[proofplan] The hypothesis $\deg D > 2g$ is just enough degree to ensure that both $D$ and $D - p - q$ lie in the "large-degree regime" $\deg \geq 2g - 1$ where Riemann–Roch simplifies: the higher cohomology $\ell(K_C - \cdot)$ vanishes, and $\ell$ is a linear function of the degree. Applying [Riemann–Roch for Large Degree Divisors](/theorems/2188) to both $D$ and $D - p - q$ — and verifying $\deg(D - p - q) \geq 2g - 1$ from the hypothesis $\deg D > 2g$ — yields $\ell(D) - \ell(D - p - q) = 2$ by direct subtraction. [/proofplan]

[step:Verify the large-degree hypothesis for both $D$ and $D - p - q$] Let $p, q \in C$ be closed points (possibly equal). By hypothesis $\deg D > 2g$, hence \begin{align*} \deg D \geq 2g + 1, \end{align*} since both sides are integers. Consequently \begin{align*} \deg(D - p - q) = \deg D - 2 \geq 2g + 1 - 2 = 2g - 1. \end{align*} Thus both $D$ and $D - p - q$ have degree at least $2g - 1$, so [Riemann–Roch for Large Degree Divisors](/theorems/2188) applies to both. [/step]

[step:Compute $\ell(D)$ via Riemann–Roch] By [Riemann–Roch for Large Degree Divisors](/theorems/2188) — whose hypothesis $\deg D \geq 2g - 1$ is verified in Step 1 (in fact $\deg D \geq 2g + 1$) — we obtain $\ell(K_C - D) = 0$ and \begin{align*} \ell(D) = \deg D - g + 1. \end{align*} [/step]

[step:Compute $\ell(D - p - q)$ via Riemann–Roch] Apply [Riemann–Roch for Large Degree Divisors](/theorems/2188) to the divisor $D - p - q$, using the verification $\deg(D - p - q) \geq 2g - 1$ from Step 1. This yields $\ell(K_C - (D - p - q)) = 0$ and \begin{align*} \ell(D - p - q) = \deg(D - p - q) - g + 1 = (\deg D - 2) - g + 1 = \deg D - g - 1. \end{align*} [/step]

[step:Subtract to obtain condition $(*)$]Combining the formulas of Steps 2 and 3: \begin{align*} \ell(D) - \ell(D - p - q) = (\deg D - g + 1) - (\deg D - g - 1) = 2. \end{align*} That is, \begin{align*} \ell(D - p - q) = \ell(D) - 2, \end{align*} which is condition $(*)$ at the pair $\{p, q\}$. Since $p, q \in C$ were arbitrary, $D$ satisfies condition $(*)$ in full.[/step]

[guided]The proof is a direct degree count: when both $D$ and $D - p - q$ are in the large-degree regime, Riemann–Roch reduces to a linear identity, and the difference $\ell(D) - \ell(D - p - q)$ equals the difference of degrees, which is $2$. **Why the threshold $2g$ rather than $2g - 1$?** Condition $(*)$ involves $\ell(D - p - q)$ for *all* pairs $p, q$, so we need [Riemann–Roch for Large Degree Divisors](/theorems/2188) to apply to $D - p - q$ as well, not just to $D$. The hypothesis $\deg D > 2g$ ensures $\deg(D - p - q) \geq 2g - 1$, which is the threshold for Riemann–Roch. The threshold for $D$ alone is $\deg D \geq 2g - 1$, but the threshold for $D - p - q$ is $\deg D \geq 2g + 1$, equivalent to $\deg D > 2g$. **The role of the cohomology vanishing.** Riemann–Roch in its full form reads $\ell(D) - \ell(K_C - D) = \deg D - g + 1$. The "correction term" $\ell(K_C - D)$ measures global sections of the complementary bundle and is hard to compute in general. However, when $\deg D \geq 2g - 1$, we have $\deg(K_C - D) \leq (2g - 2) - (2g - 1) = -1 < 0$, and a divisor of negative degree on a smooth projective curve has no global sections — that is, $\ell(K_C - D) = 0$. So the correction vanishes and $\ell(D) = \deg D - g + 1$. This is the simplification packaged into [Riemann–Roch for Large Degree Divisors](/theorems/2188). **Why does the difference $\ell(D) - \ell(D - p - q)$ reduce to a degree count?** Once both terms are in the linear regime $\ell(\cdot) = \deg(\cdot) - g + 1$, the genus and the constant cancel: \begin{align*} \ell(D) - \ell(D - p - q) = (\deg D - g + 1) - (\deg D - 2 - g + 1) = 2. \end{align*} Outside the linear regime, the correction terms $\ell(K_C - D)$ and $\ell(K_C - (D - p - q))$ would not cancel, and the difference would not simply equal $2$. **Why "$> 2g$" and not "$\geq 2g$"?** Take $\deg D = 2g$ (not strictly greater). Then $\deg(D - p - q) = 2g - 2$, which is the degree of $K_C$. Riemann–Roch for $D - p - q$ then needs the correction term $\ell(K_C - (D - p - q)) = \ell(0) = 1$, which is nonzero — so the linear formula $\ell(D - p - q) = \deg(D - p - q) - g + 1 = g - 1$ might not hold. In particular, condition $(*)$ can fail at $\deg D = 2g$: the canonical embedding theorem [Canonical Embedding Theorem](/theorems/2197) shows that for hyperelliptic curves the canonical divisor $K_C$ — of degree $2g - 2$, which is not in the regime $> 2g$ — fails condition $(*)$. **Strategic role of this theorem.** Together with [Embedding Criterion](/theorems/2195), this gives the standard projective-embedding existence theorem: every smooth projective curve admits an embedding into $\mathbb{P}^N_k$ for some $N$, by taking $D$ to be any divisor of degree $> 2g$ — for instance, $D = (2g+1)p_0$ for a chosen point $p_0$. The dimension of the projective space is $N = \ell(D) - 1 = (2g+1) - g = g + 1$ in this concrete choice, giving an embedding $C \hookrightarrow \mathbb{P}^{g+1}_k$. Smaller embedding dimensions are achieved by careful choice of $D$ — the canonical embedding (when applicable) gives $C \hookrightarrow \mathbb{P}^{g - 1}_k$.[/guided]