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I have to show that $\mathbb{Z}[i]/\langle 1+4i\rangle\simeq \mathbb{Z}_{17}$.

Attempt:I know that I have to use the isomorphism $\phi (a+bi)=[a]_{17}+4[b]_{17}$ but I don't know how to show that $\ker(\phi)\subseteq\langle 1+4i\rangle$.

Any ideas?

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Assume $\phi(a+bi)\equiv_{17}0$ then $$a+4b\equiv_{17}0$$ solving for $b$ we get $b=4a+17\cdot k$ for $k\in\mathbb{Z}$. So $a+bi=a+(4a+17\cdot k)i=a(1+4i)+17\cdot ki$. Using that $17=(1+4i)(1-4i)$ we get $$a+bi=(1+4i)(a+ki-4ki)$$ So $\ker(\phi)\subseteq (1+4i)$

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  • $\begingroup$ @giannispapav But of course! That is why this site exists! :) $\endgroup$ – cansomeonehelpmeout Jan 31 '18 at 15:57
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If $a+bi\in\ker\phi$, then $a+4b=17c$, for some integer $c$. Then $a+4b=(1+4i)(1-4i)c$ and $$ a+bi=a+4b+i(1+4i)b $$

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One common trick is to take $\Bbb Z[i]/\langle 1+4i\rangle$, and show that the map from this ring to $\Bbb Z_{17}$ induced by $\phi$ (which is well-defined because $\langle 1+4i\rangle\subseteq \ker\phi$) is injective.

It will probably help to note that in this ring, any element has a unique representative from $\Bbb Z[i]$ with real part $0$, and imaginary part between $0$ and $16$, inclusive.

This approach will give exactly the same calculations as trying to prove that $\ker\phi\subseteq \langle 1+4i\rangle$ directly, but framing it this way makes it a lot easier to keep track of what's what and how to proceed, in my opinion.

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we have $\Bbb Z[i]/(1+4i)\cong \Bbb Z[x]/(x^2+1,4x+1) $.

And $17=16(x^2+1)-(x-1)(4x+1)$, then $(x^2+1,4x+1)=(17,x^2+1,4x+1)$, so

$$\Bbb Z[x]/(17,x^2+1,4x+1) \cong \Bbb Z_{17}[x]/(x^2+1,4x+1) \cong \Bbb Z_{17}[x]/(x^2-16,4x-16)\cong \Bbb Z_{17}[x]/(x-4)(x+4,4) \cong \Bbb Z_{17}[x]/(x-4)(1) \cong \Bbb Z_{17} $$

In general $\Bbb Z[i]/(1+4i)\cong \Bbb Z/N(1+4i)\Bbb Z=\Bbb Z/17\Bbb Z ;N$ is normal of the element $1+4i$.

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