4
$\begingroup$

Show that if $a+bi$ is irreducible over $\mathbb{Z} [i]$ then $a^2 + b^2$ is prime over $\mathbb{Z} $ . (a,b are non-zero integers)

I know that $\mathbb{Z} [i]$ is a Euclidean domain and so $a+bi$ is prime in it. Can I say that as the norm of $a+bi$ is $a^2 + b^2$ that $a^2 + b^2$ must be prime over $\mathbb{Z} $? This doesn't seem correct to me but it's the best try I have been able to come up with.

$\endgroup$
3
  • 2
    $\begingroup$ it fails with a = 3, b =0 since 3 is irreducible but its norm is not a prime. $\endgroup$ Mar 10, 2016 at 3:07
  • 1
    $\begingroup$ This works only if $ab\neq 0$. $\endgroup$
    – Macavity
    Mar 10, 2016 at 3:09
  • $\begingroup$ Edited to add that $a,b$ aren't $0$ ! $\endgroup$ Mar 10, 2016 at 3:18

1 Answer 1

2
$\begingroup$

The statement is not true for the reasons mentioned in the comments. The actual theorem says that $a^2+b^2$ is either a prime or the square of one. Now note that in the Gaussian integers, irreducible elements are also prime since it is a unique factorization domain. Suppose $a+bi$ is a Gaussian prime and $=a^2+b^2=p_1p_2...p_n$ is a prime factorization of its norm in $\mathbb{Z}$. Factorization in the Gaussian integers is unique. This implies that we only have 2 prime factors in $\mathbb{Z} \subset \mathbb{Z}[i]$ (at most as many as in $\mathbb{Z}[i])$ If there is only one prime factor then $a^2+b^2$ is prime. If there are 2 , then the prime factors $p_1$ and $p_2$ must be an associates of each of (WLOG) $a+bi$ and $a-bi$ respectively. Therefore $a+bi=\pm p_1$ or $a+bi=\pm ip_1$. In any case we have that $a-bi$ is also an associate of $p_1$. And thus we have that $(a+bi)(a-bi)=a^2+b^2=p_1^2$ (up to units).

$\endgroup$
1
  • $\begingroup$ Would you mind giving a specific example for when $a^2 + b^2 = p^2$? I find examples help me immensely in understanding! $\endgroup$ Mar 10, 2016 at 3:30

You must log in to answer this question.

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