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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 $x+\sqrt{-5}$?

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    $\begingroup$ Look at $1+\sqrt{-5}$ and its product with its conjugate. $\endgroup$
    – Lubin
    Feb 10, 2015 at 0:26
  • $\begingroup$ 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. $\endgroup$
    – David
    Feb 10, 2015 at 0:58
  • $\begingroup$ Factorizations into irreducibles may not be unique. $\endgroup$
    – orangeskid
    Feb 10, 2015 at 10:31

2 Answers 2

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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.

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

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  • $\begingroup$ 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. $\endgroup$
    – ggg
    Oct 30, 2017 at 15:33
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    $\begingroup$ @gw Please read in detail my explanation: I wrote the norms are not the product of two norms, not that they are irreducible. $\endgroup$
    – Bernard
    Oct 30, 2017 at 17:27
  • $\begingroup$ Ahh, my mistake! $\endgroup$
    – ggg
    Oct 30, 2017 at 17:31

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