Is $A=\Bbb Z[i] /(3+i)$ finite?

Question

The quotient ring contains the classes of the remainders of divisions $\frac {a+ib}{3+i},$ $a,b\in \Bbb Z $. That ratio is equal to $\frac {3a+b}{10} +i\frac {3b-a}{10} $. Does this tell me what the classes look like?

Anyway, the given answer says that every class contains a representative with norm smaller than $10$, and since the elements of $\Bbb Z[i]$ that have such a norm are finitely many, this shows that $A $ is finite. How do we deduce the first statement?

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Also, I had tried a more convoluted approach: $ \Bbb Z[i] $ is a PID so $3+i$ not being irreducible implies $(3+i) $ not being maximal, and thus $A $ is not a field. In turn, this means that $A$ is not a finite integral domain (since it obviously has more than $1$ element). I guess $A $ is finite, so what is an example of a zero-divisor of it?

Answer 1

$\newcommand{\Zi}{\Bbb Z[i]}$ As Bill Dubuque points out, $10=(3+i)(3-i)$. The size of $A$ is the number of equivalence classes of $\Zi$ under the relation $\alpha\sim \beta$ whenever $\alpha-\beta$ is a multiple of $3+i$. But if $\alpha-\beta$ is a multiple of $10$ then $\alpha-\beta$ is a multiple of $3+i$. If we could show that $\Zi$ had finitely many equivalence classes under the relation $\alpha\sim'\beta$ whenever $\alpha-\beta$ is a multiple of $10$ then we would be in business.

For the question's given answer, this follows from the fact that if we round $\alpha/(3+i)$ to the nearest element $\gamma$ of $\Zi$ then $\alpha\sim\alpha-(3+i)\gamma$ and the norm of $\alpha-(3+i)\gamma$ is $<10$.

Answer 2

In this case the ideal $(3+i)$ is not so complicated hence it’s relatively easy to find a combination such as $10=(3+i)(3-i)$. But this might not always be the case; at times when its difficult to guess such a combination you can always work with the ideals in the following manner.

\begin{align} \Bbb Z[i]/(3+i) &\cong \frac{\Bbb Z[x]/(x^2+1)}{(3+x)} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{Since} \; \Bbb Z[i]\cong\Bbb Z[x]/(x^2+1)\\ &=\frac{\Bbb Z[x]}{(x^2+1,3+x)}\\ &\cong\frac{\Bbb Z[x]/(3+x)}{(x^2+1,3+x)/(3+x)} \;\;\;\;\;\;\; [\text{by Third Isomorphism Theorem for rings}]\\ & \end{align}

Now $\Bbb Z[x]/(3+x)\cong\Bbb Z$ by First Isomorphism Theorem for rings.

Also in the quotient $(x^2+1,3+x)/(3+x)$,

$3+x=0$

and $x^2+1=(3+x)(-3+x)+10\implies x^2+1=10$

Hence $(x^2+1,3+x)/(3+x)=10\Bbb Z$

Finally, $$\Bbb Z[i]/(3+i)= \Bbb Z/10\Bbb Z$$

So the quotient in fact has exactly $10$ elements.

Answer 3

$h\!: \Bbb Z \to \overbrace{\Bbb Z[i]/(3\!+\!i)}^{\textstyle A}\color{#0a0}{ \ {\rm is\ surjective\ (onto)}}\,$ by $\!\bmod\, 3\!+\!i\!:\ \, i\equiv -3\,\Rightarrow\, a\!+\!bi\equiv a\!-\!3b\in\Bbb Z$

$\color{#c00}{I := \ker h = 10\,\Bbb Z}\ $ follows immediately by means of $\,\rm\color{#48f}{rationalizing}\,$ a denominator

$$ n\in I\iff 3\!+\!i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\iff \dfrac{n}{3\!+\!i}\in \Bbb Z[i]\!\!\color{#48f}{\overset{\large \rm\ rat}\iff}\! \dfrac{3n\!-\!\color{#c00}ni}{\color{#c00}{10}}\in\Bbb Z[i]\iff \color{#c00}{10\mid n}\ $$

Thus $\, \color{#0a0}{A = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{10\,\Bbb Z}\ $ by the First Isomorphism Theorem.

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Remark $ $ By above we obtain an isomorphism $\ f(a+bi) = a-3b\,\bmod 10,\,$ which could also be directly verified. It is instructive to examine why it is multiplicative

$$\begin{align} f((a+bi)(c+di)) &= f(a+bi)f(c+di)\\[.3em] {\rm by}\ \ \ ac\color{#c00}{-1}bd-3(ad+bc) &\equiv (a\color{#c00}{-3}b)(c\color{#c00}{-3}d)\pmod{\!10}\\[.3em] {\rm by}\ \ \ \color{#c00}{{-}1} &\equiv \color{#c00}{(-3)^2}\!\!\pmod{\!10} \end{align}$$

The isomorphism can be viewed as arising from a generalization of the division algorithm, namely reduction modulo a Hermite normal form basis, e.g. see this answer.