P\oplus (Q\oplus R)$ map to $\iota (P)+\iota (Q)+\iota (R)$, and $\iota$ is injective; commutativity follows in the same way. Thus this is an abelian group structure on $E(k)$, obtained directly by the elliptic curve group law from Riemann-Roch. The identity choice is part of the construction.

An elliptic curve works over every algebraically closed field, including one algebraic over a finite field. Give it explicitly in ${\mathbb{P}}^2$ byIn the first case the only point at infinity is $O=[0:1:0]$, where the derivative with respect to $z$ is nonzero. In the affine chart $z=1$, a singular point of an algebraic variety would have $y=0$, $x^3-x=0$, and $3x^2-1=0$. The possible roots $x=0,1,-1$ cannot satisfy the last equation in characteristic different from $2$: at $0$ it is $-1$, and at $\pm 1$ it is $2$. In the second case the three homogeneous partial derivatives are $x^2,z^2,y^2$, up to harmless signs, and cannot vanish simultaneously at a projective point. Thus both cubics are smooth varieties. A reducible plane cubic would have intersecting positive-degree components by Bézout theorem, and be singular at an intersection; therefore these smooth cubics are irreducible varieties. The genus-degree formula gives $g=(3-1)(3-2)/2=1$. With the marked point $O$ they are elliptic curves.

For a smooth algebraic curve, its divisor class group is its Picard group. Its degree-zero subgroup isThe map $P\mapsto [P-O]$ gives an isomorphism of abelian groupsTo recall why it is a bijection, Riemann-Roch theorem in geometric genus one says every degree-one divisor class has exactly one independent section: a canonical divisor has degree zero, so the dual degree-$-1$ term has no section. Thus the class contains one effective divisor on an algebraic curve of degree one, namely a point $P$, and that point is unique. Translating by $-O$ gives the claimed bijection; this is precisely the chord-and-tangent group law on the elliptic curve.

Choose a prime integer $\ell$ different from $\mathrm{char}k$. The multiplication-by-n morphism $[\ell ]:E\to E$ has derivative $\ell$ times the identity at $O$, so it is nonconstant. A nonconstant morphism of algebraic varieties between smooth projective curves is surjective: its image is closed by proper morphism and cannot have algebraic dimension zero. Since $k$ is an algebraically closed field, this makes $[\ell ]$ surjective on $k$-points. ThereforeYet $E(k)$ is infinite. For each $x\in k$, the affine equation has a solution for $y$ in the algebraically closed field; since $k$ is infinite, this gives infinitely many points.

By the Fundamental theorem of finitely generated abelian groups, a finitely generated abelian group $A$ satisfying $A=\ell A$ has no free part and is a finite torsion group of order prime to $\ell$. The infinite group $E(k)$ cannot have this property while being finitely generated. Thus ${\mathrm{Pic}}^0(E)$ is not finitely generated. A subgroup of a finitely generated abelian group is finitely generated, soThis proof uses divisibility and infinitude, rather than uncountability, so it remains valid over countable algebraically closed fields.

For a smooth projective curve $C$, let $g=h^1(C,{\mathcal{O}}_C)$ be its geometric genus, let $K_C$ be a canonical divisor, and write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ for the dimension of a vector space of the Riemann-Roch space. The Riemann-Roch theorem saysfor every divisor on an algebraic curve $D$ on $C$.

Here is the cohomological proof of Riemann-Roch for curves. At a closed point $P$, the local ring is a discrete valuation ring. A local uniformizer shows that increasing the allowed pole order by one gives the short exact sequence of sheavesThe last term is a length-one skyscraper sheaf; its precise identification with $k$ is noncanonical, but its sheaf cohomology has $h^0=1$ and $h^1=0$. All the relevant groups of sheaf cohomology are finite-dimensional because $C$ is a projective variety, and coherent sheaf cohomology above degree one vanishes on a algebraic curve. The long exact sequence in sheaf cohomology therefore givesSince $k$ is an algebraically closed field, every closed point has degree one. Iterating this identity for both positive and negative coefficients of $D$ provesEvery global regular function on a projective variety which is an irreducible variety is constant; hence $h^0(C,{\mathcal{O}}_C)=1$ and $\chi ({\mathcal{O}}_C)=1-g$.

Finally apply Serre duality for a smooth projective curve:Taking dimensions of a vector space and inserting this in the Euler characteristic of a coherent sheaf identity proves the displayed Riemann-Roch theorem. The substantial inputs from sheaf cohomology are finiteness and vanishing for coherent sheaves on a projective curve, and Serre duality; the change of Euler characteristic of a coherent sheaf is proved directly by the point exact sequence. In particular the argument handles negative as well as effective divisors on an algebraic curve.

For a smooth projective curve $C$ of geometric genus $g$ and a divisor on an algebraic curve $D$, write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ and let $K_C$ be a canonical divisor. The curve form of the Riemann-Roch theorem isHere $\mathrm{deg}{K}_C=2g-2$ and $\ell (K_C)=g$. In line bundle form,The usual curve convention is a smooth geometrically integral projective curve; geometric embedding tests may be checked after extending the ground field to an algebraic closure.

Put $O=0_E$. Choose rational projective coordinates in which $O=(0:1:0)$ and its tangent is $Z=0$. Because $O$ is an inflection point of a plane cubic, that tangent intersects the cubic in the divisor $3O$. Consequently its cubic equation has the formHere $a\ne 0$, since the restriction to $Z=0$ is a nonzero multiple of $X^3$, and $b\ne 0$, since smoothness at $O$ requires a nonzero $Z$-derivative there. Completing the square in $Y$, by the rational linear change $Y^{\prime }=Y+(cX+eZ)/(2b)$, gives on $Z=1$Set $x^{\prime }=c_{0}x$ and $y^{\prime \prime }=c_0{y}^{\prime }$. ThenThese are invertible rational projective coordinate changes, and $O$ remains the unique point at infinity. Smoothness ensures that the resulting monic cubic has no repeated root. Thus this is a Weierstrass equation of an elliptic curve, without requiring any extraction of a square or cube root in $\mathbb{Q}$.

On this model write $-(x,y)=(x,-y)$ and $-O=O$. To add $P,Q$, take their joining line, or the tangent if $P=Q$, and let $R$ be the third intersection counted with multiplicity. DefineFor a vertical line its third intersection is $O$, so $P+(-P)=O$. The line through $O$ and an affine point is vertical, giving $P+O=P$. The tangent at $O$ has triple contact, giving $O+O=O$. This is the chord-and-tangent group law.

If $P,Q$ are rational, their joining line or tangent is rational. Its residual intersection point is rational: substituting the line in the cubic leaves a cubic whose other two roots, counted with multiplicity, are rational. Thus addition preserves $E(\mathbb{Q})$. The construction is symmetric in $P,Q$, so it is commutative.

For associativity, use divisor classes. A smooth plane cubic has geometric genus $1$, by the genus-degree formula. The Riemann-Roch theorem says that a divisor of degree $1$ on a genus-one curve has a one-dimensional space of sections. Therefore every degree-zero divisor class is uniquely represented by $P-O$: add $O$, take its unique effective degree-one representative, and then subtract $O$. Uniqueness also follows because a nonconstant function with its only pole a simple pole at $Q$ would contradict $\ell (Q)=1$.

Every line section is linearly equivalent to the tangent section $3O$. Thus, whenever $P,Q,R$ are collinear, including tangencies,A vertical section gives $[(-R)-O]=-[R-O]$. Hence the chord-and-tangent group law becomes ordinary addition in the degree-zero Picard group under the injective correspondence $P\mapsto [P-O]$. Associativity follows from associativity in that abelian group. Together with closure, identity and inverses already checked, this proves that $E(\mathbb{Q})$ is an abelian group.

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.

For a divisor on an algebraic curve $D$ on a smooth projective curve $C$ of geometric genus $g$, and a canonical divisor $K_C$, the Riemann-Roch theorem states

PutBy the Jacobian criterion, a projective plane curve is singular only at a common projective zero of $F$ and all its partial derivatives. HereTheir simultaneous vanishing forces $X_0=X_1=X_2=0$, which is not a point of projective space. Hence the Fermat quartic $C$ is nonsingular. In particular it is a smooth plane quartic and has geometric genus $(4-1)(4-2)/2=3$.

A minimal algebraic surface is a smooth projective surface containing no exceptional curve of the first kind, namely no smooth rational curve of self-intersection $-1$. An abelian surface contains no rational curve because every morphism from ${\mathbb{P}}^1$ to an abelian variety is constant. It therefore has no $(-1)$-curve and is minimal.

For every $n\ge 2$, the Minimal Hirzebruch surface ${\mathbb{F}}_n$ is a rational minimal surface. These surfaces are pairwise nonisomorphic: the negative section is the unique irreducible curve of negative self-intersection and has square $-n$, so an isomorphism would recover $n$. Thus there are infinitely many nonisomorphic minimal rational surfaces.

A K3 surface is a smooth projective surface $X$ with $K_X\cong {\mathcal{O}}_X$ and $H^1(X,{\mathcal{O}}_X)=0$. For a smooth curve $C\subset X$ of geometric genus $g$, the adjunction formula givesand hence

An elliptic surface is a smooth projective surface with a morphism to a smooth curve whose generic fiber is a smooth genus-one curve. If $E$ is an elliptic curve, projectionis an elliptic fibration. Its canonical bundle is pulled back from $K_{{\mathbb{P}}^1}$, so every positive pluricanonical space vanishes and the Kodaira dimension is $-\infty$. This supplies the requested negative-Kodaira-dimension example.

For an elliptically fibered K3 surface, choose a smooth quartic $X\subseteq {\mathbb{P}}^3$ containing a line $L$. The canonical bundle of a smooth projective hypersurface formula makes $K_X$ trivial, and the standard cohomology sequence gives $H^1(X,{\mathcal{O}}_X)=0$, so $X$ is K3. The pencil of planes through $L$ cuts $X$ into $L$ plus a residual plane cubic. The residual linear system $|H-L|$ is basepoint-free, has square zero and defines a morphism $X\to {\mathbb{P}}^1$ whose generic fiber is a smooth plane cubic. This is the Elliptic K3 surface from a quartic containing a line.

A surface of general type is a smooth projective surface of Kodaira dimension two. Let $B\subseteq {\mathbb{P}}^2$ be a smooth plane curve of degree eight and letbe the degree-two cover branched along $B$. The branch-cover canonical-bundle formula givesThis divisor is ample, so $X$ is a double plane of general type and $\pi$ is the required finite morphism of degree two.

Let $X$ be an irreducible variety of dimension $d$. A point $p\in X$ is nonsingular, or smooth, whenwhere $T_{p}X$ is the Zariski tangent space; it is a singular point when $\dim T_{p}X>d$. Equivalently, the local ring ${\mathcal{O}}_{X,p}$ is regular exactly at a nonsingular point.

For an irreducible affine variety $X\subseteq {\mathbb{A}}^N$ over a perfect field, the Jacobian criterion expresses the singular locus by the vanishing of the relevant Jacobian minors, so it is Zariski closed. At a point where the tangent-space dimension is minimal it equals $d$; equivalently, some $(N-d)$-rowed Jacobian minor is nonzero. Its nonvanishing locus is therefore a nonempty Zariski-open set contained in the smooth locus. Every nonempty open subset of an irreducible topological space is dense, proving the density of the smooth locus.

Assume the ground field has characteristic other than two. Forthe partial derivatives are $2X_0,\dots ,2X_{n-1},0$. They vanish simultaneously at the unique projective pointThus this point is the singular locus of the projective quadric cone.

For varieties $X,Y$,If $Y$ is smooth of dimension $m$, the product point $(x,y)$ is singular exactly when $x$ is singular. HenceandThis is the singular locus of a product with a smooth variety.

To construct the requested examples, put $r=n-k>0$. If $r\ge 2$, let $Q_r\subseteq {\mathbb{P}}^{r+1}$ be an irreducible quadric cone of dimension $r$ with one singular vertex. If $r=1$, use instead the irreducible cuspidal cubicwhose only singular point is $[0:0:1]$. Let $W$ denote the chosen $r$-dimensional variety and embedin projective space by the Segre embedding. It is irreducible and has dimension $r+k=n$, while its singular locus is the vertex or cusp point times ${\mathbb{P}}^k$, which is nonempty and has dimension exactly $k$. This is the projective variety with a prescribed-dimensional singular locus construction.

Finally, suppose that the irreducible plane curve $C\subseteq {\mathbb{P}}^2$ were smooth of degree $d$. Since it is birational to a smooth projective curve of genus two, its geometric genus would be two. But the genus of a smooth plane curve isand no integer $d$ makes this number equal to two. Therefore $C$ cannot be smooth and must contain a singular point.

For a divisor on an algebraic curve $D={\sum }_p{a}_{p}p$ over an algebraically closed field, its degree isFor a nonzero rational function $f$, its principal divisor iswhere zeros have positive order and poles have negative order.

Now let $D={\sum }_i{a}_i{p}_i$ have degree zero on the projective line. Choose a homogeneous linear form $L_i(X_0,X_1)$ whose zero is $p_i$. Since ${\sum }_i{a}_i=0$, the expressionis homogeneous of degree zero and therefore defines a rational function on ${\mathbb{P}}^1$. Each $L_i$ has one simple zero, and the cancellation of total degree removes any common scaling ambiguity, soThus every degree-zero divisor on ${\mathbb{P}}^1$ is principal. If $E$ and $E^{\prime }$ have the same degree, then $E-E^{\prime }$ is principal, whenceThis is the divisor class on the projective line.

Write $p_{\infty }=[1:0]$. The coordinate $t=X_0/X_1$ has no critical zero or pole in the finite chart. Near infinity use $s=X_1/X_0=t^{-1}$; thenHence the rational differential has one double pole at infinity and no other zero or pole:which represents the canonical divisor of the projective line.

If $D\sim m{K}_{{\mathbb{P}}^1}$, then $D\sim -2m{p}_{\infty }$, and linear equivalence of divisors gives $L(D)\cong L(-2m{p}_{\infty })$. A function in the latter space has no finite poles, so it is a polynomial, and its degree is at most $-2m$. ThereforeThis calculation uses only the rational functions on ${\mathbb{P}}^1$, rather than the Riemann-Roch theorem.

Finally, suppose distinct $p,q\in C$ satisfy $p-q=(f)$. The function $f$ has one simple pole and therefore defines a degree-one finite morphism $f:C\to {\mathbb{P}}^1$. A degree-one finite morphism between smooth projective curves is an isomorphism. It would follow that $C\cong {\mathbb{P}}^1$ and hence that $C$ has genus zero, contrary to the hypothesis. Thus the principal divisor with one simple zero and one simple pole cannot occur on $C$:

For a smooth projective curve $C$ of geometric genus one, the Riemann-Roch theorem saysThe canonical divisor has degree zero and is principal because a nonzero regular differential has no zeros. Hence $K_C\sim 0$, so equivalentlyIn particular, $\ell (D)=\mathrm{deg}D$ when $\mathrm{deg}D>0$.

The group ${\mathrm{Pic}}^0(E)$ is the group of degree-zero divisor classes on $E$, with addition induced by addition of divisors. ConsiderFor surjectivity, let $D$ have degree zero. Since $\mathrm{deg}(D+(O_E))=1$, Riemann--Roch gives $\ell (D+(O_E))=1$. A nonzero element of this space makes $D+(O_E)$ linearly equivalent to an effective divisor of degree one, necessarily $(P)$ for some point $P$. Thus $[D]=[(P)-(O_E)]$.

For injectivity, suppose $(P)-(O_E)$ is a principal divisor. If $P\ne O_E$, its defining function would be nonconstant and would have at most one simple pole, whereas Riemann--Roch gives $\ell ((O_E))=1$, so every such function is constant. Therefore $P=O_E$, and $\iota$ is a bijection.

For a divisor on an algebraic curve $D$ on a smooth projective curve $X$ of genus $g$, the Riemann-Roch theorem stateswhere $K_X$ is a canonical divisor. Taking $D=0$ gives $1-\ell (K_X)=1-g$, so $\ell (K_X)=g$. Taking $D=K_X$ then gives

To obtain a uniform projective embedding, choose a divisor $D$ of degree $2g+1$. Since every divisor appearing below has degree greater than $2g-2$, Riemann--Roch givesThe first two equalities show that the complete linear system of a divisor $|D|$ has no base point. The strict drops in the last two comparisons show respectively that its sections separate distinct points $P,Q$ and tangent directions at $P$. Thus $D$ is a very ample divisor, in accordance with the general fact that a high-degree divisor is very ample on a smooth projective curve, and its sections define a closed embeddingThe ambient dimension therefore depends only on $g$.

The Riemann-Hurwitz formula for a nonconstant morphism $h:C_1\to C_2$ of degree $n$ isChoose a smooth plane quartic $C$, so $g(C)=3$, and form the product of projective varieties $S=C\times C$. This is a smooth projective variety of dimension two. If $B\subset S$ is an irreducible curve, pass to its normalization $\tilde{B}$. At least one coordinate projection $\tilde{B}\to C$ is nonconstant, since otherwise $B$ would be a point. For that projection, Riemann--Hurwitz givesso the geometric genus of $B$ is at least three. Hence $S$ is the required surface; this is the product surface without low-genus curves construction.

Finally let $X\subset {\mathbb{P}}^2$ be a smooth plane curve of degree $d$ and let $p\notin X$. After a projective change of coordinates, take $p=[0:0:1]$. Projection away from $p$ isIts two homogeneous coordinate functions cannot vanish simultaneously on $X$, because their common zero in ${\mathbb{P}}^2$ is $p$. The criterion for a morphism of algebraic varieties therefore shows that the restrictionis a morphism. A fibre is the intersection with a line through $p$, and a general such line meets $X$ in $d$ points counted with multiplicity. Thus the projection of a plane curve from an exterior point has degree $d$.

By the genus of a smooth plane curve, $g(X)=(d-1)(d-2)/2$. Applying Riemann--Hurwitz to $\pi$ and its ramification divisor $R_{\pi }$ givesEvery ramification point contributes at least one to this degree, so

Two irreducible varieties $X$ and $Y$ are birational if they contain nonempty Zariski-open subsets $U\subseteq X$ and $V\subseteq Y$ that are isomorphic. Equivalently, their function fields are isomorphic:For an irreducible variety,An isomorphism of function fields preserves transcendence degree, so $\dim X=\dim Y$. This is the dimension from the function field.

Now let $K/\mathbb{C}$ be a finitely generated field extension. Choose field generators $a_1,\dots ,a_m$ and setThen $A$ is a finitely generated integral $\mathbb{C}$-algebra and $\mathrm{Frac}(A)=K$. The affine irreducible varietyhas function field $K$. Embed $X_0$ as a closed affine subvariety of some affine space ${\mathbb{A}}^N$, identify that affine space with the standard open chart of projective space ${\mathbb{P}}^N$, and take the projective closure $X$ of $X_0$. The closure of an irreducible topological space is irreducible, and $X_0$ is dense and open in $X$. Thus $X$ is a projective variety andThis constructs the projective model of a finitely generated field.

For the curveputIn its function field, the defining equation givesHence $\mathbb{C}(X)=\mathbb{C}(t)$, so $X$ is rational. Equivalently, this is the normalization of y squared equals x times x minus one squared, whose smooth projective normalization is ${\mathbb{P}}^1$ and has geometric genus zero.

A smooth projective curve in the projective plane of degree $d$ has, by the genus of a smooth plane curve formula,Its genus can be zero only for $d=1$ or $d=2$. Conversely, a projective line and every smooth conic over $\mathbb{C}$ are rational, so each is birational to $X$. Therefore the required degrees are exactly

Finally, in ${\mathbb{A}}_{x,y,z}^3$ takeThe polynomial $F$ is irreducible: as a quadratic in $y$, it could factor only if $x(x-1{)}^2$ were a square polynomial, which it is not. Thus $S$ is an irreducible affine hypersurface of dimension two. Its function field isso $S$ is birational to ${\mathbb{A}}^2$.

The partial derivatives areSolving $F=F_x=F_y=0$ gives $x=1$, $y=0$, with $z$ arbitrary. By the Jacobian criterion,This is an irreducible subvariety of dimension one, giving the requested singular cylinder over a nodal curve.