Evaluate an expression as an element of $F$ Let $f(x)=2x^{2}+2x+1\in \mathbb{Z}_{3}[x]$
The first part of the problem i'm trying to solve was to prove that $$\frac{\mathbb{Z}_{3}[x]}{f(x)\mathbb{Z}_{3}[x]}$$ is a field. (stating clearly any results you assume)
My proof follows as: 
$f(x)=2x^{2}+2x+1 \in \mathbb{Z}_{3}[x]$ Now $f(0)=1\ne 0$, $f(1)=5=2\ne 0$,$f(2)=13=1\ne 0$ , hence $f(x)$ has no roots in $\mathbb{Z}_{3}[x]\implies f(x)$ has no factors in $\mathbb{Z}_{3}[x]$, and since $\deg f(x)\le 3\implies f(x)$ is irreducible in $\mathbb{Z}_{3}[x] \implies \frac{\mathbb{Z}_{3}[x]}{f(x)\mathbb{Z}_{3}[x]}$ is a field as required. 
The second part of the problem is as follows; Now let $\alpha$ be a root of $f(x)$ and let $F=\mathbb{Z}_{3}[\alpha]$. Evaluate $(2+2\alpha)^{2}-(1+2\alpha)$ as an element of $F$. 
This is the part i'm unsure on what to do, i would appreciate it if anyone could explain what i need to do here and a hint on how to do so. what result am i assuming for the first part of the problem, is this if $f(x)$ is irreducible in $F$ then $\frac{F}{f(x)F}$ is a field?
Thanks for taking time to read through the question
 A: Let's talk about the first part of the problem first:
Given the quadratic polynomial
$f(x) = 2x^2 + 2x + 1 \in \Bbb Z_3[x], \tag 1$
we might as well get rid of the coefficient $2 \in \Bbb Z_3$ of the $x^2$ term, which we can do by taking
$g(x) = 2f(x) = 2\cdot 2 x^2 + 2 \cdot 2 x + 2 \cdot 1 = 1 x^2 + 1 x + 2 = x^2 + x + 2, \tag 2$
where we have used the fact that $2^2 = 2 \cdot 2 = 1$ in $\Bbb Z_3$, which also incidentally yields
$f(x) = 1 \cdot f(x) = 2 \cdot 2 f(x) = 2g(x); \tag 3$ 
it is clear that $f(x)$ is irreducible if and only if $g(x)$ is, and furthermore that their principle ideals are the same,
$\langle f(x) \rangle = f(x) \Bbb Z_3[x] = g(x) \Bbb Z_3[x] = \langle g(x) \rangle \tag 4$
which readily follows from the fact that $2 \in \Bbb Z_3[x]$.  So if we can look ahead a little and realize that we will eventually be taking $f(x)\Bbb Z_3[x]$ and/or $g(x) \Bbb Z_3[x]$, we can divide oout the factor of $2 \in \Bbb Z_3$ and deal instead with the monic polynomial $g(x)$; it just makes things a little easier.
So, then, why is $\Bbb Z_3[x] / g(x) \Bbb Z_3[x]$ a field?  Well, since $g(x)$ is a quadratic, it is reducible if and only if it has a zero $\alpha \in \Bbb Z_3$, for then we may write
$g(x) = (x - \alpha)(x - \beta)\tag 5$
for some $\alpha, \beta \in \Bbb Z_3$; but we have
$g(0) = 2, \; g(1) = 1, \; g(2) = 2, \tag 6$
and thus $g(x)$, having no zeroes, is irreducible.
Now I take it as well-known that $F[x]$, for any field $F$, is a principal ideal domain; that is, every non-trivial ideal in $F[x]$ is of the form $\langle q(x) \rangle = q(x)F[x]$ for some $0 \ne q(x) \in F[x]$; we apply this notion to show that $\langle q(x) \rangle$ is maximal when $q(x)$ is irreducible; for if $\langle q(x) \rangle$ were not maximal, then we would have some proper ideal $P$ with
$\langle q(x) \rangle \subsetneq P \subsetneq F[x]; \tag 7$
since $P$ must be principal, we also have
$P = \langle p(x) \rangle = p(x)F[x] \tag 8$
for some $0 \ne p(x) \in F[x]$; then by (7),
$q(x) \in p(x)F[x], \tag 9$
whence
$q(x) = p(x)d(x) \tag{10}$
for some $d(x) \in F[x]$; we observe that $p(x)$ cannot be a mere constant, since with $p(x) = p_0 \ne 0 \in F$ we would have, for any $r(x) \in F[x]$,
$r(x) = (p_0^{-1}p_0)r(x) = (p_0^{-1}r(x))p_0 \in P \Longrightarrow F[x] \subset P \Longrightarrow P = F[x], \tag{11}$, 
and $P$ could not be proper; thus $\deg p(x) \ge 1$; similarly we must have $\deg d(x) \ge 1$ since otherwise $d(x) = d_0 \ne 0$ a constant and then from (10) 
$q(x) = d_0 p(x), \tag{12}$
or
$p(x) = d_0^{-1} q(x) \in \langle q(x) \rangle, \tag{13}$
whence
$P \subset  \langle q(x) \rangle, \tag{14}$
also contradicting (7); thus we have $\deg d(x) \ge 1$ and  $q(x)$ is reducible in $F[x]$, also contrary to our hypotheses.  The only other option is that the ideal $\langle q(x) \rangle$ is indeed maximal in $F[x]$, and thus our assertion is proved.  Furthermore, the maximality of $q(x)$ implies that $F[x] / \langle q(x) \rangle$ is a field.
Since we have seen in quite general terms that $q(x)$ is irreducible in $F[x]$ implies that $\langle q(x) \rangle = q(x)F[x]$ is maximal, in the present case we immediately see that the irreducibility of $g(x)$ in $\Bbb Z_3[x]$ implies $\langle g(x) \rangle$ is a maximal ideal; hence $\Bbb Z_3[x] / \langle g(x) \rangle$ is in fact a field.
So what we have seen in the above is that yes, the general assumption that $F[x] / \langle q(x) \rangle$ is a field for irreducible $q(x)$ has been exploited in the text of the question; this is in fact a very well-known and elementary result, and we see that it needn't really be assumed if we are willing to execute a fairly simple "logical detour" as embodied ca. (7)-(14); on the other hand, we have also assumed, in executing this proof, that an ideal $M$ in a commutative ring $R$ is maximal if and only if $R/M$ is a field, which follows easily from the definitions; finally, we have made the assumption that $F[x]$ is a principle ideal domain for $F$ a field, which itself easy to prove once one grants the polynomial form of the Euclidean division algorithm, so I guess we could move all our assumptions back to this fact, or even go one step further and say that Euclidean division more or less follows from some elementary arithmetic operations in $F[x]$ and mathematical induction; or even go further back and validate every arithmetic operation in $F[x]$ from a set of very basic axioms, taking sort of a Russell-Whitehead approach as in Principia Mathematica; but of course, that is not really what is intended here.  Indeed, I think it probably suffices, in terms of the intent of our OP Gibberish, to point out that $q(x)$ irreducible implies $F[x]/\langle q(x) \rangle$ is a field.
So much for the first part; as for the second, we have, for a root $\alpha$ of $g(x)$ in $F = \Bbb Z_3 [\alpha]$,
$\alpha^2 + \alpha + 2 = 0; \tag{15}$
thus
$\alpha^2 = -\alpha - 2; \tag{16}$
then
$(2 + 2\alpha)^2 - (1 + 2\alpha) = 2^2(1 + \alpha)^2 - (1 + 2\alpha)$
$= 1(1 + 2\alpha + \alpha^2) - (1 + 2\alpha) = \alpha^2 = -\alpha - 2.\tag {17}$.    
A: Hint: Divide $(2+2x)^{2}-(1+2x)$ by $2x^{2}+2x+1$ and consider the remainder.
