Proving that $mathbb{Z}[i]/langle 2+3irangle $ is a finite field

Question 1: We have $Bbb Z[i]simeq Bbb Z[x]/langle x^2+1rangle$. 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 $langle x^2+1,2+3xrangle$ 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 $langle x-8rangle$ contains both $x^2+1$ and $2+3x$. Since each ideal contains the generators of the other ideal, the two ideals must be equal.

Question 1:

Yes, this is correct.

Question 2:

You have a typo here. It should be

$$ mathbb{Z}[x]/langle 13,1+x^2,2+3xrangle cong mathbb{Z}_{13}[x]/langle 1+x^2,2+3xrangle. $$ And this is proved by the isomorphism (verify this is an isomorphism): $$ fleft(p(x)+langle 13,1+x^2,2+3xrangleright):=bar p(x)+langle1+x^2,2+3xrangle, $$ 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 $mathbb Zrightarrowmathbb Z_{13}$.

Question 3:

We verify these two ideals are equal: $langle1+x^2,2+3xrangle=langle x-8ranglesubseteqmathbb Z_{13}[x]$.

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 $langle1+x^2,2+3xranglesubseteqlangle x-8rangle$.

Conversely, $x-8=27x+18=9(3x+2)$ in $mathbb Z_{13}[x]$, so $langle1+x^2,2+3xranglesupseteqlangle x-8rangle$.

Therefore the two ideals are equal.

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Hope this helps.