0

Consider $\mathbb{Z}[\sqrt{10}] = \{x + \sqrt{10}y; x,y \in \mathbb{Z} \}$.

We want to show that there is no uniqueness of factorisation.

I think that it's better to consider $9 = 3 \cdot 3 = (\sqrt{10}-1)(\sqrt{10}+1)$. But now we need to show that $3$ and $\sqrt{10}-1$ are irreducibles in this ring.

There is a problem with this. My attempt : let's say $3 = (a+\sqrt{10}b)(c+\sqrt{10}d) = \sqrt{10}(ad + bc) + (ac +10bd)$, now we have that

$ad + bc = 0$ , hence $3 = (a-\sqrt{10}b)(c-\sqrt{10}d)$ is appropriate too. So we have that $9 = (a^2 - 10 b^2)(c^2 - 10d^3)$ which give us 4 possibilities : $9 = 1 \cdot 9 = (-1)\cdot(-9) = 3 \cdot 3 = (-3) \cdot (-3)$. It's easy to see that the las two cases are invalid.

But what about first scenario. There will be some solutions for $a^2 - 10b^2 = 1$ and $c^2 - 10d^2 = 9$, but how can I show that these solutions doesn't hold equation $ad + bc = 0$? Any hints?

openspace
  • 6,470
  • 4
    If $9 = 3\cdot 3 = (\sqrt{10}-1)(\sqrt{10}+1)$ is really going to prove that this is not a UFD, then you cannot show that $3$ and $\sqrt{10}\pm 1$ are prime. Recall that $p$ is a prime if and only if it is not a unit and whenever $p|xy$, you must have $p|x$ and $p|y$. If this were a factorization into primes, then it would be unique. What you need to show is that they are irreducible, not that they are prime. – Arturo Magidin Sep 10 '19 at 21:10
  • If $a^2-10b=1$, then $(a+\sqrt{10}b)(a-\sqrt{10}b) = 1$, so $a+\sqrt{10}b$ is a unit. – Arturo Magidin Sep 10 '19 at 21:11
  • @ArturoMagidin i need to show that there is no uniqueness of prime factorisation – openspace Sep 10 '19 at 21:13
  • No, you need to show that there is no uniqueness of factorization into irreducibles. – Arturo Magidin Sep 10 '19 at 21:14
  • @ArturoMagidin I want to show that Fundamental theorem of arithmetics doesnt hold. – openspace Sep 10 '19 at 21:18
  • Yes, and that's about factorization into irreducibles. You need to keep the distinction between "prime" and "irreducible" straight. An element $p$ is prime if and only if it is not a unit, and whenever $p|xy$, either $p|x$ or $p|y$. An element $m$ is irreducible if it is not a unit, not zero, and whenever $m=xy$, either $x$ is a unit or $y$ is a unit. They are not the same concept in general. – Arturo Magidin Sep 10 '19 at 21:18
  • Unique factorization says that every element can be written as a product of irreducibles, and that this product is unique up to order and associates. From this one can deduce that every irreducible is prime (in general, nonzero prime implies irreducible but not conversely). – Arturo Magidin Sep 10 '19 at 21:20
  • @ArturoMagidin okay, I understand. So the main idea is $a + b\sqrt{10}$ is $1$ in this ring? – openspace Sep 10 '19 at 21:21
  • No, it's not $1$ in that ring. It's a unit (if $a^2-10b^2=1$). – Arturo Magidin Sep 10 '19 at 21:21

0 Answers0