Is the ideal $(x,3)$ of $\mathbb{Z}[x]$ prime or maximal? I can't solve this question. I know that (x,3) is maximal ideal if $\mathbb{Z}[x]/ (x,3)$  is a field and (x,3) is a prime ideal if $\mathbb{Z}[x]/(x,3)$ is a domain. I know that there are isomorphism theorems about it but I`m totally confused. Could some one help me please?
Thanks!
 A: Apply the isomorphism theorems for rings to justify the following
$$\left(\Bbb Z[x]\right)/\left(\langle x\,,\,3\rangle\right)\cong\left(\Bbb Z[x]/\langle x\rangle\right)/\left(\langle x,3\rangle/\langle x\rangle\right)\cong\Bbb Z/\langle 3\rangle=:\Bbb F_3$$
So...
Added as punishment for myself for being uncareful:
$$\left(\Bbb Z[x]\right)/\left(\langle x\,,\,3\rangle\right)\cong\left(\Bbb Z[x]/\langle 3\rangle\right)/\left(\langle x,3\rangle/\langle 3\rangle\right)\cong\Bbb F_3[x]/\langle x\rangle\cong\Bbb F_3$$
A: The method of @DonAntonio is best, but here’s a rough and ready argument:
To factor out by the ideal $(x,3)$ is to treat computations in the original ring $\mathbb Z[x]$ as if both $x$ and $3$ are zero. So polynomials immediately lose all their nonconstant terms, and even these behave as integers modulo $3$.
Alternatively, just construct a ring homomorphism from $\mathbb Z[x]$ onto $\mathbb Z/(3)$. What you want is $f(x)\mapsto \bigl(f(0)\pmod3\bigr)$. Now check that the kernel is $(x,3)$.
