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I am wondering how to show that $\mathbb{Z}[\sqrt {10}]$ is not a UFD.

My only idea is to show that there are two factorizations of $10$, say, $ab, uv$ such that $a$ is not a unit times $u$ or $v$. In this ring $10=2\cdot5=\sqrt {10}\cdot \sqrt {10}$, so it suffices to show $2$ is not a unit times $\sqrt {10}$. Suppose $2=\sqrt {10}(a+b\sqrt{10})=a\sqrt{10}+10b$. Then $a=0$ since $\sqrt{10}$ is not rational. So $10b=2$, which has no integer solutions. So $\mathbb{Z}[\sqrt {10}]$ is not a UFD.

Is my reasoning correct? What are the flaws?

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    $\begingroup$ I don't think this quite works. You show that $2$ is not a unit times $\sqrt{10}$, but why couldn't it factor further, or factor as something else times $\sqrt{10}$? I think a better approach would be to prove that $2$ is irreducible using a norm argument, note that $2$ must also be a prime if the ring is a UFD, then show it does not divide $\sqrt{10}$. $\endgroup$ Commented Sep 24, 2015 at 1:36
  • $\begingroup$ Related. $\endgroup$ Commented Sep 24, 2015 at 3:19
  • $\begingroup$ If you can show that $2,5$ are irreducibles, then you're done, since clearly neither of $2,5$ are associates with $\sqrt{10}$, so $\sqrt{10}\cdot\sqrt{10}$ is itself a factorization of irreducibles or can be factorized into a product of at least $4$ irreducibles. Alternatively, look at $9=3\cdot3=(\sqrt{10}+1)(\sqrt{10}-1)$. Since $\pm3$ are quadratic nonresidues, $3,\sqrt{10}\pm1$ are irreducibles, $\endgroup$ Commented Nov 24 at 3:27

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There is a small problem with your solution, since you don't know that $\sqrt{10}$ is irreducible, and so $2$ and $\sqrt{10}$ could have a common factor. The standard approach to deal with this is to use norms.

Define the norm $N(a+b\sqrt{10})=(a+b\sqrt{10})(a-b\sqrt{10})=a^2-10b^2$. The norm satisfies the property that $N(xy)=N(x)N(y)$.

The first observation is that $x$ is a unit if and only if $N(x)=\pm 1$, because $N(x)$ is a multiple of $x$ and $1=N(1)=N(xx^{-1})=N(x)N(x^{-1})$, and the only integer divisors of $1$ are $\pm 1$.

With this in mind, we can calculate $N(2)=4$ and $N(\sqrt{10})=-10$, and since the norms aren't multiples of each other, the numbers can't be multiples of each other. If they had a common non-unit factor, though, it would have to have norm $\pm 2$. So let us show that there are no elements with norm $\pm 2$.

Suppse $a^2-10b^2=\pm 2$. Reducing mod 10, we get $a^2\equiv \pm 2 \pmod{10}$, but no perfect square ends with a 2 or an 8, so this has no solutions.

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  • $\begingroup$ I believe you showed that 2, $\sqrt {10}$ are irreducible. Can you also apply this argument to show 5 is irreducible? $N(5)=25$, so I tried showing no norm $a^2 - 10b^2$ can equal 5. The most I can conclude is that $a$ is a multiple of 5, but since the coefficient on $b^2$ is a multiple of 5, I can't say anything about $b$. $\endgroup$ Commented Jun 15, 2016 at 21:03
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    $\begingroup$ While the argument I made wasn't specifically showing that those two numbers were irreducible, it follows almost immediately from what I wrote (instead of looking for common factors, just look for factors of the numbers individually). For $5$, if you reduce your equation modulo $5$, you get that $a=5k$, and so $5k^2-2b^2=\pm 1$. But then $b^2\cong \pm 2 \pmod 5$, and this has no solutions, so $5$ is indeed irreducible. $\endgroup$ Commented Jun 15, 2016 at 21:49
  • $\begingroup$ @mechanodroid I am confused by your comment , as we are not working in the ring $\mathbb Z[i]$, and so we do not have access to factorizations using elements from that ring. Whether an element is irreducible or not is dependent on the ring you are in, not an absolute proposition. $\endgroup$ Commented Apr 10, 2018 at 12:25
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    $\begingroup$ Also it suffices once you get down to $a^2 - 10b^2 = \pm 2$ to then reduce mod $10$ as there are no integer solutions to $a^2 \equiv 2$ mod $10$ or $a^2 \equiv 8$ mod $10$. $\endgroup$ Commented Oct 3, 2020 at 13:37
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    $\begingroup$ @SamanthaWyler Nice catch, thanks. $\endgroup$ Commented Oct 8, 2020 at 23:03

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