In Lemma 4.7 of AKS, why do distinct polynomials map to distinct elements modulo h(x)? I am reading Primes is in P by Agrawal, Kayal and Saxena, and I can't understand part of the proof of Lemma 4.7 (already the subject of two questions here: PRIMES in P paper - Lemma 4.7 - why are the polynomials $X^m$ distinct in $F$? and Primes is in P, proof of hendrik Lenstra Jr. lemma).
Let $\DeclareMathOperator{\ord}{ord}\ord_r(p)$ mean the order of $p$ modulo $r$, i.e. the least $k$ such that $p^k \equiv 1 \pmod r$.
We have $p$ a prime and $r$ an integer, $p > r$,
and a polynomial $h(x)$ which is an irreducible factor of the $r$th cyclotomic polynomial $Q_r(x)$ over the finite field $F_p$; the degree of $h(x)$ is $\ord_r(p)$.
Some other numbers involved are $n$, $t$, and $\ell$; I'd hope you don't need to worry about their details. But in case you do, $n$ is a multiple of $p$ with $\ord_r(n) > \log(n)^2$, where $\log$ is the binary log;
$\ell = \lfloor{\sqrt{\phi(r)}\log(n)}\rfloor$, where $\phi$ is Euler's totient function;
and $t$ is the number of elements of $I := \{ (\frac{n}{p})^i p^j \mid i,j \geq 0\}$ which are distinct modulo $r$; these $t$ residues form a group $G$.
Let $P$ be the set of all polynomials of the form $\prod_{a=0}^\ell (x + a)^{e_a}$, with powers $e_a \geq 0$.
The statement I am confused about says that any two distinct polynomials in $P$ of degree less than $t$ map to different elements in the field $F := F_p[x]/(h(x))$.
I'll reproduce the beginning of the proof. "$m$ is introspective for $f$" means $f(x)^m = f(x^m) \pmod{x^r-1,p}$.

First note that since $h(x)$ is a factor of the cyclotomic polynomial
  $Q_r(x)$, $x$ is a primitive $r$th root of unity in $F$.
We now show that any two distinct polynomials of degree less than $t$
  in $P$ will map to different elements in F. Let $f(x)$ and $g(x)$ be two such
  polynomials in $P$. Suppose $f(x) = g(x)$ in the field $F$. Let $m \in I$. We also
  have $f(x)^m = g(x)^m$ in $F$. Since $m$ is introspective for both $f$ and $g$, and
  $h(x)$ divides $x^r − 1$, we get:
  \[
  f(x^m) = g(x^m)
  \]
  in $F$. This implies that $x^m$ is a root of the polynomial $Q(Y) = f(Y) − g(Y)$
  for every $m \in G$. Since $(m, r)=1$ ($G$ is a subgroup of $\mathbb{Z}_r^*$), each such $x^m$ is a primitive $r$th root of unity. Hence there will be $|G| = t$ distinct roots of $Q(Y)$
  in $F$. However, the degree of $Q(Y)$ is less than $t$ by the choice of $f$ and $g$. This
  is a contradiction and therefore, $f(x) \neq g(x)$ in $F$.

But, isn't $Q(Y)$ just identically zero, since $f(x) = g(x)$, so $f(Y) = g(Y)$?
 A: Reading through the paper suggested by Will Jagy in the comments, It is easy to determine whether a given integer is prime by Andrew Granville, seems to have clarified this.
This part of AKS Lemma 4.7 is addressed in Granville's Lemma 4.3 on page 19, which tackles the contrapositive statement:
If $f(x) \equiv g(x) \mod (p, h(x))$, where $f$ and $g$ in $P$ have degree less than $t$, then $f(x) \equiv g(x) \mod p$.
The difference of the two polynomials is now called $\Delta(y)$, and I take it that it's understood to live in $\mathbb{F}[y]$.
I think the part I was missing is the reinterpretation of $f$ and $g$ here: 
Rather than a sum of coefficients times powers of $x$ the element of $\mathbb{F}$
(an $r$th root of unity),
we are going back to the polynomials with integer coefficients, which we can
evaluate at arbitrary elements of $F$.
When evaluated at the $t$ elements of $\mathbb{F}$ of the form $x^k$ for $k \in G$, $\Delta(x^k)$ is zero, so $\Delta$ must be the zero polynomial modulo $(p, h(x))$.
Then it says "which implies that $\Delta(y) \equiv 0 \pmod p$ since its coefficients are independent of $x$"—that is, because all its coefficients are integers.
[As a side note, I really wish people would stop saying "$f(x)$" when they just mean the polynomial $f$ itself, not an example of its output. That alone would help to make this situation clearer.]
