A question about unique factorization domain

Question

Show that $\mathbb{Z}[i]$ is a unique factorization domain

Since $\mathbb{Z}[i]$ is example of a Euclidean domain so it both P.I.D and U.F.D

can any one explain why $10=(3+i)(3-i)=2\times 5$ does not contradict unique factorization in $\mathbb{Z}[i]$.

Is any way to prove that $\mathbb{Z}[i]$ is UFD from the definition of UFD

thank you.....

Answer 1

$2=(1+i)(1-i)$ is not a prime in $\Bbb Z[i]$. Also $5=(2+i)(2-i)$. The prime factorization of $10$ is $(1+i)(1-i)(2+i)(2-i)$ and $(1+i)(2-i)=3+i$

Answer 2

The easiest way to show that $\mathbb{Z}[i]$ is a UFD from the definitions, is to show that $\mathbb{Z}[i]$ has a Euclidean division algorithm, and hence is a PID and a UFD, using the definition of a UFD. I believe that every reasonable proof anyway will use the Euclidean norm $N(x)=x\overline{x}=a^2+b^2$ for $x=a+bi$. So I don't see a good reason to avoid the standard proof with the division algorithm and the Euclidean norm.