$\mathbb{Z}[i] / (2+3i)$ has 13 elements

Question

I am self studying algebra, and have come across this exercise:

Show that $\mathbb{Z}[i]/(2+3i)$ has 13 elements and is a field.

My proof: Since Z[i] is a euclidian domain with euclidian norm function $f(c+id)= c^2+d^2$, for any element $a+bi \in Z[i]$, we can "divide" it by $2+3i$ to get $a+bi= q (2+3i) + r$ where either $r=0$ or $r=c+di$ for some integers $c,d$ not both $0$ with $c^2 + d^2=f(c+di) < f(2+3i)= 13$. But then $a+bi$ belongs to the coset $ r + (2+3i) $. So every element in the quotient ring belongs to some $r + (2+3i)$ for $r=0$ or $r=c+di$ with $c^2 + d^2<13$. Now we can count all the possible pairs of integers satisfying this condition, ie $(0,0),(0,1),(0,2),(0,3), ..., (3,1)$ to get 13 and so there are 13 elements in the quotient ring.

Is this a correct argument? Also is there an elegant way of showing it is a field rather than showing its an integral domain by showing each element has an inverse by brute force? Thanks in advance!

Answer 1

$\mathbb Z[i]/(2+3i) \cong \mathbb Z/N(2+3i)\mathbb Z =\mathbb Z/13 \mathbb Z$

or $\mathbb Z[i]/(2+3i) \cong \mathbb Z[x]/(x^2+1,2+3x)\cong \mathbb Z[x]/(13,2+3x)$ where $(x^2+1,2+3x)=(13,2+3x)$, so $\mathbb Z[x]/(13,2+3x)\cong \mathbb Z_{13}[x]/(2+3x)\cong \mathbb Z_{13} $

Answer 2

You can do this, but counting one-by-one the elements of the quotient ring would not be practical if you replaced $2+3i$ by say $2323419+3803616i$.

Note that $(2+3i)(2-3i)=13$ so that $(2+3i)\subseteq (13)$ as ideals. Therefore $R/(2+3i)$ is a quotient of the ring $R/(13)$ where $R=\Bbb Z[i]$. But $|R/(13)|=13^2$ since $a+bi+(13)=c+di+(13)$ iff $a\equiv c$ and $b\equiv d\pmod{13}$. So the size of $R/(2+3i)$ is a factor of $13^2$, so is $1$, $13$ or $13^2$.

$|R/(2+3i)|$ cannot be $1$ since that would entail $(2+3i)\mid 1$ in $R$, which is false. $|R/(2+3i)|$ cannot be $13^2$ since that would entail $2+3i\in( 13)$ which is false. So $|R/(2+3i)|=13$.

As $13$ is prime, $R/(2+3i)$ equals its own prime subring, which is a field.

Answer 3

I probably posted this before, but here it goes. You can determine the size of $\mathbb Z[i]/(a+bi)$ using the Smith normal form by considering it as a quotient of $\mathbb Z\oplus \mathbb Z$ by the subgroup generated by $(a,b)$ and $(-b,a)$. This has corresponding matrix $\begin{pmatrix} a&b\\ -b &a\end{pmatrix}$ whose determinant is $a^2+b^2$. This gives that the quotient has this size, which in your case is $2^2+3^2 = 4+9=13$. The fact the quotient is a field follows immediately since there is only one ring of prime size $13$, which is the field $\mathbb Z/13$.