What is the structure of $\mathbb Z[i]/\mathfrak p$ where $\mathfrak p$ is a prime.

Question

Initially, I was trying to prove both the isomorphism of $\mathbb Z[i]/\mathfrak p\cong \mathbb{Z}[x]/(p,x^2+1)\cong \mathbb{F}_p[x]/(x^2+1)$, where $\mathfrak p$ is a prime in $\mathbb Z[i]$ for some $p $ prime in $\Bbb Z.$ (Later on from the comment I got that the first isomorphism is not true for all prime in integers. So I am trying to ask, What is the structure of $\mathbb Z[i]/\mathfrak p$ where $\mathfrak p$ is a prime?

I have proved that for $ \phi: R \to S$ homomorphism the inverse image of a prime is prime. Now $\Bbb Z$ has inclusion in $\Bbb Z[i]$ that $i: \Bbb Z \to \Bbb Z[i]$ and hence the inverse, $i^{-1}(\mathfrak p)$ in $\mathbb Z$ is a prime and I can assume that $<p>$ is the corresponding prime ideal in $\Bbb Z$. I also know that $\mathbb Z[i]$ is a ED and hence $\mathfrak p=<P>$ where $P=a+bi$ a prime in $\mathbb Z[i]$ and also the norm of $P$, $N(P)=a^2+b^2$ is a prime. Am I going correct? Let me know if there is any other way.

So, I have a lot of pieces but I can't bring that together and for the 2nd isomorphism If I just consider the map $\psi:\mathbb{Z}[x]/(x^2+1)\to \mathbb{F}_p[x]/(x^2+1)$ s.t $f(x)\mapsto f(x)(\mod p)$ and then calculate the kernel, would that be enough? It is my begining in algebra, so I wan't check every details minutely. Any help would be appreciated. Even if I get some stepwise hint in stead of answer, that will help me in learning!

Edit From leoli's comment I can see that $\mathbb Z[i]/\mathfrak p$ can be a field as well. How to prove all the cases from the scratch as I am not getting any idea for that.

Here is a hint: If $R$ is a commutative ring, and $a,b \in R$, then the generated ideal $(b+(a))$ of $R/(a)$ by $b+(a)$ is $(a,b)/(a)$, so the third isomorphism theorem tells us that $$(R/(a))/(b+(a)) \cong R/(a,b).$$ — azif00, Sep 20 '21 at 05:54
What you are trying to prove is wrong. Consider a prime $\mathfrak p$ lying above $5$ (or any other rational prime $\equiv 1\pmod 4$). Then $\Bbb Z[i]/\mathfrak p$ is a field while $\Bbb F_5[x]/(x^2+1)$ is not. — leoli1, Sep 20 '21 at 08:02
@GerryMyerson Haha, $j: \Bbb Z \to \Bbb Z[i]$ s.t $j(a)=a$ is a ring homomorphism and hence the inverse, $i^{-1}(\mathfrak p)$. Will that work? — Ri-Li, Sep 20 '21 at 18:06
@leoli1 OMG. Can you please help me giving a detailed proof or hint on how to proceed? — Ri-Li, Sep 20 '21 at 18:10
While $\mathbb{Z}[x]/(p,x^2+1)\cong \mathbb{F}_p[x]/(x^2+1)$, the other isomorphism does not hold. One example is $\mathfrak p=(1+i)\mathbb Z[i]$. In this case $\mathbb Z[i]/(1+i)$ is a field with 2 elements, while $\mathbb{F}_p[x]/(x^2+1)$ has $p^2$ elements, so no chance to have $2=p^2$. — user26857, Sep 20 '21 at 18:53
Yeah, I have changed my question, after seeing the comment of @leoli1. So how does the set $\Bbb Z[i]/\mathfrak p$ look like? This is my question now. — Ri-Li, Sep 20 '21 at 19:04
If $\mathfrak p=(a+bi)$ with $a,b\ne0$, then $a^2+b^2$ is a prime integer, say $p$. In this case $\mathbb Z[i]/(a+bi)\simeq\mathbb F_p$. — user26857, Sep 20 '21 at 19:06
For a more general result (that is, $a,b$ coprime) see here: https://math.stackexchange.com/a/23379/121097 — user26857, Sep 20 '21 at 19:25
@Ri-Li I have posted another answer; see here: https://math.stackexchange.com/a/4255815/121097. — user26857, Sep 20 '21 at 21:37
So by "inverse" you mean "inverse image under $j$". But $j$ is not onto, so that's a little strange. — Gerry Myerson, Sep 21 '21 at 05:18

score 3 · Accepted Answer · answered Sep 20 '21 at 22:27

Let $\mathfrak p=(a+bi)$. There are two cases:

$ab=0$. In this case $\mathfrak p=(0)$, or $\mathfrak p=(p)$ with $p\in\mathbb Z$ a prime number, $p=4k+3$. Obviously, $\mathbb Z[i]/(0)=\mathbb Z[i]$. On the other side, $\mathbb Z[i]/(p)\simeq\mathbb F_p[X]/(X^2+1)$ which is a field with $p^2$ elements.
$ab\ne0$. In this case $p=a^2+b^2$ is a prime number. It follows that $\gcd(a,b)=1$ and we have $\mathbb Z[i]/\mathfrak p\simeq\mathbb F_p$. (For the last isomorphism see here.)

score 1 · Answer 2 · answered Sep 20 '21 at 18:18

1

Hint : Consider the morphism \begin{align*} \varphi : \quad& \mathbb{Z}[i] \to \mathbb{F}_p[x]/(x^2+1)\\ & a+ib \mapsto \overline{\bar{a}+\bar{b}x}. \end{align*}

(where $\bar{a}$ and $\bar{b}$ denote the projections of the integers $a$ and $b$ in $\mathbb{F}_p$, and the big overline denotes the projection of $\bar{a}+\bar{b}x$ in the quotient).

Prove that this morphism is surjective, and that its kernel is exactly $(p)$. Then apply the first isomorphism theorem to get that $$\mathbb{Z}[i]/(p) \simeq \mathbb{F}_p[x]/(x^2+1) $$

answered Sep 20 '21 at 18:18

TheSilverDoe

29,720

From where do we get that $p$ which is prime in integers in stead of a prime in gaussian integer? This might be a very silly question but I am stuck on it. – Ri-Li Sep 20 '21 at 19:01
@Ri-Li This answer provides an isomorphism which holds for any prime integer $p$, not that $p$ is prime in $\mathbb Z[i]$. – user26857 Sep 20 '21 at 19:03
Yes, but my original question is how do I prove $\Bbb Z[i]/\mathfrak p \cong \Bbb F_p[x]/(x^2+1) $? – Ri-Li Sep 20 '21 at 19:06
I think I should make it more precise. – Ri-Li Sep 20 '21 at 19:07
@Ri-Li You can't prove a wrong isomorphism. – user26857 Sep 20 '21 at 19:22
So I am trying to ask, What is the structure of $\mathbb Z[i]/\mathfrak p$ where $\mathfrak p$ is a prime? From some comments i got that the first isomorphsim is true and for the other primes it is a field. My question is how to prove that. – Ri-Li Sep 20 '21 at 19:25
You know the structure of $\mathbb Z[i]/(p)$, from this answer. If $\mathfrak p$ contains $(p)$, then $\mathbb Z[i]/\mathfrak p$ is a quotient of this ring by the third isomorphism theorem. – David A. Craven Sep 20 '21 at 21:19

What is the structure of $\mathbb Z[i]/\mathfrak p$ where $\mathfrak p$ is a prime.

2 Answers2