# If $\operatorname{Deg} D=0$ and $\dim L(D)=1$, then $D$ is principal

Let $X$ a compact Riemann surface and $D$ a divisor with $\operatorname{Deg} D=0$ and $\dim L(D)=1$. Why is then $D$ principal?

I already have seen the comments to this answer, but somehow I don't get it.

My attempt so far: If $\dim L(D)=1$, then there is some meromorphic function $f$ such that $\operatorname{div} f \ge -D$, and every other element of $L(D)$ is a multiple of it.
Probably, $D=\operatorname{div} \frac 1 f$, which would follow if $\operatorname{div} f = -D$, but why is that?

## 1 Answer

If $f\ne 0\in L(D)$, then $\operatorname{div}f + D\ge 0$. However, both $\operatorname{div}f$ and $D$ have degree 0, so this implies that in fact $\operatorname{div}f+D=0$, or $\operatorname{div}f = -D$. Then since $X$ is compact, meromorphic functions are determined up to scaling by their divisors on $X$, so $L(D)=\langle f\rangle$. In particular, this implies that $L(D)$ is one dimensional if it is nonzero.

Summarizing, when $\deg D=0$, if $L(D)\ne 0$, then for any $f\ne 0\in L(D)$, $D=\operatorname{div} 1/f$, and $f$ is a basis for $L(D)$.

• Where did you use that degree of $D$ is zero? – klirk Aug 10 '17 at 21:03
• When I said div f and D have degree zero, because that implies div f +D has degree zero, which I then used to conclude that because div f+D is effective and degree zero it is in fact identically zero – jgon Aug 10 '17 at 21:06
• Sorry, but I am unfamiliar with the notion effective, how is that defined? – klirk Aug 10 '17 at 21:08
• Effective means greater than or equal to zero – jgon Aug 10 '17 at 21:09
• No problem. I get it now. For some reason I did not think of $A:=\operatorname{div} f +D$ as a divisor itself. But yes, if $A\ge 0$ and $\operatorname{Deg} A=0$ then $A=0$. Thank you very much. – klirk Aug 10 '17 at 21:15