The principal ideal $(1+i)$ in $\mathbb{Z}[i]$ is maximal.

Question

I'm trying to prove that the principal ideal $(1+i)$ in $\mathbb{Z}[i]$ is maximal.

My approach: $m+in \equiv (m-n)$ mod $(1+i), \forall m,n \in \mathbb{Z}$ and $2 \equiv 0$ mod $(1+i)$ then what?

Answer 1

Yes, $\Bbb Z[i]/(1+i) \simeq \Bbb Z/(2)$. You've figured out that

1. All elements of $\Bbb Z[i]/(1+i)$ has a representative that is real
2. For real representatives, only whether it's odd or even matters

and that is enough. Technically you have to show that $1 \notin (1+i)$ as well (i.e. that the ideal is proper), but that's just a matter of checking that $\frac{1}{1+i} \in \Bbb C$ is not in $\Bbb Z[i]$.

Answer 2

An ideal $I$ in a ring $R$ is maximal if and only if $R/I$ is a field.

Since $\mathbb{Z}[i]/(1+i)\cong \mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}_2$ is a field, we have that $(1+i)$ is maximal.

Answer 3

$ \mathbb{Z}[i] $ is a principal ideal domain, which means that maximal ideals are those generated by irreducible elements of $ \mathbb{Z}[i] $. Now, $ N(1+i) = 2 $ which is prime in $ \mathbb{Z} $, therefore $ 1 + i $ is irreducible in $ \mathbb{Z}[i] $, and therefore generates a maximal ideal.

Answer 4

If you know that $\mathbb{Z}[i]$ is a principal ideal domain, then the fact that $1+i$ is irreducible tells you that $(1+i)$ is maximal. Indeed, if $(1+i)=(a+bi)(c+di)$, then also $(1-i)=(a-bi)(c-di)$ and multiplying the two relations gives $$ 2=(a^2+b^2)(c^2+d^2) $$ forcing either $a^2+b^2=1$ or $c^2+d^2=1$; therefore one of the factors is invertible, being one in $\{1,-1,i,-i\}$.

Otherwise you can go the hard way and prove that any element not in $(1+i)$ has an inverse modulo $(1+i)$. This means that, if $a+bi\notin(1+i)$, then there exists $c+di$ such that $(a+bi)(c+di)-1\in(1+i)$.

Suppose $a+bi\notin(1+i)$. Since $2=(1+i)(1-i)$, one among $a$ and $b$ is not even. Multiplying by $i$ essentially exchanges the real and imaginary part, so we can assume $a=2x+1$ is odd. Therefore $$ a+bi=1+bi+(1+i)(1-i)x\equiv 1+bi\pmod{(1+i)} $$ and it's not restrictive to assume $a=1$. However, if $b=2y+1$ is odd, we have $$ 1+bi=1+i+(1+i)(1-i)y\equiv0\pmod{(i+1)} $$ contrary to our assumption. Thus $b=2y$ is even and therefore $$ a+bi\equiv1\pmod{(1+i)} $$ so we are done.