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Task:

In the ring $R: = Z [i]$ we have to show:

Find the divisors of $2$ in $R$. (Hint: Use the norm $N (α) = a^2 + b^2$ for $α = a + bi ∈ R$.)

Attempt:

So I can already determine the number of divisors:

In $Z [i]$, $2$ has the decomposition $2 = (1 + i) (1-i)$.

Then the number of divisors is $4 * (1 + 1) * (1 + 1) = 16$. The $4$ indicates the number of units in $Z [i] = {1, -1, i, -i}$.

So how do I determine all $16$ divisors with the hint? I'm not getting any further.

Vek
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    Careful: $1-i=-i(1+i)$… – Aphelli Jul 08 '21 at 09:05
  • Hm now i got it to 8 divisors maybe... so we see $(1-i) = -i(1+i)$ and $(1+i)=i(1-i)$ so we can rewrite $2 = (1+i)(1-i) = i(1-i)(1-i) = i(1-i)^2 = 1i(1-i)^2$ Divisors are $1,i,(1-i),(1-i)^2$, same for $(1-i)=-i(1+i)$ gives us $-1,-i,(1+i),(1+i)^2$ ? – Vek Jul 08 '21 at 09:24
  • @Vek You say $(1+i)^2$ and $(1-i)^2$. It is easier, both for you and for us, if you call them $2i$ and $-2i$. Those are more obviously divisors of $2$, and also hint at a pattern. – Arthur Jul 08 '21 at 09:30

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So, you are finding only twelve divisors of $2$ when you try to count them. That's because there are actually twelve divisors of $2$, not sixteen.

When you render $2=(1+i)(1-i)$ and then assume divisors having the form

$u(1+i)^a(1-i)^b; a,b\in\{0,1\}, u\in\{\text{units}\}$

you miss the fact that all the factors do not combine independently: there is a unit $u$ other than the identity such that $(1+i)^1=u(1-i)^1$ (can you see what this unit $u$ is)? That means you are double-counting some factors. We say that the factorization of $2$ into $(1+i)(1-i)$ ramifies it.

To get around this ramification use a factorization that avoids using both elements of the ramified pair. For instance, you may combine $2=(1+i)(1-i)$ with $1+i=i(1-i)$ to get $\color{blue}{2=i(1-i)^2}$. Now (since the factor $i$ is a unit and $1-i$ is the only prime appearing) the divisors of $2$ are counted out by the set

$u(1-i)^b; b\in\{0,1,2\}, u\in\{\text{units}\}$

which properly gives twelve divisors with no double-counting due to ramification.

Oscar Lanzi
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