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Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements

The field portion I understand. $\mathbb{Z}[i]$ is a PID and because $q$ is congruent to 3 mod 4 it is irreducible thus prime. and hence the ideal $(q)$ is prime and therefore is maximal because $\mathbb{Z}[i]$ is a pid. Thus $\mathbb{Z}[i]/(q)$ is a field

I am not totally clear on the number of elements being $q^2$. I think perhaps the division algorithm would help but I am still not totally clear on how to use it in a case like this.

oliverjones
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2 Answers2

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You can get this field from $\mathbb{Z}[x]$ by first killing $x^2+1$ and then killing $q$, as you have shown above. But you could also get there by killing $q$ first, and then $x^2+1$. In this way, we get that the field is isomorphic to $F_p[x]$ mod $x^2+1$. How many elements does this field have?

TomGrubb
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  • not sure what you mean by "killing $x^2+1$ then killing $q$ – oliverjones Apr 23 '15 at 04:24
  • bburGsamohT means that $\mathbb{Z}[i]$ is $\mathbb{Z}[x]/(x^2 + 1)$, and hence $\mathbb{Z}[i]/(q)$ is $\mathbb{Z}[x]/(x^2 + 1, q)$. Now, we think of this quotient as $\mathbb{Z}[i]/(q)$ by quotienting by $x^2 + 1$ first, but instead we can quotient by $q$ first, and we get that this quotient is also $\mathbb{F}_q[x]/(x^2 + 1)$. – Qiaochu Yuan Apr 23 '15 at 05:30
  • @QiaochuYuan: In what $q\equiv 3\pmod 4$ is important ? I don't see why if $q$ is a prime unspecified then $\mathbb Z[i]/(q)$ is not necessarily a field. – user386627 Mar 22 '17 at 18:42
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I think computing the quotient (as sugggested in the other answer) is the easiest way to solve this. Nevertheless here is an alternative:

If you are fine about $\mathbb Z[i]/(q)$ being a field, you can find the number of elements by looking only at the abelian group structure.

As an $\mathbb Z[i]$ is generated by $1,i$, hence so is $G := \mathbb Z[i]/(q)$. $1$ and $i$ have order $q$ in $\mathbb Z[i]/(q)$. So we have two subgroups $A = \langle 1 \rangle, B =\langle i \rangle$ of order $q$ with $G=AB$ and $A \cap B=0$ (since they have both prime order, they are either equal or have trivial intersection), which implies $G = A \times B \cong \mathbb Z/q\mathbb Z \times \mathbb Z/q\mathbb Z$.

MooS
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  • How does this show that $G$ is a field? Don't we have a theorem which states that if $R$ is a ring and $S$ is a ring, then $R \times S$ is never a field. Clearly, $\mathbb{Z}/q\mathbb{Z}$ is a ring so why is the direct product a field? – Junaid Aftab Oct 14 '17 at 03:53
  • Or have you just counted the elements? If so, this is a nice argument. I have been trying to find a ring homomorphism but I can't seem to find one. – Junaid Aftab Oct 14 '17 at 04:01