Showing that $Z_p[\alpha]$ has $p^2$ elements if $\alpha$ is a root I'm trying to show that $Z_p[\alpha]$ has $p^2$ elements if $\alpha$ is a root. But I'm not sure I understand how this could be possible since $Z_p$ itself has only $p$ elements.  
Is there a sense in which it is possible to show this, or am I right in that the question as stated is nonsense?
 A: Ok, let me try to interpret your question. Are you asking whether for any polynomial $f$, $Z_p[x]/(f(x))=Z_p[\alpha]$ is the finite field with $p^2$ elements? 
In general, that is not correct. For instance, adjoining a root of $x^3+x+1$ to $Z_2$ will actually give you the finite field with $8$ elements since this polynomial is irreducible in $Z_2[x]$.
In general, there exist irreducible degree $k$ polynomials and these will generate the finite field with $p^k$ elements.
A: I assume that $\alpha$ is a square root of some element since that is the only way that your question makes sense to me, also I assume that $p$ is a prime number. I also assume that $\mathbb{Z}_p[\alpha]$ is a ring that is obtained by adding $\alpha$ to it.
So, we know that $\mathbb{Z}_p$ is a field, therefore it makes sense to talk about Euclidean division in $\mathbb{Z}_p[x]$. So you need to identify $\mathbb{Z}_p[x]/(p(x))$ where $p(\alpha)=0$ and $\deg(p(x))=2$.
If $f(x) + (p(x)) \in \mathbb{Z}_p[x]$ then you can divide $f(x)$ by $p(x)$ and get $f(x)=p(x)q(x)+r(x)$ where $\deg(r(x))<\deg(p(x))$.
But $p(x)q(x) \in (p(x))$ so we can say
$f(x) - r(x) = p(x)q(x) \in (p(x)) \implies f(x)+(p(x))=r(x)+(p(x))$
That means everything in $\mathbb{Z}_p[x]/(p(x))$ can be identified by a polynomial of degree less than $2$. How may polynomials of degree less than $2$ you can have in $\mathbb{Z}_p[x]$? All such polynomials will be of the form $ax+b$ for some $a,b \in \mathbb{Z}_p$. A simple counting argument shows that you'll have $p^2$ because you have $p$ choices for $a$ and $p$ choices for $b$ and that answers your question.
