Polynomials in $Z_p[x]/f(x)$ For shorthand, suppose $K=\mathbb{Z}_p[x]/f(x)$, $p$ a prime, and $\deg(f)=n$ where $f\in \mathbb{Z}_p[x]$.
Then, how do we show that (1) $K$ can be written as $\mathbb{Z}_p[\theta]$, where $\theta$ is the class of $x$, (2) $f(\theta)=0$ in $K$ and if $\deg(g)<n$, then $g(\theta)\neq 0$, (3) $K$ is a field if and only if $f$ is prime in $\mathbb{Z}_p[x]$, in which case $b^{p^n}=b$ for every $b\in K$.
Any help is greatly appreciated.
 A: 1) If $g \in \mathbb{Z}_p[x]$ suppose $g(x) = \sum_{i=0}^{i=n} a_ix^i $ then $$g(x) + (f(x)) = \sum_{i=0}^{i=n}  a_i (x^i + (f(x)) ) = \sum_{i=0}^{i=n}  a_i (x + (f(x)) )^i =\sum_{i=0}^{i=n}  a_i\theta^i $$ with $\theta = $ class of $x$
2)I denote the class of $h(x) \in \mathbb{Z}_p[x]$ with $\overline{h(x)}$. Thus in $K$ $$f(\theta) = f(\overline{x}) = \overline{f(x)}$$ but clearly $\overline{f(x)} = 0$ in $Z_p[x]/(f(x))$
3) $\mathbb{Z}_p $ is a field, so $\mathbb{Z}_p[x]$ is a PID, and so $Z_p[x]/(f(x))$ is a field $\Leftrightarrow (f(x))$ is a maximal ideal $\Leftrightarrow$  $f(x)$ is prime in $\mathbb{Z}_p[x]$.
In this case $\mathbb{Z}_p \subset \mathbb{Z}_p[x]/(f(x))$ and $\mathbb{Z}_p[x]/(f(x))$ is a vector space over $\mathbb{Z}_p $.
The dimension of this vector space is the degree of $f$, i.e. $n$. 
Thus $$|\ \mathbb{Z}_p[x]/(f(x)) \ | = p^n$$ But $ \mathbb{Z}_p[x]/(f(x)) $ is also a field, and so the elements $\neq 0 $ in this field constitutes a multiplicative group of cardinality $p^n -1$ , therefore for every $ b \neq 0$ in $K$ $$b^{p^n -1} = 1$$ Including $0$ we have $$b^{p^n} = b \ \ \ \forall b \in K$$
