0

If $N(z)=a^2+b^2 =\text{ prime }\in \Bbb{Z}$ show $$\frac{\Bbb{Z}[i]}{(a+bi)}$$ is a field of prime order.

from Prove if $z \in \mathbb{Z}{[i]}$ such that $N(z)$ is a prime number then $z$ is irreducible.

$a+bi$ is irreducible. $\Bbb{Z}[i]$ is an integral domain and $a+bi$ is irreducible in $\Bbb{Z}[i]$ so somehow its a field by a theorem (which theorem I am not sure, need to fix this)

$$ \begin{aligned} a+bi &\equiv0 \mod(a+bi) \\ a & \equiv b *i \mod(a+bi) \\ [a] &= [bi] \end{aligned}$$

so $z\in \Bbb{Z}[i]$ $z=\alpha+\beta i$

$$\begin{aligned} z+(a+bi)\equiv [z] =[\alpha + \beta i]=\text{ stuff I need to think more} \end{aligned} $$

somehow that is $\Bbb{Z}_p$ so $\frac{\Bbb{Z}[i]}{a+bi} \cong \Bbb{Z}_p$ being a field.

Teddy38
  • 3,309
Tiger Blood
  • 1,940
  • THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM - FALL SESSION 2006 110.401 - ADVANCED ALGEBRA I. they just use it. I dont have a proof and could prob use an isomorphism theorem of rings to prove it – Tiger Blood Nov 14 '17 at 02:47
  • Don't they teach you TeX at THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences? – Teddy38 Nov 14 '17 at 08:36
  • Here are two more duplicates: 1, 2. As in the linked posts, you can show that $\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}/(a^2 + b^2)$ and then note that $\mathbb{Z}/p\mathbb{Z}$ is a field with $p$ elements. – Viktor Vaughn Nov 14 '17 at 16:13

0 Answers0