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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.

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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.

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    $\begingroup$ is it just isomorphic to Z ? $\endgroup$ – Nicole Mar 14 '15 at 21:19
  • $\begingroup$ @Nicole yup! Do you know the first isomorphism theorem for rings? Consider the homomorphism $\mathbb{Z}[x]/(x) \to \mathbb{Z}$ given by $f(x) \mapsto f(0)$. $\endgroup$ – hunter Mar 14 '15 at 21:20
  • $\begingroup$ yay thanks! Hmm not sure how to connect the isomorphism theorem here.. Is it not enough to say since Z[x]/<x> is isomorphic to Z. I know Z is an integral domain, it is NOT a field. And thus I know now that I have a prime ideal and not a maximal $\endgroup$ – Nicole Mar 14 '15 at 21:22
  • $\begingroup$ @Nicole That is the idea, yes. $\endgroup$ – rschwieb Mar 14 '15 at 23:30
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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).

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