Why does genus 0 imply that the function field has one generator? I have read that given a genus 0 modular curve $X$ defined over a base field  $k$. The corresponding function field k(X) can be generated by one modular function.
I have also read that converse, that given a function field has a single generator, then the curve the function field is defined over must have genus zero.
I cant seem to find references or a good explanation of this, or even if my understanding is okay. Could someone give an explanation on the intuition or provide some reference? I have Diamond and Silverman texts.
 A: I assume that "genus" in your post means "geometric genus".
Let $C$ be an irreducible projective algebraic curve without singularities, defined over the field $k$. Let $F$ be the function field of $C$ and let $K$ be the algebraic closure of $k$ in $F$.
Assume $g(C)=0$ for the genus of $C$ and that $C$ possesses a $k$-rational point $P$. Since the residue field $K(P)$ at the point $P$ is an algebraic extension of $k$, one concludes that $K=k$. Then by the Riemann-Roch-theorem
$\ell(P)=\deg(P)+[K:k]=1+1=2$,
where $\ell(P)$ is the dimension of the $k$-vector space $L(P)$ of rational functions $f\in F$, such that $(f)+P\geq 0$ for the divisor $(f)$ of $f$. There is thus a non-constant function $x\in L(P)$, which then must have $P$ as its only pole, where the pole order is $1$.
In general it is known that the field degree $[F:k(f)]$ equals the degree of the pole divisor of $f$.
In the present case one thus gets $[F:k(x)]=1$, that is $K(C)=k(x)$.
Curves of genus $0$ without a $k$-rational point can have a function field, that cannot be generated by a single element. An example is given by the projective closure of the curve $x^2+y^2=-1$ defined over the reals. Another example is $F:=\mathbb{Q}(x,i)$, $i^2=-1$, which corresponds to the projective line over $\mathbb{Q}(i)$ considered as a curve over $\mathbb{Q}$.
Assume on the other side, that $K(C)=k(x)$ for some rational function $x$. Let $P$ be the pole of $x$, which has degree $1$. For sufficiently large $m$ the Riemann-Roch-theorem gives
$\ell(mP)=\deg(mP)-g(C)+1=m-g(C)+1$
The vector space $L(mP)$ has the basis $1,x,x^2,\ldots ,x^{m}$, hence $\ell(mP)=m+1$ and thus $g(C)=0$.
