Ideal generated by $1+2i$ is the same as $(a+bi|a, b \in \mathbb{Z}, 2a=b mod 5)$ in the ring $\mathbb{Z}[i]$ = ($a+bi|a, b \in \mathbb{Z}$

Question

Show that in the ring $\mathbb{Z}[i]$ = ($a+bi|a, b \in \mathbb{Z}$ of Gaussian integers, the ideal generated by $1+2i$ is the same as the set $(a+bi|a, b \in \mathbb{Z}, 2a=b \mod 5)$.

I'm pretty confused by this question. How would you show the ideal generated by $1+2i$ in the first ring, and then be able to equate it to the second set? Would appreciate any detailed responses. Many thanks.

Answer 1

Denote by $I$ the ideal generated by $1 +2 i$ and by $J$ the set $\left\{a +b i : a, b \in \mathbb{Z}; 2 a \equiv b \pmod 5 \right\}$. We want to prove that $I = J$.

• Then $J$ is an ideal of $\mathbb{Z}[i]$.
  Indeed, it is clearly a subgroup of $\mathbb{Z}[i]$ and $$\forall a +b i \in J, \forall c +d i \in \mathbb{Z}[i], (a +b i) (c +d i) = (a c -b d) +(a d +b c) i \in J$$ since $2 (a c -b d) \equiv a d +b c \pmod 5$ because $2 a \equiv b \pmod 5$.
  Therefore, since $1 +2 i \in J$, $I \subset J$.
• Conversely, for all $a +b i \in J$, there exists $k \in \mathbb{Z}$ such that $b = 2 a +5 k$ and hence, $a +b i = (1 +2 i) (a +2 k + k i) \in I$.
  Thus, $J \subset I$.

Answer 2

Consider $a+bi=(1+2i)(c+di)$ with $c$, $d\in\Bbb Z$. Then $a=c-2d$ and $b=2c+d$. But $2a=2c-4d\equiv 2c+d\equiv b\pmod5$.

Now suppose $a\equiv2b\pmod5$. Then $(1-2i)(a+bi)=(a+2b)+(b-2a)i$. Then $b-2a\equiv0\pmod 5$ and $a+2b\equiv a+4a\equiv0\pmod5$. So $(1-2i)(a+bi)=5c+5di$ with $c$, $d$ integers, and $a+bi=(1+2i)(c+di)$.