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An abelian surface is a two-dimensional abelian variety. It contains no rational curve: a morphism from ${\mathbb{P}}^1$ to an abelian variety is constant. Consequently every abelian surface is a minimal algebraic surface.

An abelian variety of dimension $g$ has complex multiplication when its rational endomorphism ring contains a commutative semisimple $\mathbb{Q}$-algebra of dimension $2g$. For an elliptic curve in characteristic zero, this means its geometric endomorphism ring of an elliptic curve is an order in an imaginary quadratic field. The resulting rank-one Tate module representations connect torsion points to class field theory.

The dual of an abelian variety represents algebraically trivial line bundles rigidified at the origin. Its group law is tensor product. A group homomorphism $f:A\to B$ induces $\hat{f}:\hat{B}\to \hat{A}$ by pullback of these line bundles.

An isogeny is a surjective morphism of algebraic groups with finite kernel. For abelian varieties it is a finite surjective group morphism. An isogeny of elliptic curves is its one-dimensional abelian-variety specialization.

For a line bundle $L$ on an abelian variety $A$, this closed subgroup consists of $a$ such that $T_a^{\ast }L\cong L$. It is a stabilizer of a Picard group class, which need not be the same as the stabilizer of an individual Weil divisor representing that class.

For a commutative group scheme $G$, the multiplication-by-$n$ morphism $[n{]}_G:G\to G$ adds a point to itself $n$ times. On an abelian variety, it is surjective for every nonzero integer $n$.

For a symmetric ample line bundle $L$ on an abelian variety, the canonical height is ${\hat{h}}_L(P)={\lim }_{n\to \infty }{4}^{-n}{h}_L([2^n]P)$. It is quadratic and differs from the associated logarithmic Weil height by a bounded function. The canonical height of an elliptic curve specializes this construction to a one-dimensional abelian variety.

Let $X$ be a connected compact Riemann surface of genus $g$. Choose an oriented 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 complex differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The Riemann bilinear relations begin with the identityTo prove it, cut $X$ along the chosen cycles to obtain a fundamental polygon $P$. On this simply connected polygon, a closed differential form has a primitive $F$, with $dF=\alpha$. Stokes theorem gives ${\int }_P\alpha \wedge \beta ={\int }_{\partial P}F\beta$. The two copies of each edge have opposite orientations, and their values of $F$ differ by the period acquired in passing between them. Pairing the copies of the $a_i$ and $b_i$ edges gives, respectively, $-B_i(\alpha )A_i(\beta )$ and $A_i(\alpha )B_i(\beta )$. Summing proves the identity; the convention $a_i\cdot b_i=1$ fixes its sign.

For holomorphic one-forms $\omega ,\eta$, their wedge product is zero because locally both are multiples of $dz$. Thus the first holomorphic Riemann bilinear relations isFor a nonzero holomorphic one-form $\omega =f(z)dz$, the second relation follows by applying the bilinear identity to $\omega ,\overline{\omega }$:In particular, a holomorphic one-form with all $a$-periods zero must vanish. Since $\dim H^0(X,K_X)=g$, the map taking a holomorphic one-form to its $a$-periods is an isomorphism. We can therefore choose normalized holomorphic one-forms ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_i}{\omega }_j={\delta }_{ij}$. Put $B_{ij}={\int }_{b_i}{\omega }_j$. The first relation applied to ${\omega }_j,{\omega }_k$ gives $B_{jk}=B_{kj}$, and the second applied to ${\sum }_j{c}_j{\omega }_j$ gives $2{\overline{c}}^{,t}(\mathrm{Im}B)c>0$ for $c\ne 0$. ConsequentlyThese conclusions make the construction of the Jacobian variety work. Integration identifies the image of $H_1(X,\mathbb{Z})$ in $H^0(X,K_X{)}^{\ast }$ with the period lattice $\Lambda ={\mathbb{Z}}^g+B{\mathbb{Z}}^g$. The real-linear map $(m,n)\mapsto m+Bn$ is invertible: its imaginary part is $Yn$, and $Y$ is invertible. Hence $\Lambda$ is a discrete lattice of full real rank $2g$, andis a compact complex torus, rather than a non-Hausdorff quotient. Changing the symplectic basis changes the displayed coordinates but not this intrinsic quotient.

The Riemann bilinear relations also supply its polarization of a complex torus. Use the Hermitian form $H(z,w)=z^t{Y}^{-1}\overline{w}$, linear in the first argument, and set $E=-\mathrm{Im}H$. Symmetry of $B$ givesThus $E$ is an integral positive Riemann form on a complex torus. Its matrix on the displayed lattice basis is the unimodular symplectic matrix. The corresponding positive holomorphic line bundle, equivalently the one constructed from theta automorphy in Solution 4, gives a principal polarization of a complex torus. By the Kodaira embedding theorem this makes the complex torus projective, hence an abelian variety. The normalization $E=-\mathrm{Im}H$ here is important because $H$ was chosen linear in its first argument. If $g=0$, the spaces of holomorphic one-forms and periods are zero and the Jacobian variety is a point.

Choose a projective embedding of the abelian surface $X$. By Bertini's theorem, a general hyperplane section $C$ is a smooth projective curve that is connected. It is ample, so $C^2>0$. Choose $p\in C$ and translate the inclusion so it sends $p$ to the identity of $X$. The universal property of the Jacobian variety then gives a homomorphism of abelian varietieswhose composition with the based Abel map of an algebraic curve is this translated inclusion.

The image is a closed connected algebraic subgroup, since $J(C)$ is projective, and it contains a translate of $C$. If the image had dimension one, that irreducible curve would be the image subgroup itself. A distinct coset is disjoint from it and has the same numerical divisor class: translation acts trivially on cohomology, being homotopic to the identity on the complex torus. Its self-intersection would therefore be zero, contradicting $C^2>0$. The image has dimension two and is all of $X$. Thus $\Phi$ is surjective, as expressed by the principle that an ample curve on an abelian surface generates the surface.

Throughout the algebraic questions, $C$ is a smooth projective connected algebraic curve over the algebraically closed field $k$. This is the curve convention needed for the Jacobian variety; for an arbitrary singular or nonproper curve the assertions need not hold. Choose $P_0\in C(k)$ and write $\pi :C\times T\to T$.

The degree-$d$ Picard functor of a curve iswith pullback as its action on morphisms. Equivalently, normalize $\mathcal{L}$ by tensoring with ${\pi }^{\ast }(\mathcal{L}{|}_{P_0\times T}{)}^{-1}$ and remember the resulting trivialization along $P_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C\times T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit; preservation of this trivialization forces that unit to be $1$. Thus normalized line bundles have effective descent and the displayed functor is already the appropriate sheaf, rather than just a list of classes over fields.

Here is a construction of its representing scheme. The symmetric product of a curve $C^{(r)}$ represents relative effective degree-$r$ Cartier divisors. It is projective, and its completed local ring at $\sum m_i{Q}_i$ is a power-series ring in the elementary symmetric coefficients of $m_i$ local parameters at each $Q_i$. Thus it is smooth of dimension $r$, in every characteristic. PutThis is open by the semicontinuity theorem for coherent cohomology. It is nonempty: choose $g$ points successively imposing independent vanishing conditions on $H^0(C,K_C)$, so that $H^0(C,K_C-D)=0$, and use Serre duality. The Riemann-Roch theorem gives $h^0({\mathcal{O}}_C(D))=1$ on $U$.

For any fixed line bundle $B$ of degree $g-d$, use a copy $U_B$ of $U$ with the universal normalized line bundle $\mathcal{O}(D)\otimes B^{-1}$. This represents exactly the open subfunctor on which $H^1({\mathcal{L}}_t\otimes B)=0$. Indeed, cohomology and base change for line bundles on a curve makes ${\pi }_{\ast }(\mathcal{L}\otimes B)$ a rank-one vector bundle. The evaluation morphismhas a unique relative zero Cartier divisor of degree $g$ and recovers the point of $U_B$. No choice of a global generator on $T$ is needed. Conversely that zero Cartier divisor recovers $\mathcal{L}$ modulo a bundle from $T$.

These charts cover: given any degree-$d$ line bundle $L$, choose a nonspecial effective degree-$g$ Cartier divisor $D$ and take $B=\mathcal{O}(D)\otimes L^{-1}$. On the overlap of $U_B$ and $U_{B^{\prime }}$, the unique effective representative of $\mathcal{O}(D)\otimes B^{-1}\otimes B^{\prime }$ is constructed by the same rank-one evaluation map. It therefore gives a regular transition morphism; uniqueness gives both the inverse and the cocycle identity. Gluing gives a scheme $J^d$ representing ${\mathcal{P}}^d$, with a universal normalized line bundle.

The construction is of finite type, not an infinite union of indispensable charts. Choose $m\ge 2g-1$ and a fixed line bundle $F$ of degree $m-d$. Every $L\otimes F$ has a nonzero section by the Riemann-Roch theorem, so the universal effective Cartier divisor gives a surjection $C^{(m)}\to J^d$. The inverse images of the charts cover the projective, hence quasi-compact, symmetric product of a curve $C^{(m)}$. A finite subcover gives finitely many charts for $J^d$. This surjection also makes $J^d$ irreducible. The charts show that it is smooth of dimension $g$.

For properness, apply the valuative criterion for properness. Let $R$ be a discrete valuation $k$-algebra with fraction field $K$, and let $L_K$ be a degree-$d$ line bundle on $C_K$. A rational section represents it by a Cartier divisor. Its closure in the regular surface $C_R$ is again a Cartier divisor, since regular local rings are factorial. Thus $L_K$ extends to a line bundle on $C_R$, whose fibre degree remains $d$ by constancy of line bundle degree in a family. Two extensions with the same generic class differ by a vertical Cartier divisor. The special fibre is irreducible, so that Cartier divisor is an integer multiple of the entire fibre, the principal Cartier divisor of a power of a uniformizer. The extensions therefore give the same normalized Picard class. This proves existence and uniqueness in the valuative criterion, hence separatedness and properness of the finite-type scheme $J^d$.

Consequently ${\mathrm{Pic}}_C^d={\mathrm{Jac}}_C^d$ is a smooth proper $k$-variety of dimension $g$. Tensoring by ${\mathcal{O}}_C(-d{P}_0)$ identifies it with $J^0$. On $J^0$ the tensor product and inverse of line bundles induce the algebraic group operations; it is the Jacobian variety, an abelian variety.

We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A={\mathrm{Pic}}^0(C)$, an abelian variety of dimension $g$.

First compute the prime-to-characteristic torsion of an abelian variety. The differential of $[N]:A\to A$ at the origin is $N\mathrm{id}$, and translation gives the same assertion everywhere. Consequently $[N]$ is an étale morphism, its fibres have dimension zero, and properness makes it finite. Its image is closed of dimension $g$, hence all of $A$. Choose an ample line bundle $L_0$ on $A$ and put $L=L_0\otimes [-1{]}^{\ast }{L}_0$, which is symmetric and ample. The Theorem of the Cube gives the recurrenceAfter fixing a trivialization at the origin, induction from $n=0,1$ gives $[n{]}^{\ast }L\cong L^{\otimes n^2}$. Comparing top intersection products under the finite map $[N]$ givesThe intersection number is positive because $L$ is ample, so $\mathrm{deg}[N]=N^{2g}$. Since $[N]$ is étale and $k$ is algebraically closed field, its kernel has precisely $N^{2g}$ points.

For a prime power ${\ell }^r\mid N$, write the finite abelian group $A[{\ell }^r]$ as a product of cyclic groups of orders ${\ell }^{a_j}$, with $1\le a_j\le r$. Its subgroup killed by $\ell$ is $A[\ell ]$, so the number of factors is $2g$. Its total order is ${\ell }^{2gr}$, so $\sum a_j=2gr$ forces every $a_j=r$. Combining the prime-power factors by the Chinese remainder theorem provesThe same reasoning works for the dual abelian variety $\hat{A}$, also of dimension $g$.

To define the Weil pairing for an abelian variety on two copies of $A[N]$, we first justify the required theta identification with $\hat{A}$. Identify $X={\mathrm{Pic}}^{g-1}(C)$ with an $A$-torsor and use ordinary translation $T_q(x)=x+q$. The Theorem of the square, a consequence of the Theorem of the Cube, makesa group homomorphism $A\to \hat{A}$. Fix the negative Abel map of an algebraic curve $j:C\to A$, $P\mapsto [P_0-P]$, and let $\rho :\hat{A}\to {\mathrm{Pic}}^0(C)=A$ be pullback by $j$. Pullback by a translate of $j$ gives the same map: algebraically trivial line bundles on an abelian variety are translation-invariant as Picard classes, by the Theorem of the square.

For any $q\in A$, choose a degree-$g$ class $D$ such that both $D$ and $D+q$ have one-dimensional section spaces. This is possible because the nonspecial degree-$g$ classes form a nonempty open subset of the irreducible Picard scheme of a curve, and its translate has a nonempty intersection with it. Since $T_q{\varphi }_D={\varphi }_{D+q}$, the restriction of the theta divisor to an Abel curve proved above givesEach ${\varphi }_D$ differs from $j$ by a translation. Thus $\rho \lambda ={\mathrm{id}}_A$. This equality on closed points is equality of morphisms, since $A$ is reduced and the target is separated. In particular $\lambda$ is a closed immersion, being a section of the separated morphism $\rho$. Its image has dimension $g=\dim \hat{A}$, so it is all of the reduced irreducible abelian variety $\hat{A}$. Therefore $\lambda$ is an isomorphism of schemes. This proves the needed theta autoduality of a Jacobian, including possible infinitesimal issues in positive characteristic.

Here is a definition of the Weil pairing for an abelian variety that also proves nondegeneracy. For $M\in \hat{A}[N]$, the Theorem of the Cube gives $[N{]}^{\ast }M\cong M^{\otimes N}\cong {\mathcal{O}}_A$. Choose a trivialization of $[N{]}^{\ast }M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N{]}^{\ast }M$ becomes multiplication by scalars under this trivialization: these scalars are constant because every invertible regular function on the proper connected abelian variety $A$ is constant. They form a characterwhose values lie in ${\mu }_N$, since $NP=0$ for every $P\in G$. Rescaling the trivialization changes none of these scalars. Translation gives multiplicativity in $P$, and tensor product gives ${\chi }_{M\otimes M^{\prime }}={\chi }_M{\chi }_{M^{\prime }}$. DefineThis fixes a convention for the Weil pairing for an abelian variety; reversing the convention for the descent action inverts all values and changes no nondegeneracy assertion.

If ${\chi }_M=1$, the trivialization is equivariant for the deck group. Descent then makes $M$ the trivial line bundle on the target. Thusis injective. The source and target both have $N^{2g}$ elements, since $k$ contains all $N$ distinct roots of unity and $A[N]\cong (\mathbb{Z}/N{)}^{2g}$. Hence it is an isomorphism. Characters of this group separate its points, so the Weil pairing for an abelian variety is nondegenerate in both variables. Composing with the isomorphism $\lambda$ proves the pairing on $A[N]$ is perfect.

Pull $L$ back by the morphism of varietiesThe fiber of this line bundle over $x$ is ${\phi }^{\ast }{T}_x^{\ast }L$. An abelian variety is connected, so the preceding constancy of line bundle degree in a family applies. Consequently $\boxed{\mathrm{deg}({\phi }^{\ast }{T}_x^{\ast }L)=\mathrm{deg}({\phi }^{\ast }L)}$ for every $x\in X$.

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.

The irregularity of an algebraic surface and the geometric genus of an algebraic surface are respectively

If $\pi :X^{\prime }\to X$ is the blowup at a point with exceptional divisor $E$, thenA holomorphic two-form on $X$ pulls back to one on $X^{\prime }$. Conversely, a holomorphic two-form on $X^{\prime }$ descends across $E$: locally it is a form on the punctured smooth surface $X\setminus \{p\}$, and its coefficients extend over the codimension-two point $p$. Equivalently, ${\pi }_{\ast }{K}_{X^{\prime }}=K_X$. Pullback is therefore an isomorphismwhich proves the birational invariance of the geometric genus of a surface in this case.

Let $X=C\times D$, where both smooth projective curves have positive genus. The Künneth theorem gives the irregularity of a product of curvesIf a surface is a hypersurface in projective space, dimension forces it to be a smooth hypersurface $Y\subseteq {\mathbb{P}}^3$. Fromand the intermediate cohomology vanishing for line bundles on projective space, one obtains $H^1(Y,{\mathcal{O}}_Y)=0$. Thus every such hypersurface has irregularity zero, whereas $X$ has positive irregularity. Hence $C\times D$ is not isomorphic to a hypersurface. This is the product of positive-genus curves is not a projective hypersurface obstruction.

The Albanese variety $\mathrm{Alb}(X)$ is the universal abelian variety receiving a pointed morphism from $X$, andLet the smooth image curve of the Albanese morphism be $C$ of genus $g$. Pullback of holomorphic one-forms along the dominant map $X\to C$ is injective, giving $g\le q(X)$. On the other hand, the universal property of the Jacobian extends $C\to \mathrm{Alb}(X)$ to a homomorphismThe Albanese image generates the entire Albanese variety, so this homomorphism is surjective and $q(X)\le \dim \mathrm{Jac}(C)=g$. Therefore the Irregularity from a smooth Albanese curve image is

For an abelian variety $X$, consider the commutator morphismIt is the identity whenever either coordinate is the identity. The Mumford rigidity lemma applied successively to the two complete connected factors makes $c$ constant everywhere; its value at $(e,e)$ is $e$. Therefore every pair of points commutes, so the group law is commutative.

The Theorem of the square says that for an abelian variety $X$, a line bundle $L$, and $x,y\in X(k)$,where $T_x$ denotes translation by $x$.

The Theorem of the square states that for a line bundle $L$ on an abelian variety $X$ and $x,y\in X(k)$,