# Kummer theory and ramified covers

Let $X \to Y$ be a cyclic cover of algebraic varieties, with $Y$ smooth and $X$ normal, say over the complex numbers. Let $G$ denote the Galois group and let $\chi$ be a character. By Kummer theory (aka Hilbert 90), the extension $K(X) / K(Y)$ is generated by a function $f$ such that $f^\sigma / f=\chi(\sigma)$.

Is it true that $f$ does not vanish outside the ramification locus?

The picture I have in mind is $y^n=(x-a_1)\cdots (x-a_r)$ where this is the case, but maybe more complicated things can happen.

The answer is no in general, as shown by the following example.

Let $\pi \colon X \to Y$ be an étale (i.e., unramified) double cover of projective curves over $\mathbb{C}$, with $g(X) \geq 3$ (for instance, we can take $g(X)=3$, $g(Y)=2$).

Then the ramification locus is empty. However, the generator $f$ of the $\mathbb Z / 2$-extension $K(X)/K(Y)$ must necessarily vanish somewhere on $X$.

Indeed, since $X$ is compact, there exist no non-constant holomorphic functions on it, hence $f$ has at least one pole. But the Residue Theorem implies that the number of zeroes of $f$ equals the number of poles (if counted with multiplicities), so $f$ has at least one zero on $X$.

• Thanks for your answer, Francesco. Of course, you are right. But, if I a finite set of points is fixed, is it possible to find $f$ not having zeros nor poles on these points?
– zyx
Jun 21, 2015 at 13:32
• I guess this should be possible modifying with functions in $K(Y)$, am I right?
– zyx
Jun 23, 2015 at 14:33
• Assume that your generator $f$ vanishes only at the ramification points of $\pi \colon X \to Y$. If $g \in K(Y)$ is such that $g$ does not vanish on the branch locus of $\pi$, then $f + \pi^*g$ is a generator of the extension that does not vanish on the ramification locus of $\pi$. Jun 23, 2015 at 18:50
• Sorry, I think the question was not clear. I fix a set of points outside the ramification locus of the morphism. Is it possible to find a generator $f$ not vanishing at these points?
– zyx
Jun 24, 2015 at 3:48
• You can use the same procedure, cant'you? Call $B$ your set of points and let $f$ be a generator of the extension. If $f$ does not vanish at $B$ you are done; otherwise, replace $f$ with $f+\pi^*g$, where $g \in K(Y)$ is any rational function non vanishing at the points of the set $\pi(B)$. In fact, if $b \in B$ we have $$(f+\pi^*g)(p) = f(p) + g(\pi(b)) = g(\pi(b)) \neq 0.$$ Jun 24, 2015 at 5:58