What does it mean if an element generates $F^\times/(F^\times)^n$ for a field $F$? I encountered this phrase in the following excerpt from Milne's Fields and Galois Theory (page 72):

Could someone explain what the underlined part actually means?
 A: Consider an actual example: if $F$ contains a primitive 4th root of unity and $a, b \in F^\times$ then $F(\sqrt[4]{a}) = F(\sqrt[4]{b})$ if and only if $a = bc^4$ or $a = b^3c^4$ for some $c \in F^\times$.  This is when $a$ and $b$ viewed in $F^\times/(F^\times)^4$ generate the same subgroup.
Why is it important that $F$ contains a primitive 4th root of unity here?  For example, isn't it true that from $a = bc^4$ or $a = b^3c^4$ we get $F(\sqrt[4]{a}) = F(\sqrt[4]{b})$ whether or not there is a primitive 4th root of unity in $F$? The subtlety is that if $F$ doesn't contain a primitive 4th root of unity, the notation $F(\sqrt[4]{d})$ for $d \in F^\times$ is possibly ambiguous: there could be a field extension of $F$ containing two different fields that deserve to be described as $F(\sqrt[4]{d})$.  For example, inside of $\mathbf C$ the numbers $\sqrt[4]{2}$ (positive real 4th root of 2) and $i\sqrt[4]{2}$ are each a fourth root of 2 and thus $\mathbf Q(\sqrt[4]{2})$ and $\mathbf Q(i\sqrt[4]{2})$ are isomorphic but distinct subfields of $\mathbf C$ that could be considered $F(\sqrt[4]{b})$ when $F = \mathbf Q$ and $b = 2$. Similarly, the concrete fields $\mathbf Q(\sqrt[4]{2\cdot 81})$ and $\mathbf Q(i\sqrt[4]{2\cdot 81})$ in $\mathbf C$ could be considered $F(\sqrt[4]{b})$ when $F = \mathbf Q$ and $a = 2 \cdot 81$. While $a = b \cdot 3^4$, the fields $F(\sqrt[4]{a}) = \mathbf Q(\sqrt[4]{2\cdot 81}) = \mathbf Q(\sqrt[4]{2})$ and $F(\sqrt[4]{b}) = \mathbf Q(i\sqrt[4]{2})$ are not equal.
If $F$ contains a primitive 4th root of unity, such ambiguity in notation does not occur: in any field extension of $F$ containing two different 4th roots of $a$, they generate the same extension field: if $\sqrt[4]{a}$ is one 4th root of $a$, the other 4th roots of $a$ are $-\sqrt[4]{a}$ and $\pm i\sqrt[4]{a}$, and
$$
F(\sqrt[4]{a}) = F(-\sqrt[4]{a}) = F(i\sqrt[4]{a}) = F(-i\sqrt[4]{a})
$$
since all 4th roots of 1 are in $F$.
It is also important in the theorem you quote that the extension $F(\sqrt[n]{a})$ has degree $n$ over $F$.  This is not always the case. Consider $\mathbf Q(\sqrt[4]{9})$: adjoining to $\mathbf Q$ a root of $x^4 - 9 = (x^2-3)(x^2+3)$. The 4th roots of 9 are $\pm\sqrt{3}$ and $\pm\sqrt{-3}$, and the fields $\mathbf Q(\sqrt{3})$ and $\mathbf Q(\sqrt{-3})$ have degree 4 over $\mathbf Q$, not degree 2 over $\mathbf Q$. These two interpretations of what $\mathbf Q(\sqrt[4]{9})$ means are not only unequal, but are in fact not even isomorphic (as extensions of $\mathbf Q$).
