Why is the order of $\mathbb{Z}[i]/\langle 3+i\rangle$ equal to 10?

Question

The reasoning my book uses is as follows:

Since $(3+i)(3-i) = 10$, $10 + \langle 3+i\rangle = 0 + \langle 3+i\rangle $. Also, $i + \langle 3+i\rangle = -3 + \langle 3+i\rangle = 7 + \langle 3+i\rangle$. So, $\mathbb{Z}[i]/\langle 3+i\rangle = \{k + \langle 3+i\rangle \mid k = 0,1,\ldots ,9\}$, since $1 + \langle3+i\rangle$ has additive order $10$.

I do not understand why the additive order of $1 + \langle 3+i\rangle$ is $10$, or why the fact that $(3+i)(3-i) = 10$, shows the additive order or $1 + \langle 3+i\rangle$ is $10$. If someone could please explain this I would greatly appreciate it.

Note: I apologize for not using MathJax to write this post. I intend to learn it over the coming weekend.

Answer 1

You might find clearer the following way of viewing the proof. Write $\,A = \Bbb Z[i]/(3+i)$.

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

$\color{#c00}{I := \ker h = 10\,\Bbb Z}\ $ follows immediately by $\,\rm\color{#90F}{rationalizing}\,$ a denominator (cf. simpler multiples)

$$ n\in I\iff 3\!+\!i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\iff \dfrac{n}{3\!+\!i}\in \Bbb Z[i]\!\!\color{#90f}{\overset{\large \rm\ rat}\iff}\! \dfrac{n(3\!-\!i)}{10}\in\Bbb Z[i]\iff \color{#c00}{10\mid n}\ $$

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.

Remark $ $ We tested divisibility by $\color{#90f}{\textit{rationalizing}}$ the denominator in order to reduce division by an algebraic irrational $\,3+i\,$ to a simpler division by an integer (its norm $=10)$. This is a special case of the method of simpler multiples. Same proof works for $\,a+bi$ when $\gcd(a,b)=1$.

Another view you might find illuminating arises from rewriting the ideal as a module in Hermite normal form: $\ I = (3\!+\!i) = (10,3\!+\!i) = 10\Bbb Z + (3\!+\!i)\Bbb Z.\,$ But it is trivial to test module membership given such a triangularized basis, namely $$\begin{align} a\!+\!bi = a\!-\!3b +b(3\!+\!i)&\in I = 10\Bbb Z + (3\!+\!i)\Bbb Z\\ \iff\ a\!-\!3b&\in I\\ \iff\ a\!-\!3b &\in 10\Bbb Z \iff 10\mid a\!-\!3b \end{align}$$

Further this shows that $\, a\!+\!bi\bmod I\, =\, a\!-\!3b\bmod 10.\ $

The criterion generalizes to an ideal test for modules $\rm\,[a,b\!+\!c\:\!\omega]\,$ in the ring of integers of a quadratic number field, e.g. see section 2.3 Franz Lemmermeyer's notes linked here.

This is a special case of module normal forms that generalize to higher degree number fields, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's $ $ A Course in Computational Number Theory.

Answer 2

You can argue using the Smith Normal Form. As a $\mathbb Z$-module, $\mathbb Z[i]$ is free of rank two with basis $1,i$. The ideal $\langle 3+i\rangle$ is a free submodule with basis $\{(3,1),(-1,3)\}$ (why?) and the matrix of this is simply $$\begin{pmatrix} 3 & -1 \\ 1 & 3 \end{pmatrix}$$ which is of determinant $10$. In fact, we can follow the algorithm and find its normal form:

$$\begin{pmatrix} 1 & -7 \\ 1 & 3 \end{pmatrix} \text{do $R_1\to R_1-2R_2$}$$ $$\begin{pmatrix} 1 & 0 \\ 1 & 10 \end{pmatrix}\text{do $C_2\to 7C_1+C_2$}$$ $$\begin{pmatrix} 1 & 0 \\ 0 & 10 \end{pmatrix}\text{do $R_2\to R_2-R_1$}$$

Answer 3

$\textbf{Hint: }$ You already have $10(1+\langle 3+i\rangle) = 10 + \langle 3+i\rangle = \langle 3+i\rangle$. This implies that the order is at most 10. Next, the order of $1+\langle 3+i\rangle$ is 10 iff $3+i$ does not divide any of $1,\ldots,9$. So the question is, does $3+i$ divide any of $1,\dots,9$?

Answer 4

Prop 1) $10 \equiv 0 + <3+i>$.

Pf: $a \equiv b + <3+i>$ means there is a $k \in \mathbb Z[i]$ so that $a = b + k(3+i)$.

$10 = 0 + (3-i)(3+i)$ so $10 \equiv 0 + <3_i>$.

Prop 2) $i \equiv -3 + <3+i>$

Pf: $i = -3 + (3+i)$ so $i \equiv -3 +<3+i>$.

Prop 3) If $a = j + mi; j,m \in \mathbb Z$ Then $a \equiv j - 3m + <3+i>$.

proof: $j+mi = j - 3m + m(3 + i)$.

Prop 4) If $a = j+ mi$ then $a \equiv r + <3+i>$ for some $0 \le r < 10$.

Pf: We can find some $v$ so that $j-3m = 10*v + r$ where $0 \le r < 10$. That is basic division.

So $a \equiv j - 3m \equiv r + 10v + <3 + i>\equiv r + <3+i>$.

ANd that's more or less it. Any element of $\mathbb Z[i]$ is equivalent to an integer between 0 and 9. So the set of equivalence classe $\mathbb Z[i]/<3 + i> = \{k + <3+i>| k = 0...9\}$

The only thing to show is if $j,k \in \{0...9\}$ and $j\ne k$ then $j \not \equiv k + < 3+i>$.