At the end of page 297 of Algebra by Serge Lang, he claims this:
We are working in a field k, and we let α be a root of the reducible (by hypothesis) polynomial:
Xᵖ - a
where a∈k and p is prime, then α generates an extension k(α) of degree d < p over k.
We then see that:
N(α)ᴾ = aᵈ
Where N denotes the norm of k(α) over k. The next part is what I don't understand, he says that because d and p are coprime, a is then a p-th power.
But I don't get it, I don't see why we would be allowed to take d-th roots of the norm of α and stay in k.
Any help would be appreciated.
I don’t think we are just “taking $d$th roots,” since as you suggest this doesn’t make much sense. In particular, we expect that the fact that $p$ and $d$ are coprime will play a role, and it’s not clear how that happens if we just try “taking roots.” (Actually, as you observe, it’s not even clear that it makes sense to “take roots” at all, since there may be no $d$th root to begin with.) We will need to use coprimality of $p$ and $d$ if we hope to make any progress.
With this in mind, suppose we have $a,b\in k$ and coprime integers $m,n\in\mathbb Z$ with $a^n = b^m$. Since $m$ and $n$ are coprime, there are integers $r$ and $s$ with $rm + sn = 1$. Then $$ a=a^{rm + sn}=a^{rm}(a^n)^s=a^{rm}(b^m)^s=(a^rb^s)^m. $$ So $a$ is an $m$th power in $k$.
In your case, take $a=a$, $b=N(\alpha)$, $n=d$, and $m=p$.
I suppose it’s worth mentioning that this is really a group theory problem: the only way we use the assumption that $a,b\in k$ is to be sure that $a$ and $b$ are elements of an abelian group (here $k^*$).