# $\mathbb Z[\sqrt{-5}]$ is not a UFD [duplicate]

Prove that the ring of integers of $\mathbb Q (\sqrt{-5})$ does not have unique factorisation.

Since $-5\equiv 3\pmod 4$, I know that the ring of integers of $\mathbb Q (\sqrt{-5})$ is $\mathbb Z [\sqrt{-5}]$.

I assume the way to prove that this does not have a unique factorisation is to give two different factorisations, but what exactly do I use to factorise? Do I consider the polynomial $x+\sqrt{-5}$?

• Look at $1+\sqrt{-5}$ and its product with its conjugate. Feb 10, 2015 at 0:26
• By and large this will be trial and error. There is a nice list of non-unique factorisations in ${\Bbb Q}(\sqrt{-d})$ in Stewart and Tall, Algebraic Number Theory, page 83. Feb 10, 2015 at 0:58
• Factorizations into irreducibles may not be unique. Feb 10, 2015 at 10:31

In a UFD, an element is irreducible if and only if it is prime.

Observe that $$2$$ is irreducible in $$\mathbb Z + \mathbb Z\sqrt{-5}$$: Suppose $$2 = (a + b \sqrt{-5})(c + b \sqrt{-5}),$$ taking the norm of both sides gives us $$4 = (a^2 + 5b^2)(c^2 + 5d^2)$$ which means $$a^2 + 5b^2 = 1, 2$$ or $$4$$. If $$a^2 + 5b^2 = 1$$, then $$a = 1$$ and $$b = 0$$ which means $$a + b \sqrt{-5} = 1$$ which is a unit and we're done. If $$a^2 + 5b^2 = 4$$, then $$a = 2$$ and $$b = 0$$ which means $$c + d \sqrt{-5} = \frac{2}{a + b\sqrt{-5}} = \frac{2}{2} = 1,$$ a unit, which means we're done. Notice that $$a^2 + 5b^2 = 2$$ can never happen: $$b$$ will have to be zero because if it's not, then the sum is greater than 5, which means it's greater than $$2$$, which means $$a^2 = 2$$ which only holds when $$a = \sqrt 2 \notin \mathbb Z$$, so this case can't occur. Conclude by definition that $$2$$ is irreducible in $$\mathbb Z + \mathbb Z \sqrt{-5}$$.

Observe that $$2 \mid 6 = (1 + \sqrt{-5})(1 - \sqrt{-5})$$ but $$2 \nmid (1 + \sqrt{-5}), (1 - \sqrt{-5})$$.

Say, by way of contradiction, that $$2 \mid 1 + \sqrt{-5}$$, then there exist $$a, b \in \mathbb Z$$ such that $$1 + \sqrt{-5} = 2(a + b \sqrt{-5})$$ which means $$2a = 1$$ and $$2b = 1$$ which can only happen if $$a = b = 1/2 \notin \mathbb Z$$, a contradiction. Similar reasoning works for $$1 - \sqrt{-5}$$.

Conclude that $$2$$ is not prime, but it is irreducible. Hence we're not in a unique factorization domain.

• When showing 2 is irreducible, in the step where we assume that $a^2 + 5b^2 = 4$, how can you write $\frac{2}{2}$? $\frac{1}{2} \not \in \mathbb{Z}$. May 2, 2019 at 13:56
• @Junglemath by cancellation in the integral domain $\mathbb{Z}[-\sqrt{5}]$ Feb 13, 2021 at 19:52

We can write two decompositions of $6$: $$6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$$ All factors are irreducible, since their norms can't be decomposed as the product of two norms $\neq 1$ and the factors on the rhs are not associated with the factors on the lhs, since their norms are different.

This proves $\mathbf Z[\sqrt{-5}]$ does not have unique factorisation.

• Actually $N(2)=4, N(3)=9$, and $N(1+\sqrt{-5}) = N(1-\sqrt{-5}) = 6$, so all the norms can be decomposed into (multiple) prime factors.
– ggg
Oct 30, 2017 at 15:33
• @gw Please read in detail my explanation: I wrote the norms are not the product of two norms, not that they are irreducible. Oct 30, 2017 at 17:27
• Ahh, my mistake!
– ggg
Oct 30, 2017 at 17:31