Let $p$ be a prime $\equiv 1 \pmod 4$. Prove that $\Bbb{Z}[i]/(p)$ $\cong$ $\Bbb Z_p \times\Bbb Z_p$

Question

I tried to construct a ring homomorphism from $\Bbb{Z}[i]$ to $\Bbb Z_p \times\Bbb Z_p$ but it turns out $f$($a+bi$) = $(a,b)$ is not a homomorphism because it doesn't split over multiplication. I think the requirement of $p$ $\equiv 1 \pmod 4$ has something to do with Fermat's sum of squares theorem but I can't proceed from there. Any help will be appreciated.

Answer 1

Consider $f(a+b\ i) = (a+b\ u, a-b\ u)$ where $u^2\equiv -1 \pmod p$ ($u$ exists since $p \equiv 1 \pmod 4$)

To see where this come from, observe that since $f$ is a ring homomorphism we have:

(All the following should be taken modulo $p$)

$f(a+b\ i) = f(a)+f(b)f(i) = a\ f(1) + b\ f(1)f(i) = a(1,1)+b(1,1)f(i)=(a,a)+(b,b)f(i)$

We want to find $f(i) = (x,y)$

Now, $(-1,-1)=f(-1) = f(i^2) = f(i)^2 = (x,y)^2 = (x^2,y^2)$

So $x^2=-1$ and $y^2 = -1$.

Then $x=\pm u$ and $y=\pm u$, but defining $f(i) = (u,u)$ or $f(i) = (-u,-u)$ give clearly not surjective functions, so we should have $f(i) = (u,-u)$ or $f(i) = (-u,u)$. Both of this choices works.

I left the details of proving that this is an isomorphism to you.

Answer 2

Seems like one way to go, assuming you know what the primes of $\mathbb{Z}[i]$ are, would be to apply the Chinese remainder theorem by noticing that, since $p \equiv 1 \mod 4$, there exist $(a,b) \in \mathbb{N}^2$ such that $p=a^2+b^2=(a+ib)(a−ib)$ with the factors being coprime in $\mathbb{Z}[i]$, and therefore: $$\mathbb{Z}[i]/(p)\cong \mathbb{Z}[i]/(a+ib) \times \mathbb{Z}[i]/(a−ib)$$ There only remains to show that both these quotients are of cardinal $p$, but that's true because, given a euclidean ring $R$, its euclidean norm $N$ and an element $s$ of $R$, we have $|R/(s)| = N(s)$, and here we have $N(a \pm ib) = |a \pm ib|^2 = p$.

(I find jjagmath's answer better due to being shorter, simpler, and constructive, but I figured that it wouldn't hurt to write mine.)

Answer 3

We can prove that $\Bbb{Z}[i]/(p) \cong \Bbb{Z}_p \times \Bbb{Z}_p$ using the Chinese Remainder Theorem and the fact that $p$ is a prime $\equiv 1 \pmod 4$.

First, note that $\Bbb{Z}[i]$ is a Euclidean domain with norm $N(a+bi) = a^2 + b^2$. Therefore, for any $z = a+bi \in \Bbb{Z}[i]$, we can write $z = qp + r$, where $q, r \in \Bbb{Z}[i]$ and $N(r) < N(p)$.

Now, let $\varphi : \Bbb{Z}[i] \to \Bbb{Z}_p \times \Bbb{Z}_p$ be the function defined by $\varphi(a+bi) = ([a]_p, [b]_p)$, where $[a]_p$ denotes the residue class of $a$ modulo $p$.

We claim that $\ker \varphi = (p)$. To see this, suppose $a+bi \in \ker \varphi$. Then $[a]_p = [b]_p = 0$, which means that $a$ and $b$ are both divisible by $p$. Therefore, we can write $a = mp$ and $b = np$ for some $m,n \in \Bbb{Z}$. But then $z = a+bi = (mp + ni)i$, so $N(z) = (mp)^2 + (np)^2 = p^2(m^2+n^2)$. Since $p \equiv 1 \pmod 4$, we have $m^2 + n^2 \equiv 0 \pmod 4$, which implies that $p^2$ divides $N(z)$. But since $N(z) < N(p)$, this means that $z = 0$ or $N(z) = 1$. Therefore, $\ker \varphi = (p)$.

Next, note that $\varphi$ is surjective, since for any $([a]_p, [b]_p) \in \Bbb{Z}_p \times \Bbb{Z}_p$, we can find $z = a+bi$ such that $[a]_p = [a]_p$ and $[b]_p = [b]_p$, simply by setting $a$ and $b$ to be the solutions of the system of congruences $a \equiv [a]_p \pmod p$ and $b \equiv [b]_p \pmod p$.

Therefore, by the First Isomorphism Theorem, we have $\Bbb{Z}[i]/(p) \cong \Bbb{Z}[i]/\ker \varphi \cong \operatorname{im} \varphi \cong \Bbb{Z}_p \times \Bbb{Z}_p$, as desired.