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There are examples of factorization over complex numbers with coefficients in $\mathbb{Z}[\sqrt[]{-5}]$ For example: $(1+i \sqrt[]{5})\cdot(1-i \sqrt[]{5})=6 = 3\cdot2$

What is the smallest example (if any) of non unique factorization over Gaussian integers?

Stepan
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    Gaussian integers $ \Bbb Z[i]$ have unique factorization; and did you mean $\mathbb Z[\sqrt{\color{red}-5}]$? – J. W. Tanner Jun 26 '19 at 14:19
  • Thank you. I've fixed the domain. – Stepan Jun 26 '19 at 14:25
  • I don't think you have fixed the domain! What does $\Bbb Z+\sqrt 5$ even mean? – TonyK Jun 26 '19 at 14:41
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    "Gaussian Integers" usually refers to $\mathbb{Z}[i]$ rather than $\mathbb{Z}[\sqrt[]{-5}]$ . At first, it seems that there is no unique factorization as $10 = 2 \times 5 = (3 + i) \times (3 - i)$ but the catch is that $2$ and $5$ are not primes any more as $2 = (1 + i)(1 - i)$ and similarly for $5$. Don't assume that primes in $\mathbb{Z}$ are necessarily primes in larger rings. – badjohn Jun 26 '19 at 14:58
  • @badjohn How do we know that factorization process over Gaussian primes terminates in a finite number of steps for a finite integer? Is it because p=q*r, implies ||p|| > ||q|| and ||p|| > r ? – Stepan Jul 01 '19 at 16:34
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    @Stepan Yes, you may use the norms to prove that factorization must terminate. It's better to use $a^2 + b^2$ rather than $\sqrt(a^2 + b^2)$ as this will always be an integer. Proving the uniqueness requires more work. – badjohn Jul 01 '19 at 17:11

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