Cosets of $\langle 1+i\rangle$ in $\mathbb{Z}[i]$

Question

I know the cosets of $\langle 1+i\rangle$ in $\mathbb{Z}[i]$ are $\langle 1+i\rangle + 0$ and $\langle 1+i\rangle + 1$. However, in a solution I saw, they also noted that $1+0i \not\in \langle 1+i \rangle$ and thus concluded there are only two cosets. Why is this necessary?

Answer 1

The general element in $(1+i)$ is $$ (a+bi)(1+i)=a-b+(a+b)i, \quad a,b\in\Bbb{Z}\tag1 $$ If $a-b+(a+b)i=c+di, \:c,d\in\Bbb{Z}$, then $$ a=\frac{c+d}{2}, \:b=\frac{-c+d}{2} $$ which has solution in $\Bbb{Z}$ iff $c$ and $d$ are either both even or both odd.

So there are only $2$ cosets of $(1+i)$, which are the sum of real and imaginary being either even ($(1+i)$ itself) or odd. It is easy to see that $1 \not\in (1+i)$ since $1$ is odd and $0$ is even.

Answer 2

A nice way to see this using the geometry of numbers:

Embed the ring $R=\mathbb{Z}[i]$ into $\mathbb{R}^2$ via the additive homomorphism that maps $1\mapsto (1,0)$ and $i\mapsto (0,1)$. The ring $R$ is also a $\mathbb{Z}$-module, and this view makes that structure quite apparent. Any ideal is a $\mathbb{Z}$-submodule (although the converse of this statement is not true). Under this embedding, a submodule is clearly visible as a sublattice of $\mathbb{Z}^2$.

The ideal in question shows up as the sub-lattice generated by the point $(1,1)$. Note that this lattice only has one non-trivial translation (up to equivalence) onto lattice points of $\mathbb{Z}^2$. Thus, only one coset other than itself.