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 presentation below. 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{(3\!-\!\color{#c00}i)\color{#c00}n}{\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.

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

Note that $\mathbb Z[i]\simeq \mathbb Z[x]/(x^2+1)$, hence

$$\mathbb Z[i]/(i+3)\simeq \mathbb Z[x]/(x^2+1, x+3)\simeq \frac{\mathbb Z[x]/(x+3)}{(\overline{x^2+1})}\simeq \mathbb Z/(10)$$

This should be all very intuitive. If one has to, the first two isomorphisms can be justified by suitable isomorphism theorems for rings. The last step is because the isomorphism from $\mathbb Z[x]/(x+3)$ to $\mathbb Z$ is simply sending $x$ to $-3$, hence $x^2+1$ is sent to $10$ in $\mathbb Z$.

This kind of "polynomial ring quotient calculus" is very useful in algebraic geometry and number theory for recognizing quotients.

Answer 5

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 $j \not \equiv k + < 3+i>$.