How to prove that $Z[i]/7$ is a field? I know that $Q[i]$ is a field since we can find the inverse for each $a+bi$ namely $\frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2}$. However, I am not sure how to do so for something like $Z[i]/7$ since we can only have non-negative integers less than 7 as coefficients. 
 A: Hint: You have the right formula to find the inverse of any element, the only problem is that the denominator $a^2+b^2$ might be zero (contained in the ideal generated by $7$). Why can this not happen?
A: Hint: the same method of rationalizing the denominator works over any field where $\,-1\,$ is not a square, since then $\,a^2+b^2\neq 0,\,$ for else $\,b = 0 \,\Rightarrow\,a= 0,\,$ contra $\,a+bi\neq 0,\,$ and $\, b\neq 0\,\Rightarrow\, (a/b)^2 = -1,\,$ contra hypothesis that $\, -1\,$ is not a square  (as is true in $\,\Bbb Z/7).$
More generally one can rationalize denominators of algebraic numbers by taking norms (multiplying by conjugates), or by reading off the inverse from a polynomial having it as root.
A: $\mathbb{Z}[i]/(7)$ is a field. From standard results in ring theory, $\mathbb{Z}[i]$ is a PID, so all nonzero prime ideals are maximal. Also, the quotient of a ring (for our purposes, say commutative and with identity) by a maximal ideal is a field. In our case, $(7)$ is prime in $\mathbb{Z}[i]$ hence maximal. Therefore $\mathbb{Z}[i]/(7)$ is a field. 
