I'm trying to understand the proof of the following problem:
Let $L / K$ be a field extension, and assume that $K$ has $q=p^{r}$ elements. Let $P(X)$ be a irreducible polynomial in $K[X]$. Let $M$ be a subfield of $K(\alpha)$ for $\alpha$ a root of $P(X)$, and suppose that the degree of $K(\alpha)$ over $M$ is $r .$ Show that $P(X)$ factors, in $M[X],$ as a product of $\frac{d}{r}$ irreducible polynomials of degree $r$.
Proof: We have the previous results: (a) $P\left(\alpha^{q}\right)=P(\alpha)^{q}$ for all $\alpha \in L$, (b) If $P(X)$ is irreducible, and has a root $\alpha$ in $L$, then $P(X)$ factors in $L$ as $(X-\alpha)\left(X-\alpha^{q}\right)\left(X-\alpha^{q^{2}}\right) \ldots\left(X-\alpha^{q^{d-1}}\right),$ where $d$ is the degree of $P(X)$. Now, let $Q(X)$ be an irreducible element of $M[X]$ that divides $P(X) .$ Then, since $P(X)$ factors into linear factors in $K(\alpha), Q(X)$ has a root in $K(\alpha) ;$ by (b) this root is of the form $\alpha^{q^{s}}$ for some $s$. Applying (b) with $M$ in place of $K$ and $Q(X)$ in place of $P(X)$ (and noting that $M$ has $q^{\frac{d}{r}}$ elements) we see that the other roots are $\left(\alpha^{q^{s}}\right)^{q^{k} \frac{d}{r}},$ for $0 \leq k \leq r-1,$ so $Q(X)$ has degree $r$ as claimed.
This proof feels a bit too quick for me. Could anyone explain the line "Applying (b) with $M$ in place of $K$ and $Q(X)$ in place of $P(X)$ (and noting that $M$ has $q^{\frac{d}{r}}$ elements) we see that the other roots are $\left(\alpha^{q^{s}}\right)^{q^{k} \frac{d}{r}},$ for $0 \leq k \leq r-1,$ so $Q(X)$ has degree $r$ as claimed?" More precisely,
(1) Why is $0 \le k \le r-1$, and why are the $\left(\alpha^{q^{s}}\right)^{q^{k} \frac{d}{r}}$ distinct for distinct $k$?
(2) How do we conclude $Q(X)$ has degree $r$?
Partial answers (answering only some of the questions) are very welcome! (I will upvote them too)
By the way, this is a problem from a basic algebra course. I do not know Galois Theory.
From the factorization of $P(X)$ we know that the elements $\alpha^{q^i}$, $i=0,1,\ldots,d-1$, are all distinct. It may be better to restate this in the form $$ \alpha^{q^i}=\alpha^{q^j} $$ if and only if $i\equiv j\pmod d$
If $M$ is an intermediate field, $K\subset M\subset K(\alpha)$, by the tower law of field extensions we know that $\ell:=[M:K]$ is a factor of $d=[K(\alpha):K]$. Also, $M$ has $q^\ell$ elements.
As noted, the polynomial $Q(X)$ has a root, say, $\beta=\alpha^{q^s}$. By (b) the other roots of $Q(X)$ are $$ \beta_j:=\beta^{q^{\ell j}} $$ for $j=0,1,\ldots, t-1$, $t=\deg Q(X)$. For all $i$ we have $\beta_i=\alpha^{q^{s+\ell i}}.$
The earlier observation says that $$\beta_j=\beta_i$$ if and only if $$s+\ell i\equiv s+\ell j\pmod d$$ if and only if $$ i\equiv j\pmod {d/\ell}. $$ So there are $d/\ell$ distinct elements $\beta_i$. We know that the zeros of $Q(X)$ are distinct, so we can conclude that $t=d/\ell$.
This answers both parts as we see that $\ell=d/r$.