Prove that a nonzero integer p is prime if and only if the ideal (p) is maximal in Z. I know that if (p) is maximal then Z/(p) is a field which implies its an integral domain which then implies that (p) is a prime ideal and p is prime. I'm having trouble going the forward direction assuming that p is prime and then showing it's a prime ideal. 
 A: In general, if $(a)$ is an ideal of $\mathbb Z$ then the ideals contained between $(a)$ and $\mathbb Z$ correspond to the divisors of $a$. For example, if $a=10$ then $\mathbb Z=(1)\supset(5)\supset(10)$ and $\mathbb Z=(1)\supset(2)\supset(10)$. This is a consequence of the more general fact that $(a)\supseteq(b)\iff a\mid b$. Now what are the divisors of a prime integer?
A: One proof would be to use the fact that $\mathbb{Z}/n\mathbb{Z}$ is an integral domain iff it is a field.  It is a basic fact from number theory that the units in $\mathbb{Z}/n\mathbb{Z}$ are exactly the numbers coprime to $n$ and the non units are always zero divisors $\mod n$.  
A: Hint $ $ Notice that for principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e.  $(a)\supseteq (b)\iff a\mid b,\,$ thus  
$\qquad\quad\begin{eqnarray} (p)\,\text{ is maximal} 
&\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (d)\\
&\iff&\  p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ d\\
&\iff&\  p\ \ \text{ is irreducible}\\
\end{eqnarray}$
