How to show that $\mathbb{Z}/13\mathbb{Z}$ is isomorphic to $\mathbb{Z}[i]/(3-2i)$?

Question

I know that there is a unique morphism between $\mathbb{Z}$ and $\mathbb{Z}[i]/(3-2i)$ defined by $f(x)=x+(3-2i)$. I want to use the first isomorphism theorem, but i do not know how to show that $f$ is surjective, any idea?

Answer 1

Try defining $\phi:\mathbb{Z}[i]\rightarrow \mathbb{Z}/13\mathbb{Z}$ by $$\phi(a+bi)\equiv_{13}a-5b$$

This is homomorphism since: $$\phi(a+bi)+\phi(c+di)=(a-5b)+(c-5d)=(a+c)-5(b+d)=\phi((a+c)+i(b+d))$$ $$\phi(a+bi)\phi(c+di)=(a-5b)(c-5d)=(ac-bd)-5(ad+bc)=\phi((a+bi)(c+di))$$

This works since $(-5)^2\equiv_{13} -1$, that is, it works just like $i$ in $\mathbb{Z}/13\mathbb{Z}$. Now,

• Surjectivity follows easily with $b=0$, as $a$ can be any integer.
• For injectivity we need to show that $(3-2i)\subseteq\ker(\phi)$, and $\ker(\phi)\subseteq (3-2i)$. Then it will follow that $\ker(\phi)=(3-2i)$. Note that $\phi(3-2i)=0$, moreover any multiple of $3-2i$ maps to $0$, so $(3-2i)\subseteq \ker(\phi)$. Assume that $\phi(a+bi)=a-5b\equiv_{13}0$, then $a\equiv_{13}5b$ and so $a=5b+13k$ working over the integers. Using that $5+i=(1+i)(3-2i)$ and $13=(3-2i)(3+2i)$ we can write $$a+bi=b(5+i)+(3-2i)(3+2i)=(3-2i)(b(1+i)+(3+2i))$$ and so we have $\ker(\phi)\subseteq(3-2i)$.

Answer 2

The key is to show that $i+(3-2i)$ is in the image of $f$, for then it is easy to see that every $a+bi+(3-2i)$ is. But $i+(3-2i)=8+(3-2i)$ as $(3-2i)\mid(8-i)$.