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This is the first time I'm posting a question, so I apologize beforehand if I have made any mistake. I'm trying to prove the following argument but couldn't finish ($\implies$) part. Also, I came up with a solution for ($\impliedby$) but still not sure.

Theorem: Let $p$ be an odd prime. Show that $p$ is not an irreducible element of $\mathbb{Z}[i]$ $\iff$ there exists an integer $x$ such that $x^2\equiv -1$ (mod $p$).

My attempt:

($\implies$) If $p$ is not irreducible in $\mathbb{Z}[i]$ (and $p$ is not zero nor a unit), it can be expressed as product of two non-unit elements, say $p=(a+bi)(c+di)$ for some $a,b,c,d\in \mathbb{Z}$, so $p^2=N(a+bi)N(c+di)$ but $N(a+bi)>1$ and $N(c+di)>1$ since they are not units. Then $p.p=p^2=(a^2+b^2)(c^2+d^2)$ is possible only if $p=a^2+b^2=c^2+d^2$. Observe $p\equiv 3$ (mod 4) or $p\equiv 1$ (mod $4$). First one isn't possible since $a^2+b^2\equiv 0,1$, or $2$ (mod $4$). So, we conclude $p\equiv 1$ (mod $4$)......

I don't know how to proceed from now on (though I know Legendre's symbol and using Euler's criterion I can prove the existence of such $x$, this course is about Rings & Fields and I am trying to prove the existence of $x$ for $p\equiv 1$ (mod $4$) using ring theory)

($\impliedby$) Suppose there is an $x\in\mathbb{Z}$: $x^2\equiv -1$ (mod $p$) and assume for a contradiction that $p$ is irreducible in $\mathbb{Z}[i]$. Then $x^2+1=p.t$ for some $t\in\mathbb{Z}$, ie $x^2+1^2=(x+i)(x-i)=p.t$. Since $\mathbb{Z}[i]$ is a principal ideal domain, any irreducible is also a prime. Thus, $p|(x+i)(x-i)\implies p|(x+i)$ or $p|(x-i)$. If $p|(x+i)$, then $p(k+mi)=(x+i)$ for some $k,m\in\mathbb{Z} \implies p.m=1$ which is not possible. (We reach a similar contradiction for $(x-i)$ case). Thus $p$ cannot be irreducible.

Eren
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  • Since $p \equiv 1 \text{ mod } 4$, $(-1)^{\frac{p-1}{2}} = 1$ and thus $x^2 \equiv -1 \text{ mod } p$ has a solution by Euler's criterion. You already seem to know this. You have already shown that the ring structure of $\mathbb{Z}[i]$ forces $p \equiv 1 \text{ mod } 4 $. The rest is just number theory... – Peyton Morris Mar 03 '25 at 21:17
  • Yes, but I want to use ring theory arguments not number theory. – Eren Mar 03 '25 at 21:22
  • I know. My point is that I think you are trying to use tools that don't yield results for the question you are trying to answer. You may as well forget about $p \equiv 1 \text{ mod } 4$ if you are only trying to use ring theory. Any further deduction of the answer from there is going to rely on properties of the integers, i.e. number theory. – Peyton Morris Mar 03 '25 at 22:14
  • I thought I should find a cyclic group of order $p-1=4k$ so that it has an element of order $4$ by Cauchy's Theorem but couldn't find one. If there were, I would go $x^2=-1$ but only such group that comes to my mind is $\mathbb{Z}_{p-1}$ which makes the relation mod $(p-1)$. – Eren Mar 03 '25 at 22:35
  • Every cyclic group of order $p-1$ is isomorphic to $\mathbb{Z}_{p-1}$. – Peyton Morris Mar 03 '25 at 22:46
  • The the linked dupes (and their Linked questions) for the standard proofs of this FAQ. – Bill Dubuque Mar 04 '25 at 03:31

1 Answers1

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Here's the closest I can get to a strictly "algebraic" solution.

For a prime $p$, it is a well known theorem that the group of units of the ring $\mathbb{Z}_p$ form a cyclic group of order $p-1$. Let $g$ denote the generator of this group. In a cyclic group whose order is even, there is precisely one element of order two. Since $g^{\frac{p-1}{2}}$ has order two, we must have $g^{\frac{p-1}{2}} \equiv -1 \text{ mod } p$. Hence, $x=g^{\frac{p-1}{4}}$ must be a solution to $x^2 \equiv -1 \text{ mod } p$.

You can find ten different proofs of the theorem here.

It's actually the case that $x^2 \equiv -1 \text{ mod } p$ has a solution if and only if $p \equiv 1 \text{ mod } 4$. You can show that $$x^2 \equiv -1 \text{ mod } p \implies p \equiv 1 \text{ mod } 4$$ by making use of Lagrange's theorem.

Peyton Morris
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