Is Z[x]/3 a field? Is Z[x]/3 a field? I am wondering this since a field quotiented with an ideal generated by a polynomial is a field. 
 A: Note that 
$$\mathbb{Z}[X]/(3) \cong (\mathbb{Z}/3\mathbb{Z})[X]$$
and thus $\mathbb{Z}[X]/(3)$ is not a field (since a  polynomial ring is never a field).
A: Below is a way that doesn't require  $\,\Bbb Z[x]/3\Bbb Z[x]\,\cong\, (\Bbb Z/3)[x]$ and $x$ invertible in $R[x]\!\iff\! R = 0$
$$\begin{align}  xf(x)&=1\ \ {\rm  in}\ \ \Bbb Z[x]/3\Bbb Z[x]\\  
\Rightarrow\ \ x f(x) &= 1 + 3 g(x)\ \ {\rm  in}\ \ \Bbb Z[x] \\
\Rightarrow\ \ 0 f(0) &= 1 + 3 g(0)\ \ {\rm  in}\ \ \Bbb Z \\
\Rightarrow\ \ 0 &= 1\ \ {\rm in} \ \Bbb Z/3
\end{align}\qquad$$
Remark $ $ It has an instructive universal view as in the linked post: if $x$ is invertible in $\Bbb Z[x]/3\Bbb Z[x]$ then so too is every element $\,r\,$  in every ring where $\,3 = 0,\,$  as follows simply by evaluating $\, xf(x) = 1 + 3g(x)\,$ at $\,x = r.\,$ Thus to present a counterexample it suffices to exhibit any nonunit $\,r\,$ in any ring where $\,3=0.\, $ A natural choice is the nonunit $\,\rm r=0\in \Bbb Z/3,\,$ yielding the above proof.
Strangely, these simple elementary consequences of universal properties of polynomial (and quotient) rings are often overlooked in such elementary contexts - where it is common to instead appeal to degree-based arguments. Witness to this is the fact that - embarrassingly - my trivial linked answer is one of my most popular answers. That many readers apparently were not aware of this viewpoint  seems to indicate that we need to do a better job highlighting the power of such universal properties. That's why I chose to emphasize this viewpoint. Even readers who have not yet studied such properties can understand simple proofs like that above - which may help plant the germ of the idea and provide some further motivation once one begins study of universal mapping properties.
A: We have $(3) \subset (3,X) \subset \mathbb{Z}[X]$. Since both inclusions are strict, $(3)$ is not a maximal ideal of $\mathbb{Z}[X]$.
Indeed, if $X \in (3)$, then $X=3f(X)$ and so $1=3f(1)$, a contradiction.
Similarly, if $1\in(3,X)$, then $1=3f(X)+Xg(X)$ and so $1=3f(0)$, a contradiction.
A: It is not a field, as polynomials are not invertible. Moreover you need to quotient by an irreducible polynomial to get a field. If you quotient by $x^{2}$, then $x*x=0$ in the quotient.
