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Choose $P\in X(k)$ and put $D=(g+1)P$. The Riemann-Roch theorem givesAs constants supply only a one-dimensional subspace, the Riemann-Roch space $L(D)$ contains a nonconstant rational function $f$. It has no poles except possibly at $P$. It must have a pole, since a global regular function on an irreducible projective variety is constant. Let its pole order at $P$ be $m$; then $1\le m\le g+1$.

On the open set where $f$ is regular, define $\varphi =[f:1]$. At its pole use the other affine chart, $\varphi =[1:1/f]$. The local ring of a smooth algebraic curve is a discrete valuation ring, and the positive valuation of $1/f$ at $P$ makes the latter expression regular there. The two expressions agree on their overlap, so they give a regular map $X\to {\mathbb{P}}^1$.

This map is nonconstant. It is proper because its source is projective and its target is separated, so its image is closed; a nonconstant image in the projective line must be the whole line. Each finite-value fibre consists of zeros of a nonzero rational function $f-a$, and the infinity fibre is the finite pole set. Hence every fibre is finite, giving a quasi-finite morphism. The proper quasi-finite criterion therefore makes $\varphi$ a finite morphism.

Finally its pullback of the infinity divisor is exactly the pole divisor $mP$. The degree of a morphism of curves equals the degree of this pullback, or equivalently the function-field extension degree $[k(X):k(f)]$. ThusThe argument includes inseparable maps: the degree here is the full field-extension degree, not just the separable degree. This proves the gonality bound from Riemann-Roch with an explicitly constructed finite regular map.

Here the degree-zero Picard group of a curve uses tensor product as its group operation. The map is the Abel map of a pointed smooth projective curve. If $\alpha (p)=\alpha (q)$, then $\mathcal{O}(p-q)$ is trivial. By the divisor description of line bundles, there is a nonzero rational function $f$ with principal divisor $(f)=p-q$.

If $p\ne q$, this function has exactly one pole, of order one. It defines a nonconstant morphism of algebraic varieties $X\to {\mathbb{P}}^1$: locally at a pole one uses $1/f$ as the target coordinate. The degree of a morphism of curves is the total pole degree, so it is one. The map is finite, since a nonconstant map of projective integral curves has finite fibers and is proper, and a degree-one finite map to the normal projective line is an isomorphism. Algebraically, on a target affine chart the finite coordinate algebra lies inside the common function field and is integral over the integrally closed coordinate ring, so it equals that ring. Thus $X\cong {\mathbb{P}}^1$ and its genus is zero, contradicting $g\ge 1$. ConsequentlyThis spells out why a principal divisor with one simple zero and one simple pole cannot occur on a curve of positive genus.

If $g=0$, the chosen point $p_0$ is enough to identify $X$ with ${\mathbb{P}}^1$, even over a nonclosed field. By the Riemann-Roch theorem, $h^0(\mathcal{O}(p_0))=2$, since the complementary canonical twist has degree $-3$ and no nonzero section. Besides constants there is a rational function with its sole pole a simple pole at $p_0$, giving the same degree-one isomorphism. On ${\mathbb{P}}^1$, every degree-zero divisor is principal: over an algebraically closed field write a product of powers of linear forms for its finite point coefficients, with the degree-zero condition giving the coefficient at infinity. Over a general field the same construction uses irreducible polynomial factors for closed points, weighted by their degrees. Therefore