6
$\begingroup$

I want to show that for $p$ a prime, if $p\equiv 1\pmod{4}$, then $p$ is not irreducible in $\mathbb{Z}[i]$.

I know that $x^2\equiv -1\pmod{p}$ has a solution when $p\equiv 1\pmod{4}$, so $p\mid x^2+1$. I could then prove that $\mathbb{Z}[i]$ is a PID, so Euclid's lemma holds, and then see that $p\mid (x+i)$ or $p\mid (x-i)$, using the fact that $\mathbb{Z}[i]$ is a PID, so if $p$ is irreducible, then $p$ is prime. But then this gives a contradiction since $1/p$ is not an integer, so $(x\pm i)/p\notin\mathbb{Z}[i]$.

I could do all this, but it seems like a lot of effort for the result. Is there a slicker, more elegant proof that $p\equiv 1\pmod{4}$ implies $p$ is not irreducible in the Gaussian integers?

By the way, I'm using this to finally conclude Fermat's theorem on the sum of two squares, so I don't want to refer to that in this.

  • 0
    A prime is reducible in the Gaussian integers iff it's a sum of two natural squares. So see [Proofs of Fermat's theorem on sums of two squares](http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares).2011-10-08
  • 0
    Dear Travis, The argument you outline is the standard one, and I'm not sure you can make it any shorter, since proving the reducibility statement you ask about is tantamount to proving that $\mathbb Z[i]$ is a PID. Regards,2011-10-09

4 Answers 4