Show $f:\mathbb{Z}\to\mathbb{Z}[i]/(3+2i)$ via $c\mapsto c+(3+2i)$ is surjective

Question

Write $x=a+bi+(3+2i)\in\mathbb{Z}[i]/(3+2i)$. I want to find a $c\in\mathbb{Z}$ such that $f(c)=x$. I think we need to use the fact that there exist $m,n\in\mathbb{Z}$ so that $2m+3n=1$, but I am not sure how. I'd like to do something similar to the following trick for when the codomain is $\mathbb{Z}[i]/(3-i)$: Let $z=a+bi+(3-i)$. Note $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, and thus $f(a+3b)=a+bi+(3-i)=z$. How can I adapt this example to when the codomain is a quotient of $\mathbb{Z}[i]$ by $(3+2i)$ instead of $(3-i)$?

I think it is a matter of rewriting $a+bi$ in a clever way, like adding and subtracting the same quantity to obtain a nice form. I have played around with it for an hour and keep getting stuck. Any hints?

Answer 1

Here's a way to follow your nose to find a suitable preimage. Using your notation, fixing a given $x=a+bi\in\mathbb{Z}[i]$, you want to find $c\in\mathbb{Z}$ such that $x-c\in (3+2i)$. Write $x-c=(a-c)+bi=d+bi$.

Now $d+bi\in (3+2i)$ if $\frac{d+bi}{3+2i}\in\mathbb{Z}[i]$. Computation shows $$ \frac{d+bi}{3+2i}=\frac{d+bi}{3+2i}\cdot\frac{3-2i}{3-2i}=\frac{(3d+2b)+(3b-2d)i}{13}\in\mathbb{Q}[i]. $$ Since $c$ is yet to be determined, you have freedom to adjust $d$ as you'd like, and you need $13\mid 3d+2b$ and $13\mid 3b-2d$ for the above to be in $\mathbb{Z}[i]$. If $13$ does divide both these numerators, it divides their sum, which is $d+5b$, so a natural guess for such $d$ is $d=13-5b$. Unless I've made a bad algebra error, if you set $d=13-5b$, a quick check shows both ratios are integers. This translates to saying $a-c=13-5b$, or $c=a+5b-13$. This $c$ is then a suitable preimage for $x+(3+2i)$.

Answer 2

For a Gaussian integer $c+di$, $(3+2i)(c+di)=(3c-2d)+(2c+3d)i$.

So, in order for the coset $a+bi+\langle3+2i\rangle$ (using angle brackets to avoid confusion) to contain at least one integer, there must exist integers $c$ and $d$ for which $b+2c+3d=0$, or $2c+3d=-b$. In fact, $c=b$ and $d=-b$ will work.

Now, $(a+bi)+(3+2i)(b-bi)=(a+bi)+(3b+2b)+(2b-3b)i=(a+bi)+5b-bi=a+5b$. Hence, for any Gaussian integer $a+bi$, $f(a+5b)=a+bi+\langle3+2i\rangle$, showing that $f$ is indeed surjective.