I'm trying to prove the order of a particular quotient field of $\mathbb{Z}[i]$ without using the characteristic of the field. My hesitation in using the characteristic comes from the fact that I'm working my way through Dummit and Foote's Abstract Algebra and encountered this question before the section on field theory. This question arose in the section on UFD's. There must by some way to prove my claim without using the field theoretic notion of a characteristic.
Claim: Let $q \in \mathbb{Z}$ be a prime such that $q \equiv 3$ (mod $4$). Then, $\mathbb{Z}[i]/(q)$ is a field of order $q^2$.
A few things about this claim are immediately clear. First, q is clearly an irreducible in $\mathbb{Z}[i]$ because it would only be reducible if $q=2$ or $q \equiv 1$ (mod $4$). Since $\mathbb{Z}[i]$ is a UFD, it follows that $(q)$ is prime. Moreover, it's pretty simple to show that $\mathbb{Z} \cap (q) = q\mathbb{Z}$. By the second isomorphism theorem for rings, you then recover that $\mathbb{Z} + (q)/(q) \cong \mathbb{Z}/q\mathbb{Z}$. But that seems like a dead end because I can't figure out a way to break down the rest of $\mathbb{Z}[i]$ and show that it's disjoint with $\mathbb{Z} + (q)$ and isomorphic to $\mathbb{Z}/q\mathbb{Z}$ also.
Answer
It is easy to see that $a+bi\in\mathbb{Z}[i]$ is divisible by $q$ iff both $a$ and $b$ are divisible by $q$ as integers. It follows that $a+bi\equiv c+di$ mod $q$ iff $a\equiv c$ and $b\equiv d$ mod $q$. So this gives a bijection (in fact, an additive group-isomorphism) $\mathbb{Z}[i]/(q)\to \mathbb{Z}/(q)\times\mathbb{Z}/(q)$.
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