Problem of the minimal polynomial I'm new to the minimal polynomial. Here is a problem and its solution that I translated from Korean (so it might contain some errors):

Problem
Let $n$ be a positive integer. Let $M_n(\mathbb C)$ be the vector space of all $n$-by-$n$ complex matrices. Let $A\in M_n(\mathbb C)$ satisfies $A^6 - A^3 + I = 0$.  Let $T:M_n(\mathbb C)\to M_n(\mathbb C)$ be a linear operator such that $T(B)=AB$. Show that $T$ is diagonalizable.
Solution
  
  
*
  
*From,$A^6 - A^3 + I = 0$, $(-A^3)^6=A^{18}=I$.
  Thus the minimal polynomial $f(x)$ of $A$ divides $x^{18}-1$ and has distinct complex roots.
  
*Meanwhile, for a nonnegative integer $k$, since $T^k(B)=A^kB (B \in V)$, $f(T)=0$.
  
*Conversely, for a polynomial $g(x)$ s.t $g(T)=0$, $g(A)=0$ because $g(T)B=g(A)B=0$.
  
*Thus the minimal polynomial of $T$ is $f(x)$.
  
*Since $f(x)$ has distinct complex roots, $T$ is diagonalizable.

My Question:
a) I cannot understand 1). Why does  $x^{18}-1$ have distinct roots?
Also I understood that it has 18 distinct roots. Is that correct? (Or maybe two?)
b) Why do we need 2)~4)? Isn't it trivial that the minimal polynomial of a matrix and the corresponding transformation are same?
c) So the result says that $A$ was 18-by-18 matrix, right?
Thank you.
 A: To answer your questions:
a) This is a fact about the complex numbers which is worth remembering: for any $n = 1,2,3,\dots$ the roots of the equation $x^n = 1$ are
$$
x = \exp(2 \pi ki/n), \quad k = 0,1,2,\dots,n-1
$$
(note: $\exp(i \theta) = e^{i\theta} = \cos(\theta) + i \sin \theta$).
These roots are called the "$n$th roots of unity".  All $n$ of these roots are distinct.
b) Note that $T:M_n(\Bbb C) \to M_n(\Bbb C)$ is not the transformation that one would normally associate with the matrix $A$. The transformation that you're thinking of is $L_A: \Bbb C^n \to \Bbb C^n$ given by $x \mapsto Ax$.  Yes, both transformations involve multiplication by $A$, but the spaces involved are different.  $T$ is a map between two vector spaces of dimension $n^2$ whereas $L_A$ is a transformation between spaces of dimension $n$.
c) There is no reason to suppose that $A$ was $18 \times 18$.  In fact, we could make a $6 \times 6$ matrix with the required properties, for example, by using the companion matrix associated with the given polynomial.
In fact, if we simply take $A = \omega I$ where $\omega$ is any solution to $\omega^6 - \omega^3 + 1 = 0$, then we have a satisfactory matrix which could work for any $n$.

To your comment below:
To see 2) in detail, what we've really said is that if $g$ is any polynomial such that $g(A) = 0$, then we have $g(T) = 0$.  This is true because for any $B$, 
$$
[g(T)](B) = g(A)B = 0B = 0
$$
(However, since the polynomials satisfying $g(A) = 0$ are just the multiples of $f$, it was enough to consider $f(T))$
In 3), we show that if $g$ is any polynomial such that $g(T) = 0$, then we have $g(A) = 0$.  In particular: if $g(T) = 0$, then we can say that for every $B$, we have
$$
[g(T)](B) = g(A)B = 0
$$
Now, it's up to you to show that because $g(A)B = 0$ for every $B \in M_n$, it must be the case that $g(A) = 0$.
Between parts 2) and 3), we have shown that a polynomial $g$ will satisfy $g(T) = 0$ if and only if it satisfies $g(A) = 0$.  Since we're looking at the same set of polynomials, of course the minimal polynomial (the lowest common divisor to this set) will be the same.
