[proofplan] The strategy is to prove $\beta$ is a bijection of sets first; the group law on $C$ is then defined by transport along $\beta$, and the group axioms are inherited automatically. Bijectivity rests on two consequences of the genus-$1$ Riemann–Roch theorem: every degree-$1$ divisor $D$ on $C$ has $\ell(D) = 1$, so it has a unique effective representative. Injectivity: $\beta(p) = \beta(q)$ forces $p \sim q$, hence $p$ and $q$ are both unique effective representatives of the same class, so $p = q$. Surjectivity: for $[D] \in \mathrm{Jac}(C)$, the class $[D + p_0]$ has degree $1$ and so contains a unique effective representative — necessarily a single point $r$ — giving $[D] = [r - p_0] = \beta(r)$. [/proofplan] [step:Reduce the group structure on $C$ to the group structure on $\mathrm{Jac}(C)$ via transport] The Jacobian $(\mathrm{Jac}(C), +, 0)$ is an abelian group: it is the kernel of the degree homomorphism $\deg : \mathrm{Pic}(C) \to \mathbb{Z}$, hence a subgroup of the abelian group $\mathrm{Pic}(C)$, with identity the zero divisor class $0 = [0]$. We claim $\beta(p_0) = 0 \in \mathrm{Jac}(C)$. Indeed, $\beta(p_0) = [p_0 - p_0] = [0] = 0$. Once we prove $\beta : C \to \mathrm{Jac}(C)$ is a bijection (Steps 2 and 3), we can transport the group law on $\mathrm{Jac}(C)$ to $C$: \begin{align*} p \oplus_{p_0} q := \beta^{-1}(\beta(p) + \beta(q)). \end{align*} The resulting structure $(C, \oplus_{p_0}, p_0)$ is automatically an abelian group with identity $p_0$ (because $\beta(p_0) = 0$), and $\beta$ is automatically a group isomorphism — these are formal consequences of transport of structure along a bijection. In other words: claims (i) and (ii) of the theorem reduce to bijectivity of $\beta$. [/step] [step:Establish that every degree-$1$ divisor on $C$ has $\ell = 1$] Let $E$ be any divisor on $C$ with $\deg E = 1$. The genus is $g = 1$, so $2g - 1 = 1$, and $\deg E = 1 \geq 2g - 1$. By [Riemann–Roch for Large Degree Divisors](/theorems/2188), \begin{align*} \ell(E) = \deg E - g + 1 = 1 - 1 + 1 = 1. \end{align*} Hence by [Unique Effective Representative When $\ell(D) = 1$](/theorems/2189), the divisor $E$ has a unique effective representative — a unique effective divisor $E' \in \mathrm{Div}(C)$ with $E' \sim E$. We further note: a degree-$1$ effective divisor on $C$ is necessarily a single closed point. An effective divisor $E' = \sum_p n_p \cdot p$ with $n_p \geq 0$ has $\deg E' = \sum_p n_p \cdot [\kappa(p) : k]$. Over the algebraically closed $k$, every closed point has residue field $k$, so $[\kappa(p) : k] = 1$ and $\deg E' = \sum_p n_p$. The unique solution to $\sum_p n_p = 1$ with all $n_p \geq 0$ is $n_{r} = 1$ for some $r \in C$ and $n_p = 0$ for $p \neq r$, i.e., $E' = r$ is a single point. Conclusion: every degree-$1$ divisor on $C$ is linearly equivalent to a unique point of $C$. [/step] [step:Prove $\beta$ is injective using uniqueness of the effective representative] Suppose $\beta(p) = \beta(q)$ for $p, q \in C$. Then $[p - p_0] = [q - p_0]$ in $\mathrm{Pic}(C)$, equivalently \begin{align*} (p - p_0) - (q - p_0) = p - q \end{align*} is principal, so $p \sim q$ as divisors on $C$. The divisor $p$ has degree $1$; by Step 2 it has a unique effective representative. Both $p$ and $q$ are effective divisors of degree $1$ in the class $[p]$, and the unique-effective-representative property forces $p = q$. Hence $\beta$ is injective. [/step] [step:Prove $\beta$ is surjective by lifting Jacobian classes to points] Let $[D] \in \mathrm{Jac}(C)$, so $\deg D = 0$. The shifted divisor $D + p_0$ has degree \begin{align*} \deg(D + p_0) = \deg D + \deg p_0 = 0 + 1 = 1. \end{align*} By Step 2, the class $[D + p_0]$ contains a unique effective representative, which is a single closed point $r \in C$: \begin{align*} D + p_0 \sim r. \end{align*} Subtracting $p_0$ on both sides (linear equivalence respects divisor sums): \begin{align*} D \sim r - p_0, \end{align*} i.e., $[D] = [r - p_0] = \beta(r)$ in $\mathrm{Jac}(C)$. Hence every element of $\mathrm{Jac}(C)$ is in the image of $\beta$, and $\beta$ is surjective. [guided] The crux of surjectivity is: \emph{can we always represent a degree-zero class $[D]$ in the form $[r - p_0]$ for some single point $r \in C$?} Equivalently: can we always choose a representative of $[D]$ that, after adding $p_0$, becomes a single closed point? Step 2 gives an unequivocal yes — and the mechanism is the genus-$1$ specialness of Riemann–Roch. Why does Step 2 hold? The key observation is that the genus is exactly equal to the threshold where the large-degree formula starts working: $g = 1$ implies $2g - 1 = 1$, so already at degree $1$ — the smallest positive degree — the Riemann–Roch correction $\ell(K_C - D)$ vanishes. The formula $\ell(D) = \deg D - g + 1$ then collapses to $\ell(D) = 1$ at $\deg D = 1$, and the linear system $|D|$ has exactly one effective representative. Concretely: given $[D] \in \mathrm{Jac}(C)$, form $D + p_0$. This has degree $1$. Riemann–Roch says it has a one-dimensional space of sections, and its unique effective representative — being a degree-$1$ effective divisor over an algebraically closed field — is a single point $r$. Equivalently, $D + p_0 \sim r$ for a uniquely determined $r$. Translating back: $D \sim r - p_0$, so $[D] = \beta(r)$. The point $r$ is the unique preimage of $[D]$ under $\beta$. This is special to genus $1$. In genus $g \geq 2$, the formula $\ell(D) = \deg D - g + 1$ requires $\deg D \geq 2g - 1 \geq 3$ to even apply, and the "uniqueness" picture breaks: degree-$1$ divisors typically have $\ell = 0$ (no effective representative at all in general). The bijection between the curve and its Jacobian is genuinely a genus-$1$ phenomenon. [/guided] [/step] [step:Conclude that $\beta$ is a group isomorphism and that $(C, \oplus_{p_0}, p_0)$ is an abelian group] By Steps 3 and 4, $\beta : C \to \mathrm{Jac}(C)$ is a bijection. By Step 1, $\beta(p_0) = 0$, the identity of $\mathrm{Jac}(C)$. We now define the binary operation on $C$ by transport: \begin{align*} \oplus_{p_0} : C \times C &\to C, \\ (p, q) &\mapsto \beta^{-1}(\beta(p) + \beta(q)). \end{align*} This is well-defined because $\beta$ is a bijection. The triple $(C, \oplus_{p_0}, p_0)$ is an abelian group: \begin{itemize} \item \emph{Associativity:} for $p, q, r \in C$, \begin{align*} (p \oplus_{p_0} q) \oplus_{p_0} r &= \beta^{-1}\bigl(\beta(p \oplus_{p_0} q) + \beta(r)\bigr) \\ &= \beta^{-1}\bigl((\beta(p) + \beta(q)) + \beta(r)\bigr) \\ &= \beta^{-1}\bigl(\beta(p) + (\beta(q) + \beta(r))\bigr) \\ &= p \oplus_{p_0} (q \oplus_{p_0} r), \end{align*} using associativity of $+$ on $\mathrm{Jac}(C)$ and the identity $\beta(\beta^{-1}(x)) = x$. \item \emph{Identity:} $p \oplus_{p_0} p_0 = \beta^{-1}(\beta(p) + \beta(p_0)) = \beta^{-1}(\beta(p) + 0) = \beta^{-1}(\beta(p)) = p$, similarly on the left. \item \emph{Inverses:} for $p \in C$, define $\ominus p := \beta^{-1}(-\beta(p))$. Then $p \oplus_{p_0} (\ominus p) = \beta^{-1}(\beta(p) - \beta(p)) = \beta^{-1}(0) = \beta^{-1}(\beta(p_0)) = p_0$. \item \emph{Commutativity:} immediate from commutativity of $+$ on $\mathrm{Jac}(C)$. \end{itemize} Finally, $\beta$ is a group homomorphism: \begin{align*} \beta(p \oplus_{p_0} q) = \beta(\beta^{-1}(\beta(p) + \beta(q))) = \beta(p) + \beta(q), \end{align*} using $\beta \circ \beta^{-1} = \mathrm{id}_{\mathrm{Jac}(C)}$. Combined with the bijectivity of $\beta$, this makes $\beta : (C, \oplus_{p_0}, p_0) \to (\mathrm{Jac}(C), +, 0)$ a group isomorphism, completing the proof of (i) and (ii). [/step]