Let F be a field. Can F be realized as a quotient of ℤ[i] ?

Lemma 1:

If F ≅ ℤ[i]/J for some ideal J of ℤ[i] then F is a finite field.

Proof:

It is sufficient to show that ℤ[i]/J is finite. Since ℤ[i] is not a field we can assume J not equal to zero. Since ℤ[i] is a Euclidean domain it is in particular a PID so J=(a+ib) with a and b not both zero. If (c+id)+J is an element of this quotient then dividing c+id by a+ib we get x,y,e,f in ℤ such that:

c+id=(x+iy)(a+ib)+e+if

with e²+f²≤a²+b². So that:

(c+id)+J=(e+if)+J with e²+f²≤a²+b². Since there are finite possibility for e and f we get that the quotient is finite. □

So if F can be realized as a quotient of ℤ[i] then F is finite, say |F|=pⁿ.

Lemma 2:

Let (a,b)=1. Then ℤ[i]/(a+ib)≅ℤ/(a²+b²).

Proof:

Let h be the ring homomorphism :

h:ℤ→ℤ[i]/(a+ib), n |—>n+(a+ib)

We first prove that h is surjective. Take c+id+(a+ib) in the quotient. Since (a,b)=1 we can find e,f in Z such that d=af+be. Then if n=c+fb-ea one get:

n+(e+if)(a+ib)=c+fb-ea+ea-fb+(af+eb)i=c+id

so

 c+id + (a+ib) = n+ (a+ib) =h(n)

so h is surjective. Now we prove that ker(h)=(a²+b²). Obviously h(a²+b² )=0 . To see the other inclusion take n in Ker(h) with n≠0. Then n+(a+ib)=(a+ib) so that n=(a+ib)(c+id) for some c and d in Z not both 0. So that:

c+id=n/(a+ib)=k(a-ib) for some k∈ℚ

so

n=k(a²+b²)

but since (a,b)=1 we get k∈ℤ (k=c/a=-d/b, so bc=-ad. Since a is coprime to b yet divides bc, a|c, so k∈Z). This prove the lemma. □

Lemma 3:

Let p ∈ ℤ a prime. Then p=a²+b² for some integer a and b ⇔ p=2 or p=1 mod 4

Proof:

The proof can be found in https://marcodamele.blog/wp-content/uploads/2024/07/gaussian-integer-2.pdf. □

So we are ready for the following:

Theorem:

Let F be a field. Then F is realizable as a quotient of ℤ[i] ⇔ |F|=p where p is a prime congruent 1 modulo 4 or p=2.

Proof:

If there is J ideal of ℤ[i] such that F=ℤ[i]/J then by Lemma 1 F is finite, say |F|=pⁿ for some natural number n>0. Since J is a non zero ideal of ℤ[i] we have that there is a+ib in ℤ[i] with (a,b)=1 such that:

ℤ[i]/J=ℤ[i]/(a+ib)≅ℤ/(a²+b²)

where we have the last isomorphism by lemma 2. Since F is a field we get pⁿ=a²+b²=q for some prime q. This imply n=1 and p=q. So that |F|=p and p=a²+b² so by lemma 3 p is a prime with p=2 or p=1 mod 4. Viceversa if |F|=p with p prime such that p=1 mod 4 or p=2 then p=a²+b². So, since the only field of p element is ℤp we get:

F=Z/(a²+b²)=Z[i]/(a+ib)

so that F is a quotient of ℤ[i]. □