# Find the divisors of $2$ in $R$.

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.

• Careful: $1-i=-i(1+i)$… Jul 8, 2021 at 9:05
• 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 $2 = (1+i)(1-i) = i*(1-i)*(1-i) = i(1-i)^2 = 1*i*(1-i)^2$ 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 8, 2021 at 9:24
• @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. Jul 8, 2021 at 9:30

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.