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For a smooth degree-$d$ projective plane curve $X$, the adjunction formula gives $K_X\cong {\mathcal{O}}_X(d-3)$. A hyperplane section has degree $d$, so

For a projective variety $X\subseteq {\mathbb{P}}^n$ and a projective hyperplane $H$, the hyperplane section is the scheme-theoretic intersection $X\cap H$. The defining linear equation restricts to a global section of the line bundle ${\mathcal{O}}_X(1)$.

To view its zero scheme as an effective Cartier divisor, the restricted section must be a non-zero-divisor locally. For an integral $X$, this holds when $H$ does not contain $X$. If $X\subseteq H$, the restriction is the zero global section and the intersection is all of $X$, rather than a codimension-one Cartier divisor. More generally one must avoid the associated points that would make the local equation a zero divisor.

For a projective plane curve not contained in a line $H$, intersection multiplicities record tangencies in the section. The hyperplane-section divisor of a projective plane curve has total degree equal to the degree of the curve. The restricted bundle ${\mathcal{O}}_X(1)$ exists whether or not the particular section is regular.

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.

Write $L\cong H_1\otimes H_2^{\ast }$ as in part (d). Each very ample line bundle is the pullback of $\mathcal{O}(1)$ under a projective embedding. A hyperplane section therefore supplies an effective divisor on a complex manifold $D_i$ with $H_i\cong \mathcal{O}(D_i)$. HenceEvery Picard group class is therefore in the image of the divisor-to-Picard map:

The adjunction formula gives $K_X\cong {\mathcal{O}}_X(d-3)$. Since a hyperplane section has degree $d$, the canonical degree of a smooth plane curve is

The intersection with a line $\ell$ is a hyperplane section:The canonical bundle of a smooth projective hypersurface formula, obtained from the adjunction formula, gives for a smooth degree-four curve in ${\mathbb{P}}^2$Thus ${\mathcal{O}}_C(D)\cong K_C$, so the divisor class of $D$ is the canonical divisor class. Therefore $D$ is a canonical divisor. Its degree is also $4=2g-2$, consistently with $g=3$.