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As the title states, I'm trying to either prove or disprove that $(2i-1,2i+3) \subset \mathbb{Z}[i]$ is a prime ideal.

I know that $\mathbb{Z}[i]$ is a Euclidean domain and thus a PID and hence a UFD, and that there's an isomorphism $\mathbb{Z}[i] \cong \mathbb{Z}[x]/ \langle x^2+1 \rangle$, so I believe it suffices to prove that $\mathbb{Z}[x]/ \langle x^2+1,2x-1,2x+3 \rangle$ or $\mathbb{Z}[i] / \langle 2i-1,2i+3 \rangle$ is atleast an integral domain, but I'm not sure how, since in both cases I found the algebra extremely tedious to try and verify these claims (unless I'm missing something).

Thank you.

Isochron
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    Find a single generator of the ideal and check if it is irreducible/prime. Note that the ideal contains $4$, so the generator must be a divisor of $4$. There aren't that many... – Arturo Magidin Jan 12 '23 at 03:53
  • Same methods in the linked dupe work here, e.g. as in this answer, the ideal contains the generator norms $5,13$ so it contains $(5,13) = 1\ \ $ – Bill Dubuque Jan 12 '23 at 12:56

1 Answers1

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$$ (-2i)(2i-1) + (-1)(2i+3) = 1 $$

The ideal $(2i-1,2i+3)$ is the entire ring $\mathbb{Z}[i]$, so it is by definition not a prime ideal.

aschepler
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