Let $K/F$ be a Galois extension of number fields with Galois group $G$. Let $E$ be an elliptic curve defined over $F$ and $f \in K(E)^{\times}$ be a function.
Define the trace of $f$ to be $Tr_{K/F}(f)= \sum\limits_{g \in G} g(f)$. If $f$ is nonconstant, then can its trace ever be $0$? If so, can you give me any necessary conditions for this to happen?
EDIT: I wish to assume that some nonzero coefficient of $f$ lies in $F$, although I'm not sure whether I would need one in both the numerator and denominator to prevent the trace being zero. This is to prevent cases like Jykri's example below.
I believe I have now found an answer so I shall post it here and if anyone spots any mistakes then please let me know.
As Alex pointed out above, if $f,f'$ are in the same Galois orbit then the trace of $f-f'$ is certainly zero and I believe that this is only case.
Since $K/F$ is Galois, $K(E)/F(E)$ is also a Galois extension of function fields of the same degree $n$ and isomorphic Galois groups. We may then apply the normal basis theorem which states that the exists $\alpha \in K(E)$ such that the set $\{ g(\alpha) | g \in Gal(K(E)/F(E)) \}$ form a basis for the extension $K(E)/F(E)$.
Now the image of the trace map is nonzero so has rank $1$ as an $F(E)$ vector space. This means the kernel has dimension $n-1$ by rank-nullity theorem.
Notice that the set $\{ \alpha - g(\alpha) | g \in Gal(K(E)/F(E)) \}$ is linearly independent and has size $n-1$. Since every element is in the kernel, this must be a basis and so any element of the kernel has the form $f-g(f)$ as claimed.