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?

Each $a+bi$ is congruent to some $c+di$ with $c$ and $d$ small — Hagen von Eitzen, Apr 26 '20 at 11:40
@Hagen I've thought that up to the equivalence relation, $a+bi$ and $a'+b'i$ are different if 3 does not divide their difference in $\mathbb{Z}[i]$, so $9\not| (a-a')^2 + (b-b')^2$ and thus $3$ doesn't divide $a-a'$ and 3 doesn't divide $b-b'$, so there are 3*3 elements in the quotient set. Am i correct? — deiv, Apr 26 '20 at 11:48

score 5 · Answer 1 · answered Apr 26 '20 at 11:50

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

score 3 · Accepted Answer · edited Apr 26 '20 at 21:25

3

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 $3i$.

edited Apr 26 '20 at 21:25

user26857

53,190

answered Apr 26 '20 at 11:49

thanks for your response, it is clear for me now. My approach was to see the different elements in the quotient set, seeing why their difference could not be divided by 3, but your approach is better. – deiv Apr 26 '20 at 12:06

lhf · Answer 3 · 2020-04-26T13:01:54.380

1

$\mathbb{Z}[i]/(3)$ 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 $3\cdot \bar 1 = 0 = 3\cdot \bar i$.

edited Apr 26 '20 at 13:01

answered Apr 26 '20 at 11:54

lhf

221,500

score 1 · Answer 4 · answered Apr 26 '20 at 12:01

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 $N(a+ib)=a^2+b^2$.

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

4 Answers4