Edit: I misread the text, which says before the following passage that we assume $x$ to be even. In which case the statement is true (I believe).
I am trying to understand a proof that for any [edit: even] integer $x$, then $(x + i)$ and $(x – i)$ are coprime in the Gaussian integers:
A prime element in $\mathbb{Z}[i]$ that divides both $x + i$ and $x − i$ divides their difference $2i$,
so it is up to units equal to $1 + i$.
However, $1 + i$ does not divide $x + i$ if $x$ is even,
so we conclude that $x + i$ and $x − i$ are coprime in $\mathbb{Z}[i]$.
Here is what I understand, one statement at a time:
A prime element in $\mathbb{Z}[i]$ that divides both $x + i$ and $x − i$ divides their difference $2i$,
This is a property of divisibility in general.
so it is up to units equal to $1 + i$.
This is because $2i = (1 + i)^2$ and $(1 + i)$ is irreducible, so the statement above follows from unique factorization in the Gaussian integers (which also gives us that prime = irreducible).
However, $1 + i$ does not divide $x + i$ if $x$ is even,
I checked this manually. Suppose $1 + i$ divides $x + i$, i.e. $x + i = (a + bi)(1 + i) = (a – b) + (a + b)i$. Then $a + b = 1$, so one is odd and the other is even, implying that $a – b$ is odd.
so we conclude that $x + i$ and $x − i$ are coprime in $\mathbb{Z}[i]$.
Now this is the leap that I don't understand. This is saying that the gcd of $x + i$ and $x – i$ is a unit. How does this follow from the previous statements?
Best Answer
This is not true: When $x=1$, $(1+i)(-i)=1-i$, hence the principal proper ideals $(1+i)$ and $(1-i)$ are the same.
Indeed, $x-i$ and $x+i$ are coprime in $\mathbb Z[i]$ for $x\in\mathbb Z$, iff $x$ is even. The "if" part has been established in the post. Or to put it in another way, if $x$ is even, then $(x-i)(x+i)=x^2+1$ is odd hence each prime factor of it is unramified, therefore $\langle x-i\rangle$ and $\langle x+i\rangle$ don't share any common prime factor.
When $x$ is odd, note that $$x+i = (\frac{1+x}{2} + \frac{1-x}{2}i)(1+i)$$ $$x-i=(\frac{x-1}{2} - \frac{x+1}{2}i)(1+i)$$
so they are both in the prime ideal $\langle 1+i\rangle$.