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

By applying some isomorphisms and realizing $\mathbb{Z}[i]$ as the quotient ring of $\mathbb{Z}[x]/(x^{2}+1)$, I found that:

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

However, I run into trouble as I can't divide $x^{2}+1$ by $2+3x$ in $\mathbb{Z}[x]$. What would be another way to approach the problem?

Thanks for the help.

hoyast
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2 Answers2

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Firstly, consider the norm $N:\mathbb{Z}[i]\rightarrow\mathbb{Z}[i]$ given by $a+bi\mapsto a^2+b^2$. Because $N(2+3i)=13=1(\mathrm{mod} 4)$ which is a prime element in $\mathbb{Z}$, $2+3i$ is a prime element in $\mathbb{Z}[i]$. Hence $\mathbb{Z}[i]/(2+3i)$ is an integral domain.

Secondly, $13\in(2+3i)$ and then $\mathbb{Z}[i]/(2+3i)\cong\frac{\mathbb{Z}[i]/(13)}{(2+3i)/(13)}$. Because $\mathbb{Z}[i]/(13)\cong\mathbb{F}_{13}[i]$ is finite, $\mathbb{Z}[i]/(2+3i)$ is also finite.

Therefore, $\mathbb{Z}[i]/(2+3i)$ is a finite integral domain, which is a finite field.

PS: Since $\mathbb{Z}[i]$ is a Dedekind domain, one obtains directly that any quotient ring of $\mathbb{Z}[i]$ is finite.

Hebe
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$2+3i$ is prime in $\mathbb Z[i]$ since $N(2+3i)=13$ is a prime in $\mathbb Z$. But $\mathbb Z[i]$ is a PID, so $\mathbb Z[i]/(2+3i)$ is a field.

Why is $\mathbb Z[i]/(2+3i)$ finite? Note that the elements of $\mathbb Z[i]/(2+3i)$ are the residues modulo $2+3i$, that is, the elements of $\mathbb Z[i]$ whose norm is $<13$. (Don't forget that $\mathbb Z[i]$ is Euclidean.)

user26857
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  • The 13 elements of $\mathbb Z[i]/(2+3i)$ are $\lbrace 0, \pm 1, \pm 2, \pm i, \pm 2i, \pm 1 \pm i\rbrace$. The norms of 3 and $3 + i$ are < 13, but $3 \equiv 2i \mod (2 + 3i)$ and $3+i \equiv -2 \mod (2 + 3i)$. – PM 2Ring Dec 05 '15 at 12:29
  • Sorry, my comment wasn't intended as a criticism of your answer. I was mostly just adding some additional info for future readers, in case they were confused about the status of numbers with norm < 13 that do not appear in the set I gave. – PM 2Ring Dec 05 '15 at 13:09