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I am building on a previous question, where we showed that $\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$. Now, I want to prove the statement in the title of this question.

First, let $p$ be an odd prime number. By definition, $p$ is prime in $\mathbb{Z}[i]$ if, whenever $p|ab$ for $a,b \in \mathbb{Z}[i]$, then $p|a$ or $p|b$.

I feel like I have quite a few pieces. Firstly, $$\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$$ Then, if we let $(p)$ be the ideal in $\mathbb{Z}[i]$ or $\mathbb{Z}[x]$ generated by the odd prime $p$, then we can conclude $$\mathbb{Z}[x]/(x^2+1)/(p) \cong \mathbb{Z}[i]/(p)$$ Finally by the proposition that, for any ring $R$, and $I$ and ideal of $R$, with $(I)=I[x]$ the ideal generated by $I$ in $R[x]$, then $R[x]/(I) \cong (R/I)[x]$, I want to assume that $$\mathbb{Z}[i]/(p) \cong \mathbb{Z}/n\mathbb{Z}[i]$$ or, more generally $$\mathbb{Z}[x]/(p) \cong \mathbb{Z}/n\mathbb{Z}[x]$$

I assume I need to use these pieces, but I not quite grasping how to tie the two sides together. I am having a harder time understanding how to use the right side, that $x^2+1$ does not have roots in $\mathbb{Z}/p\mathbb{Z}$. I am understanding this to mean $x^2+1 \ne 0 \mod p$. I also think I need to view $x^2+1$ as an element of $\mathbb{Z}[x]$, which would be the kernel of the map to $\mathbb{Z}[i]$. But, again, I cannot tie it all together.

jeffery_the_wind
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2 Answers2

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You're almost there: $$ \mathbb{Z}[i]/(p) \cong \mathbb{Z}[x]/(p,x^2+1) \cong (\mathbb{Z}/p\mathbb{Z})[x]/(x^2+1) $$ Then

$\mathbb{Z}[i]/(p)$ is a domain

$\quad$ iff $(\mathbb{Z}/p\mathbb{Z})[x]/(x^2+1)$ is a domain

$\quad$ iff $x^2+1$ is prime mod $p$

$\quad$ iff $x^2+1$ is irreducible mod $p$

$\quad$ iff $x^2+1$ has no roots mod $p$

(So we're actually talking about fields here.)

lhf
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    Yes, I guess the big point, that wasn't realizing to use was that for any ring $R$ and ideal $P$, $R/P$ is an integral domain iff $P$ is a prime ideal. Thank you, I will look over this. – jeffery_the_wind May 06 '20 at 13:55
  • Also, I assume that you are using the fact that $\mathbb{Z}/p\mathbb{Z}[x]$ is a unique factorization domain, so $x^2+1$ is prime iff it is irreducible. SO then the last step must be that $x^2+1$ is reducible iff it has a factor of degree 1. (I think i remember reading that for polynomials of degree 2 or 3, to be reducible it must have a factor of degree 1). But I guess I am not quite seeing why $x^2+1$ irreducible $\iff$ $x^2+1$ has no roots? – jeffery_the_wind May 06 '20 at 14:03
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    @jeffery_the_wind, $x^2+1$ reducible iff $x^2+1$ has a linear factor iff $x^2+1$ has a root – lhf May 06 '20 at 14:05
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Actually, it’s a lot of (technical) abstract nonsense, as all your pieces are right.

$x^2+1$ has no root in $\mathbb{Z}/p\mathbb{Z}$ iff $x^2+1$ is a prime element in $\mathbb{Z}/p\mathbb{Z}[x]$ (because it is a polynomial with degree 2). It occurs iff $\mathbb{Z}/p\mathbb{Z}[x]/(x^2+1)$ is an integral domain, iff $\mathbb{Z}[x]/(p,x^2+1)$ is an integral domain, iff $(\mathbb{Z}[x]/(x^2+1))/(p)$ is an integral domain, iff $\mathbb{Z}[i]/(p)$ is an integral domain, iff $p$ is a prime in $\mathbb{Z}[i]$.

If you want a less abstract proof, here it is:

Assume that $x^2+1$ has a root $r \in \mathbb{Z}/p\mathbb{Z}$. Then let $r_0$ be an integer above $r$: then $r_0^2+1=(r_0+i)(r_0-i)$ is divisible by $p$, but neither $r_0+i$ nor $r_0-i$ is divisible by $p$. This proves $\Rightarrow$.

Assume $p$ isn’t a prime in $\mathbb{Z}[i]$. Then there exists integers $a,b,c,d$ with $p$ not dividing $a+bi$ or $c+di$, but dividing $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. Then $p$ divides $(ac-bd)^2+(ad+bc)^2=(a^2+b^2)(c^2+d^2)$ thus $p$ divides eg $a^2+b^2$. As $p$ does not divide $a+bi$, there exists $b_1$ such that $p|bb_1-1$. Then $p|(ab_1)^2+1$, so that $ab_1 \in \mathbb{Z}/p\mathbb{Z}$ is a root of $x^2+1$.

Aphelli
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  • I guess the key piece I'm not quite grasping is the very first statement: "$x^2+1$ has no root in $\mathbb{Z}/p\mathbb{Z}$ iff $x^2+1$ is a prime element in $\mathbb{Z}/p\mathbb{Z}[x]$". All the rest I follow no problem. – jeffery_the_wind May 06 '20 at 14:10
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    Only $\Rightarrow$ is unclear. If $x^2+1|ab$ in $\mathbb{Z}/p\mathbb{Z}[x]$, write $a=(x^2+1)a_1+a_2$ and same for $b$ with $a_2,b_2$ with degree at most $1$. As $x^2+1$ has no root, if $q$ is a polynomial of degree $1$ in $\mathbb{Z}/p\mathbb{Z}[x]$, as $x^2+1$ has no root so $q$ and $x^2+1$ are coprime. So $a_2b_2$ is coprime with $x^2+1$ (a contradiction) unless $a_2=0$ or $b_2=0$, ie $x^2+1|a$ or $x^2+1|b$. – Aphelli May 06 '20 at 14:49