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I am practicing for my algebra qual and I would like to know if $\mathbb{Z}[2\sqrt{2}]$ is a PID. I had no intuition at first except the fact that $\mathbb{Z}[i\sqrt{2}]$ is a ED with norm $N(a+i\sqrt{2}b)=a^2+2b^2$.

I tried proving that $\mathbb{Z}[2\sqrt{2}]$ is a ED with norm $N(a+b2\sqrt{2})=a^2-8b^2$ using the standard proof for $\mathbb{Z}[i\sqrt{2}]$ and $\mathbb{Z}[i]$. This led nowhere.

This may not be a PID, but I'm not sure how to prove it. Any tips? Thanks in advance.

Justine
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    I'm not sure that $\mathbb{Z}[2\sqrt 2]$ is a PID. I did some light calculation to get $2$ and $2\sqrt 2$ are irreducible in this domain. So now consider $8$ - there are two ways to "factorize" $8$, i.e, $2^3$ or $(2\sqrt 2)^2$. So the ring is not a UFD, then definitely not a PID. I will check my calculations again.... – M. L. Nguyen May 22 '15 at 14:01

2 Answers2

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It can be proved that $2$ and $2\sqrt{2}$ are coprime in $\mathbb{Z}[2\sqrt{2}]$, but $(2,2\sqrt{2})\neq(1)$, so it cannot be a principal ideal in $\mathbb{Z}[2\sqrt{2}]$.

Censi LI
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  • @user26857 I've replaced $1$ by more standard notation $(1)$, and "coprime" means $\gcd(2,2\sqrt 2)=1$, which I think is commonly used... – Censi LI May 23 '15 at 15:15
  • @user26857 I have to say you are right. Thanks for pointing out it. – Censi LI May 23 '15 at 15:23
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If $\mathbb{Z}[2\sqrt{2}]$ is a PID, then it is a UFD.

But, $8 = 2 \times 2 \times 2 = 2\sqrt{2} \times 2\sqrt{2}$, that is, $8$ has two kinds of factorization.

Contradiction.

P.-S. Park
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    UFD is about factorization into primes, not every factorization. Have you proved that $2$ and $2\sqrt 2$ are primes in $\mathbb Z[2\sqrt 2]$? – user26857 May 23 '15 at 14:47
  • Suppose $\mathbb{Z}[2\sqrt{2}]$ is a UFD. Since it is a subring of a UFD $\mathbb{Z}[\sqrt{2}]$ and $2 = \sqrt{2} \times \sqrt{2}$ in $\mathbb{Z}[\sqrt{2}]$, $2$ should be irreducible in $\mathbb{Z}[2\sqrt{2}]$. – P.-S. Park May 25 '15 at 01:35