Let $p\in \mathbb{Z}$ be prime such that $p \equiv 3 \bmod 4$. I've read in many places that $p$ must be irreducible in $\mathbb{Z}[i]$, but I can't see why.
Could someone please explain the reason?
Determining that a certain prime integer is irreducible in the Gaussian integers
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abstract-algebra
ring-theory
divisibility
gaussian-integers
2 Answers
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Note that the norm-square $|a+bi|^2 = a^2+b^2$ is always an integer for gaussian integers, and is multiplicative in the sense that $|z|^2|w|^2 = |zw|^2$. In particular since $|p|^2 = p^2$ if $p$ were to factor in $\mathbb{Z}[i]$ it would have to factor as $p = zw$ with $|z|^2 = |w|^2 = p$.
However if we write $z=a+bi$ this means $a^2+b^2=p$. In particular, this implies that $a^2 \equiv -b^2 \bmod p$ so $(a/b)^2 \equiv -1 \bmod p$. However if $p \equiv 3 \bmod 4$ then the multiplicative group $\mathbb{Z}/p\mathbb{Z}^*$ is an abelian (in fact cyclic) group of order not divisible by four and hence can't have any nontrivial solutions to $x^4 \equiv 1$. So this is a contradiction, and therefore $p$ must remain irreducible in $\mathbb{Z}[i]$.
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This is because of an elementary fact , that $x^2 \equiv -1$ modulo p has only solutions iff $ p \equiv 1 $ modulo 4. You can see this proof in Niven Zuckermann's book of Number Theory.