Prove or disprove that $\mathbb{Z}[x] / (x^3 - 2)$ is a field. Prove or disprove that $\mathbb{Z}[x] / (x^3 - 2)$ is a field.
I am pretty sure it is not a field because if you consider $F = \mathbb{Z}[x] / (x^3 - 2)\mathbb{Z}[x]$ where $F$ is a field, then char $F = 0$.
Not sure how to write a proof for this though. 
 A: Hints:
$$\begin{align*}\bullet&\;\;\;x^3-2\in\Bbb F_7[x]\;\;\text{is irreducible}\\{}\\
\bullet&\;\;\text{As ideals in}\;\Bbb Z[x]\;\;,\;\;\langle\;x^3-2\;\rangle \lneqq\langle\;x^3-2\,,\,7\;\rangle\end{align*}$$
You may want to try to prove the general theorem: an ideal in $\;\Bbb Z[x]\;$ is maximal iff it is of the form $\;\langle f(x),p\rangle\;$ , with $\;p\;$ a prime and $\;f(x)\in\Bbb Z[x]\;$ such that $\;f(x)\pmod p\;$ irreducible in $\;\Bbb Z_p[x]\;$ .
Personally for me, the above is one of the most stunning and beautiful facts in basic abstract algebra.
A: Why do you think char 0 matters? If you can answer that question, that would give a lead to coming up with a proof.
It does turn out to be true that every quotient field of $\mathbb{Z}[x]$ has nonzero characteristic, so there is surely a proof along these lines.  It would require two parts: showing that $\mathbb{Z}[x] / (x^3-2)$ is actually characteristic 0 (under the assumption it is a field), and showing that this is a contradiction.
I don't know what particular things you know, so I'm not really sure how to suggest going about the problem along these lines.
The simplest is probably, IMO, inspired by the fact a char 0 field must contain $\mathbb{Q}$: find some integer $n$ for which you can easily prove that $n$ is both nonzero and noninvertible in $\mathbb{Z}[x] / (x^3 - 2)$. 
A: The easiest way to see your quotient is not a field is to note there is some maximal ideal containing $x^3 - 2$. One such maximal ideal is $(7,x^3 - 2)$. To see this is maximal, it is enough to observe that $x^3 - 2$ is irreducible in $\Bbb{Z}/7\Bbb{Z}[x]$. For a cubic, this amounts to showing there are no roots, and we see that $1^3 - 2 \neq 0, 2^3 - 2\neq 0$, $3^3 - 2 = 6 \neq 0$, $4^3 - 2 = 62 \neq 0$, $5^3 -2 = 4 \neq 0$, $6^3 - 2 = 4 \neq 0$ mod $7$.
A: Lots of nice ideas. Its amazing how many ways one can approach a problem. A good way for me to see this problem is most likely through a contradiction. I appreciate everyone's help. Now I am going to attempt a proof by contradiction and let me know how this sounds. This is what I have obtained:
Suppose $F = \mathbb{Z}[x] / (x^3 - 2)\mathbb{Z}[x]$.
char $f = 0 \Rightarrow 2$ should be an invertible modulus $(x^3 - 2)\mathbb{Z}[x]$. Then we have that $(d(x) + (x^3 - 2)\mathbb{Z}[x])(2 + (x^3 - 2)\mathbb{Z}[x]) = 1 + (x^3 - 2)\mathbb{Z}[x]$. Assume that deg $d(x) \leq 2$. 
Hmm how should I conclude this then?
