Prove that $\mathbb{Z}[i]/(3)$ and $\mathbb{Z}[i]/(7)$ are finite fields and find their cardinality

Question

In my rings subject's test I had to prove that $\mathbb{Z}[i]/(21)$ was decomposed as a product of two finite fields, and that was easy to prove for me because $21 = 3\cdot 7$, and $\mathbb{Z}[i]$ is a PID and $3$ and $7$ are irreducible, so by the Chinese Remainder Theorem, and because $(3)$ and $(7)$ are maximal, we have the decomposition into two fields.

What I am unable to do is to find why they are finite fields and in such case, I had to find their cardinality. Can you help me/give some hints?

Answer 1

You can view $\mathbb{Z}[i]$ as a 2D grid of points, quotient by the ideal $(3)$ makes it wrap around at 3 so you the $x$ and $y$ coordinates are limited to the set $0,1,2$. In total you have 9 points.

To make this rigorous show that you can take any $x + iy \in \mathbb Z[i]$ and then express it as $x + iy + (3) = x' + i y' + (3)$ with $x', y' \in \{0,1,2\}$. And that each of these $9$ values are distinct.

Note that in $\mathbb{Z}[i]$ the ideal $(3)$ contains values like $3i$.

Answer 2

There is a short way to see it: $$\mathbb{Z}[i]/(3) \cong \mathbb{Z}[X]/(X^2+1,3) \cong (\mathbb{Z}[X]/(3)) /(X^2+1) \cong \mathbb{F}_3 / (X^2+1)$$ Where $\mathbb{F}_3$ is the field with three elements.

As one can see easily, this is just $\mathbb{F}_3 [i]$, a field with 9 elements. The other case with 7 works out similarly, I think you can fill in the details that are still needed. Greetings,

Markus Zetto

Answer 3

$\mathbb{Z}[i]/(3)$ is finite because its additive group is a finitely generated abelian group with finite exponent (at most $3$ in this case). More precisely, $\mathbb{Z}[i]/(3)=\mathbb{Z}\bar 1 + \mathbb{Z} \bar i$ and $3\cdot \bar 1 = 0 = 3\cdot \bar i$.

Answer 4

As $\mathbf Z[i]\simeq\mathbf Z[X]/(X^2+1)$, we have that, for instance, $$\mathbf Z[i]/(3)\simeq\mathbf Z[X]/(X^2+1)\big/3\mathbf Z[X]/(X^2+1)\simeq\mathbf Z/3\mathbf Z[X]\big/(X^2+1).$$ Now $X^2+1$ is irreducible over $\mathbf Z/3\mathbf Z=\mathbf F_3$ ($3$ is inert), so we have a quadratic extension of $\mathbf F_3$, i.e., up to an isomorphism, $\mathbf F_9$.

Note: $\mathbf Z[i]$ is not only a U.F.D., but a Euclidean domain, for the euclidean function $N(a+ib)=a^2+b^2$.