Consider set $\mathbb{Z}[\sqrt{-5}] = \{a+b\sqrt{5}i : a,b \in \mathbb{Z} \}$ show that it is a ring

Question

Consider set $\mathbb{Z}[\sqrt{-5}] = \{a+b\sqrt{5}i : a,b \in \mathbb{Z} \}$.

My task is to show some features listed below:

1. Show that $\mathbb{Z}[\sqrt{-5}]$ is a ring.

I would like to show that $\mathbb{Z}[\sqrt{-5}]$ is a subring of $\mathbb{C}$.

Let $z,w \in \mathbb{Z}[\sqrt{-5}]$ and $z = a + b\sqrt{5}i, w = c+d\sqrt{5}i$, where $a,b,c,d \in \mathbb{Z}$.

It is sufficient to show that

$z+w, -z, zw \in \mathbb{Z}[\sqrt{-5}]$

$z+w = a+b\sqrt{5}i + c+d\sqrt{5}i = (a+c) + (b+d)\sqrt{5}i \in \mathbb{Z}[\sqrt{-5}]$

$-z = -(a+b\sqrt{5}i) = (-a) + (-b)\sqrt{5}i \in \mathbb{Z}[\sqrt{-5}]$

$zw = (a+b\sqrt{5}i)(c+d\sqrt{5}i) = (ac - 5bd) + (ad+bd)\sqrt{5}i \in \mathbb{Z}[\sqrt{-5}]$

Hence $\mathbb{Z}[\sqrt{-5}]$ as a subring of ring $\mathbb{C}$ then it is also a ring.

2. Let $z = 2+\sqrt{5}i$. Show that $z$ is irreducible.

I know the definition of being irreducible, it has to be non invertible and the following has to be true:

$\forall x,y \in \mathbb{Z}[\sqrt{-5}]$ if $ z = xy \Rightarrow x$ or $y$ has to be invertible.

Hint is to use that fact that norm $\lvert z\rvert = 3$. Hence all divisiors of $z$ have norm lower or equal than $3$.

I do not know how to use this hint.

3. Show that $z \lvert 3\cdot 3$ and $z$ does not divide $3$.

For the first one I was thinking about

$3 \cdot 3 = 9 = (2+\sqrt{5}i)(2-\sqrt{5}i) \Rightarrow z \lvert 3\cdot 3$

For the second one I was thinking about dividing $3$ by $z$ in $\mathbb{C}$.

Then $$\frac{3}{z} = \frac{3}{2+\sqrt{5}i} = \frac{2}{3} + (\frac{-1}{3})\sqrt{5}i$$ which is not an element of $\mathbb{Z}[\sqrt{-5}]$.

Is this point correct?

4. Show that $z$ is irreducible element, but not a prime element.

This can be obtained from points 2. and 3.

Since $z\lvert 3\cdot 3$, but $z$ does not divide $3$.

Answer 1

If, for some $x, y \in \mathbb{Z}(\sqrt{-5})$, we have $$ z = 2 + \sqrt{-5} \iota = xy, $$ then we must also have $$3 = \lvert z \rvert = \lvert x \rvert \, \lvert y \rvert, $$ and since $\lvert x \rvert, \lvert y \rvert \in \mathbb{N}$, therefore we must have $$ \lvert x \rvert, \lvert y \rvert \in \{ 1, 3 \}. $$

Thus we have the following two cases:

1. $\lvert x \rvert = 1$, $\lvert y \rvert = 3$

2. $\lvert x \rvert = 3$, $\lvert y \rvert = 1$

If $x = a + b \iota \sqrt{5}$ is such that $$ \lvert x \rvert = \sqrt{ a^2 + 5b^2 } = 1, $$ where $a, b \in \mathbb{Z}$, then we must have $$ a^2 + 5 b^2 = 1, $$ and so $b = 0$ and $a = \pm 1$, which gives $$ x = \pm 1, $$ and then $$ z = 1z \ \mbox{ or } \ z = (-1)(-z). $$

And, similarly, for the case when $\lvert x \rvert = 3$ and $\lvert y \rvert = 1$.