= Solution
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> gives
$$
\ell(D)-\ell(K-D)=\deg D+1-1=1.
$$
Since $K-D$ 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 $P=Q$. Consequently
$$
\iota:E(k)\longrightarrow\operatorname{Pic}^0(E),\qquad P\longmapsto[P-O]
$$
is 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. Define
$$
\boxed{P\oplus Q=\iota^{-1}\bigl(\iota(P)+\iota(Q)\bigr).}
$$
Equivalently, $R=P\oplus Q$ 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 $P'$ with $P+P'\sim2O$. 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 \b[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.
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