# Number of elements and units in quotients of $\mathbb Z[i]$

I was given this problem but have no idea how to solve it.

How many elements are there in $\Bbb Z[i]/(3)$? In $\left(\Bbb Z[i]\right)/(3+2i)$? In $\Bbb Z[i]/(5)$? How many units are there in each of these rings?

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What ahve you tried? –  Thomas Andrews Jul 15 '14 at 1:38
I don't really understand what to do –  david Jul 15 '14 at 1:39

The point here is to compute the size of the quotient ring, which is the index. You want to use

$$\Bbb Z[i]/3\cong \Bbb Z[x]/(x^2+1,3)\cong (\Bbb Z/3\Bbb Z)[x]/(x^2+1)$$

since $x^2+1$ is irreducible modulo 3, you get a field extension of degree 2, which means you have a field with 9 elements.

Similarly $(3+2i)(3-2i)=13$, and they are coprime as $3+2i-(3-2i)=4i$ but $2\not |(3+2i)$, so then you can use the Chinese remainder theorem to see that:

$$\Bbb Z[i]/(3+2i)\oplus \Bbb Z[i]/(3-2i)\cong \Bbb Z[i]/13\Bbb Z[i]\cong \Bbb Z[x]/(x^2+1,13)$$

the last of these is a field of size $13^2$, whence each summand must be a field of size $13$.

Finally we factor $5=(2+i)(2-i)$, since $\Bbb Z[i]$ is a Euclidean domain, these are irreducible since they have norm $5$, but then since $\Bbb Z[i]$ is a PID, every prime ideal is maximal, hence

$$\Bbb Z[i]/(5)\cong \Bbb Z[i]/(2+i)\oplus\Bbb Z[i]/(2-i)$$

each of which is a field of characteristic $5$ since both $(2+i)$ and $(2-i)$ divide $5$, and I claim each is a field of size $5$, but for that it is sufficient to note that--since the ring is Euclidean--the norm of the generating element is exactly the index of the ideal (this is true more generally, but it's easiest to see here because the ring is Euclidean), hence there are 25 elements in this ring.

For units you just need to see that most of the rings you have are fields, hence the number of units is the size of the quotient ring minus 1. The other possibility is the last one, where you need to account for the fact that $(2+i)(2-i)=0$.

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Note that the number of residue classes is equal to the norm. –  Rene Schipperus Jul 15 '14 at 3:45
So you have $N 3=9, N(3+2i)=13, N 5=25$. –  Rene Schipperus Jul 15 '14 at 3:46
Yes, but the student appears to be at a more basic level where simpler theorems are more helpful. Norm = index is a wonderful characterization, but not one I chose to lean on since it also requires proof. –  Adam Hughes Jul 15 '14 at 4:07

Hint for one of them to get you started:

Note that $\mathbb{Z}_3[i] \cong \mathbb{Z}_3[x]/\langle x^2 + 1 \rangle$.

Now since $x^2 + 1$ is irreducible in $\mathbb{Z}_3$, then $\langle x^2 + 1 \rangle$ is a maximal ideal (see here). Therefore, $\mathbb{Z}_3[x]/\langle x^2 + 1 \rangle$ is a field. (So the units are...?)

Finally, it's easy to list out the elements by hand. They will be precisely all the polynomials over $\mathbb{Z}_3$ with degree $\leq 1$. (Why?)

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