For ideals $I=(8)$ and $J=(5+5i)$ in $\mathbb{Z}[i]$, what are $IJ$, $I+J$ and $I\cap J$?

Question

Let $I=(8)$ and $J=(5+5i)$ be ideals in $\mathbb{Z}[i]$.

How do I find $x,y,z\in\mathbb{Z}[i]$ such that:
$IJ=(x)$,
$I+J=(y)$,
$I\cap J=(z)$?

Is it correct that $y=13+5i$ and $x=40+40i$?

Answer 1

To compute $IJ$, note that if $I = (x_1, \ldots, x_m)$ and $J = (y_1, \ldots, y_n)$, one can show that $$ IJ = (x_1, \ldots, x_m)(y_1, \ldots, y_n) = (x_i y_j : 1 \leq i \leq m, 1 \leq j \leq n), $$ i.e., the product of the ideals is generated by all products of their generators.

Recall that in a PID we have $(a) + (b) = (\gcd(a,b))$ and $(a) \cap (b) = (\text{lcm}(a,b))$. Since $\mathbb{Z}[i]$ is a PID, it suffices to determine the prime factorization of $8$ and $5 + 5i$ in $\mathbb{Z}[i]$. Since $$ 2 = (1 + i)(1 - i) = (1+i) (-i)(i + 1) = -i (1 + i)^2 $$ then $8 = 2^3 = i(1+i)^6$. Note that $$ 5 + 5i = 5(1+i) = (2+i)(2-i)(1+i) \, . $$ (One can show that these factors are indeed prime since their norms are prime, where $N: \mathbb{Z}[i] \to \mathbb{Z}$ is the multiplicative norm map $N(a+bi) = a^2 + b^2$.) Then $\gcd(8, 5 + 5i) = 1+i$ and $$ \text{lcm}(8, 5+5i) = \frac{8(5+5i)}{\gcd(8,5+5i)} = \frac{8(5+5i)}{1+i} = 8 \cdot 5 = 40 \, . $$