Question

Factorise 10 in two ways in Z[under-rootof -6]. Hence, show that Z[under-rootof -6] is not a UFD.

Answer 1

Answer on Question #44585 – Math – Abstract Algebra

Factorise 10 in two ways in $\mathbb{Z}[\sqrt{-6}]$. Hence, show that $\mathbb{Z}[\sqrt{-6}]$ is not a UFD.

Solution.

Note that:

$$10 = 4 - (-6) = 2^2 - (\sqrt{-6})^2 = (2 + \sqrt{-6})(2 - \sqrt{-6});$$

So:

$$10 = 2 \cdot 5 = (2 + \sqrt{-6})(2 - \sqrt{-6});$$

Consider a norm of a number $x \in \mathbb{Z}[\sqrt{-6}]$:

$$x = a + b\sqrt{-6}, \quad a, b \in \mathbb{Z} \Rightarrow N(x) = a^2 + 6b^2;$$ $$\forall x, y \in \mathbb{Z}[\sqrt{-6}]: N(xy) = N(x)N(y);$$

Prove that $2, 5, 2 + \sqrt{-6}, 2 - \sqrt{-6}$ are irreducible elements of $\mathbb{Z}[\sqrt{-6}]$.

1) 2:

$$a, b \neq 1, 2 = ab \Rightarrow 4 = N(2) = N(a)N(b) \Rightarrow N(a) = N(b) = 2;$$ $$a = m + n\sqrt{-6}, \quad N(a) = 2 \Rightarrow m^2 + 6n^2 = 2 \Rightarrow n = 0, m^2 = 2 - \text{contradiction} \quad (m \in \mathbb{Z});$$

So, 2 is irreducible.

2) 5:

$$a, b \neq 1, 5 = ab \Rightarrow 25 = N(5) = N(a)N(b) \Rightarrow N(a) = N(b) = 5;$$ $$a = m + n\sqrt{-6}, \quad N(a) = 5 \Rightarrow m^2 + 6n^2 = 5 \Rightarrow n = 0, m^2 = 5 - \text{contradiction} \quad (m \in \mathbb{Z});$$

So, 5 is irreducible.

3) $2 + \sqrt{-6}$:

$$a, b \neq 1, 2 + \sqrt{-6} = ab \Rightarrow 10 = N(2 + \sqrt{-6}) = N(a)N(b) \Rightarrow N(a) = 2, N(b) = 5;$$

We proved in (a),(b) that $\forall a \in \mathbb{Z}[\sqrt{-6}]: N(a) \neq 2, N(a) \neq 5$.

So, $2 + \sqrt{-6}$ is irreducible.

4) $2 - \sqrt{-6}$:

$$a, b \neq 1, 2 - \sqrt{-6} = ab \Rightarrow 10 = N(2 - \sqrt{-6}) = N(a)N(b) \Rightarrow N(a) = 2, N(b) = 5;$$

We proved in (a),(b) that $\forall a \in \mathbb{Z}[\sqrt{-6}]: N(a) \neq 2, N(a) \neq 5$.

So, $2 - \sqrt{-6}$ is irreducible.

Now prove that $\mathbb{Z}[\sqrt{-6}]$ is not a UFD.

Assume the contrary. We have two distinct factorizations of a number 10. So, all irreducible factors are pairwise associated. Hence:

$$\left[ \begin{array}{l} 2 \sim 2 + \sqrt {- 6} \\ 2 \sim 2 - \sqrt {- 6} \end{array} \right. \Rightarrow \left[ \begin{array}{l} N (2) = N \big (2 + \sqrt {- 6} \big) \\ N (2) = N \big (2 - \sqrt {- 6} \big) \end{array} \right. \Rightarrow 4 = 1 0 - \text {c o n t r a d i c t i o n}.$$

So, our assumption doesn't hold and $\mathbb{Z}\big[\sqrt{-6}\big]$ is not a UFD.

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