Let $F$ be a field containing all $n^{th}$ roots of unity. An $n$-Kummer Extension over $F$ is a Galois extension of the form $E=F[r_1,...,r_m]$ where $r_i^n\in F$ for all $i=1,...,m$.
From Kummer theory, the groups $$\operatorname{kum}\left(E/F\right):=\left(F^{1/n}\right)^\times/F^\times\simeq \operatorname{Gal}(E/F)$$where $$F^{1/n}=\{a\in E: a^n\in F\}.$$
But consider the subgroup $\left(\left<r_1,...,r_m\right>F^\times\right)/F^\times$ of $\operatorname{kum}(E/F)$ generated by the images of $r_1,...,r_m$, i.e. it is the image of $\left<r_1,...,r_m\right>$ under the canonical map $\left(F^{1/n}\right)^\times\to\operatorname{kum}(E/F)$.
Clearly the representative elements $r_1^{d_1}...r_m^{d_m}$ of the cosets $r_1^{d_1}...r_m^{d_m}F^\times$ for $0\leq d_i\leq n-1$ in this subgroup spans the extension $E/F$, so that $$[E:F]\leq\left|\left(\left<r_1,...,r_m\right>F^\times\right)/F^\times\right|\leq \left|\operatorname{kum}\left(E/F\right)\right|=[E:F]$$and actually the kummer group is this subgroup.
I felt this is true in same cases, but I feel wrong since I can't find any book mentioning this. I believe this subgroup is simpler than the form of the Kummer group. What is wrong with my argument?