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Let $d$ be an integer, and define $$\mathbb{Z}[\sqrt{d}]:=\{ a+b\sqrt{d}|a,b \in \mathbb{Z} \}$$ . I have an example that proves that for a value of $d=-5$, the set above with the operations of addition, subtraction, and multiplication does not obey the fundamental theorem of arithmetic. Notably that $$6 = (2)(3)=(1+\sqrt{-5})(1-\sqrt{-5})$$ and $2,3,(\pm1\pm\sqrt{5})$ do not have non-trivial factorizations, ie. they are prime numbers. The proof that $2$ is prime is given by a contradiction. Assuming $2=(a+b\sqrt{-5})(c+g\sqrt{-5})$ where neither of the factors are $\pm1$, then this implies that $$4=|a+b\sqrt{-5}|^2|c+g\sqrt{-5}|^2=(a^2+5b^2)(c^2+5g^2)\implies a^2+5b^2=c^2+5g^2=2$$ which is impossible for any integers $a,b,c,g$. So 2 has no non-trivial factors in this set, hence 2 is prime.

The question now is to check if the fundamental theorem of arithmetic is true for different values of $d\in\{ \pm2, \pm3, \pm 11 \}$.

So far from my text, the only example is to prove by counter-example that some element of the set has 2 distinct prime factorizations. So my question here is: should I expect to be able to find counter-examples for all values of $d$? If not, then how would I go about proving that indeed, there is only a unique prime factorization of any element of the set?

Just as a side note, to support a counter-example, like the one above but for the case when $d=2$($\mathbb{Z}[\sqrt{2}]$), I am having a hard time proving that indeed $3$ is a prime number in this set. Maybe it isn't. Directly using the counter-example doesn't yield the same contradiction. Namely that $a^2+2b^2=c^2+2g^2=3$ works for $a=b=c=g=1$. However, $(\pm1\pm\sqrt2)(\pm1\pm\sqrt2)$ does not seem to actually yield $3$. So you can see I am getting a bit lost here. Thanks for the help!

J. W. Tanner
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jeffery_the_wind
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    Did you suddenly switch from $\mathbb{Z}\left[\sqrt-5\right]$ to $\mathbb{Z}\left[\sqrt 2\right]$? – Gary Feb 10 '20 at 17:02
  • @Gary yes i did sorry, I just fixed it. – jeffery_the_wind Feb 10 '20 at 17:04
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    for $d=\pm2, \pm3,$ and $\pm 11$, $\mathbb Z[\sqrt{d}]$ is a Euclidean domain, hence PID, hence UFD – J. W. Tanner Feb 10 '20 at 17:05
  • Great! Now reconsider your last paragraph accordingly. – Gary Feb 10 '20 at 17:05
  • @J.W.Tanner so I guess this proof comes down to proving that each one of those is a Euclidean domain. – jeffery_the_wind Feb 10 '20 at 17:09
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    Generally, $Z[\sqrt{d}]$ satisfies the FTA iff it is a PID iff it has class number $1$. For negative $d$ there are finitely many such rings and they are all known, see this. But for positive $d$ it is still an open problem whether there are infinitely many. – Tob Ernack Feb 10 '20 at 17:14
  • There is a nice geometric proof that $\mathbb{Z}[\sqrt{-2}]$ is a UFD (in fact, Euclidean). The idea is described here – Arturo Magidin Feb 10 '20 at 17:22
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    I think you're also misusing terminology a bit: in your example about the factorization of $6$ in $\mathbb{Z}[\sqrt{-5}]$, the factors you wrote down are irreducible and that is what you correctly had in mind. However: they are not prime! I recommend to recount the definition of irreducible and prime elements. – russoo Feb 10 '20 at 17:41

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