$\mathbb Z[i]/\langle17\rangle$ is a field

Question

Is the quotient ring $\mathbb Z[i]/\langle17\rangle$ a field?

Proof:

Clearly $17$ it is not prime in $\mathbb Z[i]$, we might say it is irreducible in $\mathbb Z[i]$, thus $\mathbb Z[i]/\langle17\rangle$ is not a field.

Is the proof correct?

Can it be written in a better way (more formal)?

Answer 1

$17 = (4 + i)(4 - i)$. Therefore, $(4 + i + \langle17\rangle)\cdot(4 - i + \langle17\rangle) = 0 + \langle17\rangle\in\Bbb Z[i]/\langle17\rangle$. However, $4\pm i\not\in \langle17\rangle$, because any element of $\langle17\rangle$ is of the form $17\alpha = 17m + 17ni$, $\alpha = m + ni\in\Bbb Z[i]$. Thus, we have a product of nonzero elements in $\Bbb Z[i]/\langle17\rangle$ which is $0$, so that $\Bbb Z[i]/\langle17\rangle$ is not even an integral domain, let alone a field.

Answer 2

You contradict yourself saying that 17 is not prime and that it is irreducible ("prime" and "irreducible" are the same thing in this ring).

Whichever you settle on, it deserves an explanation of why 17 does or does not have that property. (I assume this is homework in an introduction to number fields or similar setting. There are other settings where this is sufficient explanation)