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Here's my attempt as I think I vaguely remember something similar:

We have $\phi : \mathbb{Z} \to \mathbb{Z}[i]/(7+5i)$ by $\phi(n) = n + (7+5i)$. I would like now to prove that $\ker\phi = \langle 74 \rangle$ and so by the first isomorphism theorem $$\mathbb{Z}[i]/(7+5i) \cong \mathbb{Z} / \langle 74 \rangle= \mathbb{Z}_{74}.$$

Questions:

  1. How do we prove that $\ker \phi = \langle 74 \rangle $
  2. Is the last equals sign correct?
  • 1
    you probably mean $\langle 74 \rangle$2012-06-11
  • 0
    Can you prove that $74$ is at least contained in the kernel? Do you know that this map is surjective?2012-06-11
  • 0
    Ah yes I do, I've edited that in2012-06-11
  • 1
    $\phi$ is surjective because $31$ is a square root of $-1$ modulo $74$, hence $\phi(31) \in i + \langle 74 \rangle \subseteq i + \langle 7 + 5i \rangle$.2012-06-11
  • 3
    Dear user, A very similar question was asked and answered here: http://math.stackexchange.com/q/23358/221 See [this answer](http://math.stackexchange.com/a/23379/221) in particular. Regards,2012-06-11
  • 0
    Possible duplicate of [Quotient ring of Gaussian integers](https://math.stackexchange.com/questions/23358/quotient-ring-of-gaussian-integers)2018-11-24

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