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

Solution 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 $ $ 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.

Solution 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:

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

Solution 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 $1,\dots,9$?