Proving that $\mathbb{Z}[i]/\langle 2+3i\rangle $ is a finite field Prove that $\mathbb{Z}[i]/\langle 2+3i\rangle $ is a finite field.
Hi. I can't try a few steps in the next solution
$$\mathbb{Z}[i]/\langle 2+3i\rangle \simeq \mathbb{Z}[x]/\langle 1+x^2,2+3x\rangle$$ and $9(1+x^2)+(2-3x)(2+3x)=13$ then $13\in \langle 1+x^2,2+3x\rangle $. Thus $$\mathbb{Z}[x]/\langle 1+x^2,2+3x\rangle=\mathbb{Z}/\langle 13,1+x^2,2+3x\rangle \\
\simeq \mathbb{Z}_{13}[x]/\langle 1+x^2,2+3x\rangle \simeq \mathbb{Z}_{13}$$
The last isomorphism induced by $x\mapsto 8$ (check $\langle 1+x^2,2+3x\rangle=\langle x-8\rangle $ in $\mathbb{Z}_{13}[x]$)
Therefore $\mathbb{Z}[i]/\langle 2+3i\rangle \simeq \mathbb{Z}_{13}$ finite field.
Question 1. Why $\mathbb{Z}[i]/\langle 2+3i\rangle\simeq \mathbb{Z}[x]/\langle1+x^2,2+3x\rangle$?
I have this: Let $$f:\mathbb{Z}[x]\to \mathbb{Z}[i]/\langle 2+3i\rangle $$ with $f(p(x))=p(i)+\langle 2+3i\rangle $ homomorphism with $\ker(f)=\langle 1+x^2,2+3x\rangle $ then $$\mathbb{Z}[x]/\langle 1+x^2,2+3x\rangle \simeq \mathbb{Z}[i]/\langle 2+3i\rangle $$It is correct?
Question 2. Why $\mathbb{Z}/\langle 13,1+x^2,2+3x\rangle \simeq \mathbb{Z}_{13}[x]/\langle 1+x^2,2+3x\rangle $?
Question 3. Why $\langle 1+x^2,2+3x\rangle=\langle x-8\rangle$?
 A: Question 1:
Yes, this is correct.
Question 2:
You have a typo here. It should be
$$
\mathbb{Z}[x]/\langle 13,1+x^2,2+3x\rangle \cong \mathbb{Z}_{13}[x]/\langle 1+x^2,2+3x\rangle.
$$
And this is proved by the isomorphism (verify this is an isomorphism):
$$
f\left(p(x)+\langle 13,1+x^2,2+3x\rangle\right):=\bar p(x)+\langle1+x^2,2+3x\rangle,
$$
where $\bar p(x)$ is the polynomial in $\mathbb Z_{13}[x]$ whose coefficients are the images of the coefficients of $p(x)$ under the canonical morphism $\mathbb Z\rightarrow\mathbb Z_{13}$.
Question 3:
We verify these two ideals are equal: $\langle1+x^2,2+3x\rangle=\langle x-8\rangle\subseteq\mathbb Z_{13}[x]$.
First: $1+x^2=x^2-64=(x-8)(x+8)$, and $2+3x=-24+3x=3(x-8)$, in $\mathbb Z_{13}[x]$, so $\langle1+x^2,2+3x\rangle\subseteq\langle x-8\rangle$.
Conversely, $x-8=27x+18=9(3x+2)$ in $\mathbb Z_{13}[x]$, so $\langle1+x^2,2+3x\rangle\supseteq\langle x-8\rangle$.
Therefore the two ideals are equal.

Hope this helps.
A: Question 1: We have $\Bbb Z[i]\simeq \Bbb Z[x]/\langle x^2+1\rangle$. And the third isomorphism theorem says that when we are dividing out by first $x^2+1$, and then $2+3x$, we are allowed to divide out but both of them simultaneously.
Question 2: Again justified by the third isomorphism theorem, dividing out by $13$ before the other generators.
Question 3: Here we have
$$
9(2+3x)=18+27x=-8+x
$$
So $\langle x^2+1,2+3x\rangle$ contains $x-8$. Now also note that
$$
3(x-8)=3x-24=3x+2\\
(x-8)^2+3(x-8)=x^2-16x+64 +3x-24=x^2+1
$$
So $\langle x-8\rangle$ contains both $x^2+1$ and $2+3x$. Since each ideal contains the generators of the other ideal, the two ideals must be equal.
