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I need to prove that

$\Bbb{Z}/10\Bbb{Z}$ is isomorphic to $\Bbb{Z}[i]/\langle 1+3i\rangle$.

I know I can use the third isomorphism, but I would like to use the first one. I consider a homomorphism $\Bbb{Z}$ to $\Bbb{Z}[i]/(1+3i)$ : $$\phi : \Bbb{Z}\rightarrow \Bbb{Z}[i]/\langle 1+3i\rangle,\quad x\mapsto \overline{x}.$$ The kernel is a subgroup of $\Bbb{Z}$, so it looks like $p\Bbb{Z}$ for a unique $p>0$.

Plus $\overline{x}$ means (here) that it's equal to $(a+ib)+\langle 1+3i\rangle =(a+ib)+(1+3i)(a+ib)$, isn't it ? I am not sure, I used the fact that $\Bbb{Z}[i]$ is euclidean so principal and then all ideal are is that form. On the other hand $\Bbb{Z}\cap\langle 1+3i\rangle=10\Bbb{Z}$.

So the question is how can I write, correctly, the morphism $\phi$ ?

  • Start by noting that $(1 + 3i)(1 - 3i) = 10$, and consider the obvious maps $\mathbb{Z} \to \mathbb{Z}/10\mathbb{Z}$ and $\mathbb{Z} \to \mathbb{Z}[i] \to \mathbb{Z}[i]/(1 + 3i)$. – anomaly Jun 06 '15 at 20:06
  • @anomaly Do this helps to write my morphism ? –  Jun 06 '15 at 20:11
  • $\phi$ is the composite of the natural maps $,\Bbb Z\to \Bbb Z[i]\to Z[i]/(1+3i)\ \ $ See my answer for a simple way to view the proof. – Bill Dubuque Jun 06 '15 at 20:43
  • @Gato: Yes, that was why I wrote that. – anomaly Jun 06 '15 at 21:08

2 Answers2

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As you said, we define a homomorphism $\phi:\mathbb{Z}\to\mathbb{Z}[i]/\langle 1+3i\rangle$ by $$\phi(n)=n+\langle 1+3i\rangle$$ Then obviously $\phi(n)=0 \iff n\in\langle 1+3i\rangle$, but for any integer $n$, we have $$n\in\langle a+bi\rangle\iff n\in\langle a-bi\rangle$$ because $n=\overline{n}$. Moreover, if $n\in\langle \alpha\rangle$ and $n\in\langle \beta\rangle$ for any two $\alpha,\beta\in\mathbb{Z}[i]$, then $n\in\langle \mathrm{lcm}(\alpha,\beta)\rangle$, which is true (and makes sense) because $\mathbb{Z}[i]$ is a PID. Thus $$\begin{align*} \phi(n)=0 &\iff n\in\langle 1+3i\rangle\;\text{ and }\;n\in\langle 1-3i\rangle\\ &\iff n\in\langle (1+3i)(1-3i)\rangle=\langle 10\rangle \end{align*}$$ Therefore, the kernel of $\phi$ is the ideal $10\mathbb{Z}\subset\mathbb{Z}$.

Check that $\phi$ is surjective because $\phi(1)=1+\langle 1+3i\rangle$ and $$\begin{align*} \phi(3)&=3+\langle 1+3i\rangle\\ &=3+i(1+3i)+\langle 1+3i\rangle\\ &=i+\langle 1+3i\rangle \end{align*}$$ so that $\phi(a+3b)=a+bi+\langle 1+3i\rangle$ for any $a,b\in\mathbb{Z}$.

By the first isomorphism theorem, we get an induced isomorphism $\widetilde{\phi}:\mathbb{Z}/10\mathbb{Z}\to\mathbb{Z}[i]/\langle 1+3i\rangle$.

Zev Chonoles
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  • This is really nice to read, so clean. Usually I don't make this kind of comments, but I really like reading this answer (and surely because I have some difficulties with this stuff). Thanks +1. –  Jun 06 '15 at 20:22
  • @Gato: Thanks for the kind words, I'm glad to help! – Zev Chonoles Jun 06 '15 at 20:25
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    Is that obvious that lcm here is the product of $\langle (1+3i)(1-3i) \rangle$ ? And I don't understand why you compute $\phi(1)$ and $\phi(3)$ to prove that $\phi$ is surjective ? Can you explain please ? Thanks – Kenobi Jul 09 '15 at 19:33
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Let $\, w=1\!+\!3i.\, $ $\, \phi\!:n\mapsto n+(w)\in R\!=\!\Bbb Z[i]/w\,\color{#0a0}{ \ {\rm is\ surjective\ (onto)}},\,$ since every coset has an $\rm\color{#0af}{integer}$ rep: $ $ by below $\,a+b\,\color{#0af}i\equiv a+\color{#0af}3b\pmod{\!w},\,$ so $\ a\!+\!b\,i+(w) = a\!+\!3b+(w),\,$ via $\!\bmod\:\! w\!=\!1\!+\!3i\!:\,\ 0\!\equiv\! \,w\bar w\!=\!10,\,\ {\rm so}\,\ {-}3i\equiv 1\,\overset{\times\,3}\Rightarrow\,\color{#0af}{i\equiv 3}$.

$\color{#c00}{I = \ker\phi = 10\,\Bbb Z}\ $ follows simply by rationalizing a denominator

$$ n\in I\!\iff\! 1\!+\!3i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\!\iff\! \dfrac{n}{1\!+\!3i}\in \Bbb Z[i]\!\iff\! \dfrac{n(1\!-\!3i)}{10}\in\Bbb Z[i]\!\iff\! \color{#c00}{10\mid n}\qquad$$

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

Bill Dubuque
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