Why $R/(a+bi)$ is a finite field with characteristic $p$ where $a^2+b^2=p$

Question

Suppose $R=\mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $\alpha=a+bi$ and $I=(\alpha)$. Prove $R/I$ is a finite field with characteristic p.

My attempts: First prove $\alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=\mathbb{Z}[i]$ is an Euclidean domain. Hence for $x\in \mathbb{Z}[i],x=q\alpha+r$ where $N(r)\lt N(\alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.

Thanks for your hints.

always $\mathbb Z[i]/(a+bi)\cong \mathbb Z/N(a+bi)\mathbb Z$ where $(a,b)=1$ — Mustafa, Jan 03 '19 at 17:14

score 2 · Answer 1 · answered Jan 03 '19 at 17:19

2

$p = a^2+b^2 = \bar \alpha \alpha \in I$ implies $p=0$ in $R/I$.

If you know that $R/I$ is a field, then it must have characteristic $p$.

answered Jan 03 '19 at 17:19

lhf

216,483

Why $R/(a+bi)$ is a finite field with characteristic $p$ where $a^2+b^2=p$

1 Answers1