Let $R = \mathbb{Z}[x]$ and let $I = \langle x \rangle$ be the ideal in $R$ generated by $x$. Show $I$ is a prime ideal but not a maximal. So I am trying to figure out why I have an integral domain but no field. Any direction is appreciated I'm not really sure where to go with this one.
 A: Hint:
$R/I$ is a domain iff $I$ is prime, and is a field iff $I$ is maximal. You ought to compute $\Bbb Z[x]/(x)$ and consider this.
A: Suppose $f(x) = g(x)k(x) \in \langle x\rangle$.
We can write $g(x) = a + xg_1(x),\ k(x) = b + xk_1(x)$, and $f(x) = xf_1(x)$.
Then $xf_1(x) = ab + x(ak_1(x) + ag_1(x)) + x^2(g_1(x)k_1(x))$.
Equating coefficients, we see $ab = 0$ ($f$ has no constant coefficient since $x|f(x)$).
Without loss of generality, assume $g(x) \not \in \langle x\rangle$. Then $a \neq 0$, and since $\Bbb Z$ is an integral domain, we must have $b = 0$, so that $k(x) = xk_1(x) \in \langle x\rangle$. This shows $\langle x\rangle$ is a prime ideal.
Let $J = \langle 2,x\rangle$. Clearly, $I \subseteq J$, and $2 \in J-I$, so $I$ is a PROPER subset of $J$.
Suppose $1 \in J$. This means $1 = 2s(x) + xt(x)$ for some $s(x),t(x) \in \Bbb Z[x]$.
Again, writing $s(x) = c + xs_1(x)$  we have:
$1 = 2s(x) +xt(x) = 2c + x2s_1(x) + xt(x) = 2c + x(2s_1 + t(x))$, and equating constant terms we have:
$1 = 2c$, for $c \in \Bbb Z$, a contradiction. Thus $J$ is proper. Hence $I \subset J \subset \Bbb Z[x]$, so $I$ is not maximal.
(In all fairness hunter's comment is much "cleaner"- the isomorphism $\phi:\Bbb Z[x] \to \Bbb Z$ given by $\phi(f(x)) = f(0)$ accomplishes exactly the same thing as above, without as much fuss).
