0
$\begingroup$

In $\mathbb Z[i]$, consider a relation $\alpha\beta=\epsilon\gamma^n$ for $\epsilon$ a unit and $(\alpha,\beta)=1$. Then why are each of $\alpha,\beta$ associated to nth powers $\xi^n,\eta^n$?

Obviously Bezout's identity hoolds in this ring, so write $\alpha x+\beta y=1$ for $x,y\in\mathbb Z[i]$. Then $$\epsilon\gamma^n=\alpha\beta=\alpha\frac{1-\alpha x}y$$ or $$y\epsilon\gamma^n=\alpha-\alpha^2x.$$ This doesn't seem to be very helpful (ideally, we'd use this to show that $\alpha$ is associated to an $n$th power and the same would work for $\beta$), as the problem looks more complicated.

Another more relaxing approach is to take norms and get $N(\alpha)N(\beta)=N(\gamma)^n$ so that each of $N(\alpha),N(\beta)$ divides a perfect $n$th power, but this doesn't seem much more useful.

Is there a better way?

$\endgroup$
1
  • $\begingroup$ You can use unique factorization directly. $\endgroup$ Commented Jul 30, 2015 at 0:18

1 Answer 1

1
$\begingroup$

The tool here is the fundamental theorem of algebra applied to $\Bbb Z[i]$.

Indeed, the factorization of $\gamma^n$ is a product of primes whose exponents are multiples of $n$. Each of this primes is a factor of either $\alpha$ or $\beta$ but not both. Then, each prime in the factorization of $\alpha$ is raised to an exponent which is a multiple of $n$, and the same for $\beta$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .