Exercise A0.V.1.17

Proposition A1

The ring $\mathbb{Z}[\sqrt{-5}]$ is isomorphic to $\mathbb{Z}[t]/(t^2 + 5).$

Proof

Since $t^2 + 5$ is monic, we can use division algorithm to find the value of any element $f \in \mathbb{Z}[t]/(t^2 + 5)$ modulo the relation. Hence, every such element can be written in the form $a + bt$ for $a,b \in \mathbb{Z}.$

Thus, for every element of $\mathbb{Z}[t]/(t^2 + 5)$ we can find a corresponding element of $\mathbb{Z}[\sqrt{-5}]$ by mapping $a + bt$ to $a + b\sqrt{-5}.$ This gives us a bijective set function $\tau : \mathbb{Z}[t]/(t^2 + 5) \to \mathbb{Z}[\sqrt{-5}].$

We can verify directly that this function is a homomorphism:

$$\begin{align*} \tau((a_1 + b_1t) + (a_2 + b_2t)) &= \tau((a_1 + a_2) + (b_1 + b_2)t) \\ &= (a_1 + a_2) + (b_1+b_2)\sqrt{-5} \\ &= a_1 + b_1\sqrt{-5} + a_2 +b_2\sqrt{-5} \\ &= \tau(a_1 + b_1t) + \tau(a_2 + b_2 t), \end{align*}$$ $$\begin{align*} \tau((a_1 + b_1t) \cdot (a_2 + b_2t)) & = \tau(a_1a_2 + a_1b_2t + b_1ta_2 + b_1b_2t^2) \\ &= \tau((a_1a_2 - 5b_1b_2) + (a_1b_2 + b_1a_2)t) \\ &= (a_1a_2 - 5b_1b_2) + \sqrt{-5}(a_1b_2 + b_1a_2) \\ &= a_1a_2 + a_1b_2\sqrt{-5} + b_1\sqrt{-5} a_2 + b_1b_2\left(\sqrt{-5}\right)^2 \\ &= \left(a_1 + b_1 \sqrt{-5}\right) \cdot \left(a_2 + b_2 \sqrt{-5}\right) \\ &= \tau(a_1 + b_1t) \cdot \tau(a_2 + b_2t), \end{align*}$$ $$\tau(1 + 0t) = 1 + 0\sqrt{-5} = 1.$$

A bijective ring homomorphism is an isomorphism.

$\blacksquare$

Proposition A2

$R = \mathbb{Z}[t]/(t^2 + 5)$ is a Noetherian integral domain.

Proof

By theorem 1.2 we know that $\mathbb{Z}$ being Noetherian implies that $\mathbb{Z}[t]/(t^2 + 5)$ is Noetherian.

Suppose $a, b \in R$ are such that $ab = 0.$ Transporting $a, b$ over to the isomorphic ring $\mathbb{Z}[\sqrt{-5}]$ we get $a_1 + a_2\sqrt{-5}$ and $b_1 + b_2\sqrt{-5}$ such that their product is zero. But $\mathbb{Z}[\sqrt{-5}]$ is a subring of $\mathbb{C}$ and we know that $\mathbb{C}$ is a field, hence has no zero divisors.

$\blacksquare$

Proposition A3

Let $N : \mathbb{Z}[\sqrt{-5}] \to \mathbb{Z}$ be the map $a + b\sqrt{-5} \mapsto a^2 + 5b^2.$ Then $N$ is a homomorphism of multiplicative monoids.

Proof

Let $s = \sqrt{-5}.$

Observe that $N\left(a + bs\right) = \left(a + bs\right)\left(a - bs\right) = \left(a + bs\right)\overline{\left(a + bs\right)},$ where $\overline{x}$ denotes the complex conjugate of $x.$

Let $u, v \in \mathbb{Z}[\sqrt{-5}],$ then

$$N(uv) = uv \overline{(uv)} = uv \overline{u} \overline{v} = u\overline{u} \cdot v\overline{v} = N(u)N(v).$$

Since complex conjugate preserves $1,$ we have $N(1) = 1.$

$\blacksquare$

Proposition A4

Units of $R = \mathbb{Z}[\sqrt{-5}]$ are $\pm 1.$

Proof

Monoid homomorphisms preserve units, so units of $R$ have norm $\pm 1.$ Since norm is positive, the units can only have norm $1.$

Consider $a^2 + 5b^2 = 1.$ If $b \neq 0,$ the left-hand side is at least $5$ and equality cannot hold. So all units of $R$ have zero imaginary component. The equation $a^2 = 1$ has two solutions, namely $a = 1$ and $a = -1.$

Thus, the two units of $R$ are $1$ and $-1.$

$\blacksquare$

Proposition A5

Let $R = \mathbb{Z}[\sqrt{-5}].$ The elements $2, 3, (1 + \sqrt{-5}), (1 - \sqrt{-5}) \in R$ are irreducible and pairwise nonassociate.

Proof

By direct computation we establish

1. $N(2) = 4$
2. $N(3) = 9$
3. $N(1 + \sqrt{-5}) = 6$
4. $N(1 - \sqrt{-5}) = 6.$

Observe that there exist no integer solutions for $a^2 + 5b^2 \in \{2,3\}.$ Since the norm is a homomorphism of multiplicative monoids, any failure of irreducibility among these elements would produce an element of $R$ whose norm is $2$ or $3.$ No such elements exist. Thus we have established irreducibility.

By lemma 1.5 we know that associate elements $p, q$ of an integral domain (see A2) satisfy $p = uq$ for a unit $u.$ The only units of $R$ are $-1$ and $1$ (see A4).

By direct computation we can verify that multiplying any of the elements under consideration by either of the two units never produces another, distinct element on our list.

Hence, these elements are pairwise nonassociate.

$\blacksquare$

Proposition A6

Let $R = \mathbb{Z}[\sqrt{-5}].$ The elements $2, 3, (1 + \sqrt{-5}), (1 - \sqrt{-5}) \in R$ are not prime.

Proof

Observe that $(1 + \sqrt{-5})(1 - \sqrt{-5}) = 6$ and $6 \in (2), 6 \in (3).$ Elements of $(2)$ have norm divisible by $4,$ which $(1 \pm \sqrt{-5})$ doesn’t have. Elements of $(3)$ have norm divisible by $9,$ which $(1 \pm \sqrt{-5})$ doesn’t have. So $(2)$ and $(3)$ are not prime ideals.

Observe that $6 \in (1 + \sqrt{-5}).$ Elements of $(1 + \sqrt{-5})$ have the norm divisible by $6,$ which neither $2$ nor $3$ have. So $(1 + \sqrt{-5})$ is not prime.

Observe that $6 \in (1 - \sqrt{-5}).$ Elements of $(1 - \sqrt{-5})$ have the norm divisible by $6,$ which neither $2$ nor $3$ have. So $(1 - \sqrt{-5})$ is not prime.

$\blacksquare$

Proposition A7

Let $R = \mathbb{Z}[\sqrt{-5}]$. Then $R$ is not a UFD.

Proof

By A4 we know that $2, 3, (1 + \sqrt{-5}), (1 - \sqrt{-5}) \in R$ are non-units. By A5 we know that these elements are irreducible and pairwise nonassociate. We also know that $6 \in R$ is a non-unit.

There are multiple ways to factor $6$ into irreducibles:

$$2 \cdot 3 = 6 = (1 + \sqrt{-5}) \cdot (1 - \sqrt{-5}),$$

hence $R$ cannot be a UFD.

$\blacksquare$