We were asked to show where the following reasoning goes wrong. Since $1+i$ and $1-i$ are prime elements in $\mathbb{Z}[i]$, the equation $$(-i)(1+i)^2=(1+i)(1-i)=2$$ show that unique prime factorization fails in $\mathbb{Z}[i]$. Of course, since $\mathbb{Z}[i]$ is a UFD there is something wrong, but I don't really see where. I know that the unique factorization is unique up until the order and a unit. Is that just it?
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Yes, this is it. Note that $1-i=(-i)(1+i)$, i.e these primes differ by a unit. So the two factorizations you wrote are equivalent.
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unique factorization is unique up until the order and a unit
Clearly, the difference between the two factorisations isn't just ordering. So then it must be about the units, mustn't it?
Note that $(-i)(1+i)^2$ and $(1+i)(1-i)$ have a factor in common. We may cancel that factor to reduce noise, before we even begin to do any real thinking. This results in the equality $$ (-i)(1+i)=(1-i) $$ The right-hand side is an element you know is a prime (or more concretely, an irreducible element). So one of the factors on the left-hand side must be a unit. Considering the definition of $i$, the factor $(-i)$ sticks out as a good candidate. And indeed, we have $(-i)i=1$, which makes $(-i)$ a unit.
So yes, up to units and ordering, the two given factorisations of $2$ are equal.
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1$\begingroup$ It is analogous to how $2\cdot 3=(-1)\cdot (-2)\cdot3$ are two different factorisations of $6$, and how that seemingly flies in the face of the fundamental theorem of arithmetic. Only your example is more hidden. $\endgroup$– ArthurCommented Apr 7 at 16:37