What are the elements of this factor ring

Question

$Z [i]/(-1+2i)$ what are the elements of this ring ?

My tries $$ Z [i]/(-1+2i)=\{0+I,1+I,2+I,3+I,4+I,i+I,i+1+I,i+2+I,i+3+I,4+i+I \}$$

is it true?

Answer 1

More generally, let $\alpha \in \mathbb Z[i]$.

Then the ideal generated by $\alpha$ is equal to $\mathbb Z \alpha + \mathbb Z i\alpha$, an additive subgroup of $\mathbb Z^2 = \mathbb Z[i]$.

Draw the parallelogram defined by $\alpha$ and $i\alpha$ in the complex plane.

Then every element in $\mathbb Z[i]$ has a representative mod $\alpha$ inside this parallelogram.

Do this for $\alpha=-1+2i$ and you'll find the elements you need:

The elements inside the parallelogram are $0, -1, -2, -1-i, -2-i$.

Answer 2

Hint: $\Bbb{Z}[i]/(a+bi)$ has $N(a+bi)=a^2+b^2$ elements. We have $N(-1+2i)=5$. You can find these elements as follows:

Find all elements of quotient ring

Remark: For $gcd(a,b)=1$ we have $$\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2+b^2}$$

$\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2+b^2}$ if $(a,b)=1$