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I'm trying to solve the following exercise.

Let $a$ and $b$ be relatively prime integers, and $i^{2}=-1 .$ Prove that the natural map $\mathbb{Z} \rightarrow \mathbb{Z}[i] /\langle a+b i\rangle$ induces an isomorphism of $\mathbb{Z} /\left(a^{2}+b^{2}\right) \mathbb{Z}$ with $\mathbb{Z}[i] /\langle a+b i\rangle$.

And I'm totally lost. How should I proceed? I have a feeling that I should use the universal property, and take the quotient by the kernel of the natural map. This would induce an isomorphism with the image of the natural map. But I don't know what the kernel would be, and I honestly don't know how to work with rings like $\mathbb{Z}[i] /\langle a+b i\rangle$. Would anyone have a rigorous (and readable!) explanation?

Albert
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1 Answers1

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Some hints. If $\pi\colon\mathbb{Z}\to\mathbb{Z}[i]/\langle a+bi\rangle$ is the canonical map $\pi(z)=z+\langle a+bi\rangle$, then $$ z\in\ker\pi \iff \frac{az}{a^2+b^2}\in\mathbb{Z}\text{ and }\frac{bz}{a^2+b^2}\in\mathbb{Z} $$ Obviously $\langle a^2+b^2\rangle\subseteq\ker\pi$. Now you need to prove equality and that the map $\pi$ is surjective.

egreg
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