If p (mod 4) = 3 and p is a Gaussian Prime. How to show that Z[i]/(p) is equal to GF(p^2)/(x^2+1)?

Question

I currently work with Gaussian Integer. I try to use prime Gaussian Integer field for Elliptic Curve instead of prime field. . We know that every finite field isomorphic to polynomial field with modulus irreducible polynomial and all finite fields of the same size are isomorphic. . When I counting the order of the elliptic curve group, I found that Z[i]/(p) and GF(p^2)/(x^2+1) give the same output (same order). But if I change the modulus with another irreducible polynomial, the output is not same. . Why? Is any special thing that I missed? . Thankyou, and sorry for my bad english.

Answer 1

(This is a late answer, just found this question while looking for something specific related to elliptic curves.)

We have the sequence of isomorphisms of rings / fields. (I will use $=$ instead of $\cong$ to mark isomorphic objects, thus working the corresponding category.) I will also use the notation $\Bbb F_q$ for "the" field with $q$ elements instead of $GF(q)$. $$ \begin{aligned} \Bbb Z[i]/p &=(\Bbb Z[x]/(x^2+1))\ /\ (p)\\ &=\Bbb Z[x]\ /\ (x^2+1,\ p)\\ &=(\Bbb Z[x]/(p))\ /\ (x^2+1)\\ &=\Bbb F_p[x]\ /\ (x^2+1)\\ &=\Bbb F_{p^2}\ , \end{aligned} $$ where at the last step we use $p=-1$ modulo four, so that $-1$ is not a quadratic residue modulo $p$, i.e. $(x^2+1)$ has no roots in $\Bbb F_p$, i.e. it is an irreducible = prime polynomial.

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Let $i$ be the class of $x$ in the field $\Bbb F_{q}$, $q=p^2$, obtained above, so $i^2=-1$.

Then note that $GF(p^2)[y]/(y^2+1)$ is not a field, since the polynomial $y^2+1=(y+i)(y-i)$ splits in two factors over $GF(p^2)=GF(q)=\Bbb F_q$. This is related to the question of Ihf in the comments. The field with $q=p^2$ elements is the one mentioned in the above chain at forelast place, $\Bbb F_p[x]\ /\ (x^2+1)$.