TITLE: Why do the gaussian integers have only 2 congruence classes mod 1+i?
QUESTION [8 upvotes]: If we consider $Z[i]$ modulo $1+i$, why are there only two congruence classes?

REPLY [4 votes]: Let's reduce the problem to integer linear algebra.
$\mathbb{Z}[i]$ is two-dimensional, with basis $\{1, i\}$.
The ideal $(1+i)$ consists of all multiples of $1+i$.
The multiples of $1+i$ are spanned by $\{1 \cdot (1+i), i \cdot (1+i) \} = \{ 1+i, -1+i \}$. Writing coordinates relative to our chosen basis above, this is the rowspace of the matrix
$$ \left( \begin{matrix} 1 & 1 \\ -1 & 1 \end{matrix} \right) $$
We can row reduce this matrix to
$$ \left( \begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix} \right) $$
Now, if we consider any vector $(x,y)$, it's clear that we can normalize this to either $(0,0)$ or $(0,1)$ by adding in elements of the ideal $(1+i)$; i.e. by adding in linear combinations of the rows of the simplified matrix.
Thus, two equivalence classes.