For the past week, I've been trying to dissect and make rigorous my understanding of Lang's theorem, which says that:
Theorem 9.1 Let $k$ be a field and n an integer $\geq$ 2. Let $a \in k, a \neq 0.$ Assume that for all prime numbers $p$ such that $p \mid n$ we have that $a \notin k^p,$ and if $4 \mid n$ then $a \notin -4k^4.$ Then $X^n - a$ is irreducible in $k[x].$
What follows is a portion of Lang's proof. I cut the argument off where my understanding broke down.
Our first assumption means that $a$ is not a $p$-th power in $k.$ We shall reduce our theorem to the case when $n$ is a prime power, by induction. Write $n = p^rm$ with $p$ prime to $m,$ and $p$ odd. Let
$$x^m - a = \prod^m_{v = 1}(x - \alpha_v)$$
be the factorization of $x^m - a$ into linear factors, and say $\alpha > = \alpha_1$. Substituting $x^{p^r}$ for $x$ we get
$$x^n - a = x^{p^rm} - a = \prod^m_{v = 1}(x^{p^r} - \alpha_v).$$
So far, there is not much to comment on about Lang's set up. His preliminary work is quite straight forward.
We may assume inductively that $x^m - a$ is irreducible in $k[x].$ We contend that $\alpha$ is not a $p$-th power in $k(\alpha).$ Otherwise, $\alpha = \beta^p, \beta \in k(\alpha).$ Let $N$ be the norm from $k(\alpha)$ to $k$ then
$$-a = (-1)^mN(\alpha) = (-1)^mN(\beta^p) = (-1)^mN(\beta)^p.$$
This is where Lang begins to lose me. I understand his field Norm argument, as it follows from substitution and the realization of the field Norm as the product of Galois conjugates. And here $-a$ coincides with this product, hence this equality follows. But I don't see why Lang begins to invoke induction here. If someone could explain how this helps us, I would be extremely grateful.
We're inducting on the number of distinct prime powers dividing $n$, in order to reduce to the case that there's exactly one prime power dividing $n$, i.e. that $n$ is itself a prime power. That case is dealt with in the part of the proof below where you're copying (starting "We now suppose that $n = p^r$ is a prime power.")
Since we're inducting (and we're spotting the base case, for now), if we write $n = p^rm$ with $m$ coprime to $p$, then $m$ has fewer prime power factors than $n$, so we may assume the theorem is already proven for $m$, that is, that $x^m - a$ is irreducible. We need this assumption because we are trying to work in the field $k(\alpha)$, but we would not know this was a field if we didn't know $x^m - a$ was irreducible over $k$. We are going to need that it's a field, because the next step in the argument is to prove that $x^{p^r} - a$ is irreducible over $k(\alpha)$ (which again will follow becuase we're spotting the base case) and use the multiplicativity of degrees of field extensions to get that $x^m - a$ was irreducible over the base.