# $1+i$ is prime in $\mathbb Z[i]$

How does one prove the above statement?

I can show that $1+i$ is irreducible using the norm function i.e $a^2-b^2n$.

I tried to show that the ideal $(1+i)$ is prime but still came up short.

Is it possible to show that $\mathbb Z[i]/(1+i)$ is an integral domain?

• The norm function is $a^2+b^2$. – lhf Jan 17 '18 at 16:13
• – lhf Jan 17 '18 at 16:13

Yes. You should be easily able to see that $\mathbb Z[i]/(1+i)\cong \mathbb Z/2\mathbb Z$.
If it helps, you can recast $\mathbb Z[i]/(1+i)$ as $\mathbb Z[x]/(x^2+1, 1+x)$.
$N(1+i)=2$, which is prime, and the norm is multiplicative. If $1+i$ were composite, its norm would be too.
To prove $1+i$ is prime, the simplest consists in proving Gauß integers are a Euclidean domain, i.e. for any $a+ib, c+id\in \mathbf Z[i]$ there exist $q, r\in \mathbf Z[i]$ such that $$a+bi=q(c+di)+r \qquad N(r)< N(c+di)$$ To prove it, you have to show there exists a Gauß' integer $q$ such that $$N\biggl(\frac{a+bi}{c+di}-q\biggr)< 1.$$ There will result $\mathbf Z[i]$ is a P.I.D., so that irreducible elements generate a prime ideal.
• But he said he knows how to show $1+i$ is irreducible, the question is how to show it's prime. – David C. Ullrich Jan 17 '18 at 15:48