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Suppose I have some ideal $I$ in $\mathbb{Z}[\sqrt{m}]$ (usually, $ m < 0 $). What, in general, would be a course of action to prove this ideal is maximal or prime? I know the former implies the latter, and an ideal $I$ being prime or maximal is equivalent to $R/I$ being a domain or field, respectively.

An example I have been given is $I = (3,1-\sqrt{-23})$ in $\mathbb{Z}[\sqrt{-23}]$. I proved this ideal is not principal as I thought that might be useful.

However, when trying to establish $R/I$ in these $\mathbb{Z}[\sqrt{m}]$ I fail to understand the course of action.

Any links to useful material would also be much appreciated.

tobias mulder
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1 Answers1

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In rings of algebraic integers, non-zero prime or maximal is the same thing.

Here, note $\mathbf Z[\sqrt{-23}]\simeq \mathbf Z[X]/(X^2+23)$, and $X^2+23=(X+1)(X-1)+8\cdot 3\in (1-X,3)$, so \begin{align*} \mathbf Z[\sqrt{-23}]/(3,1-\sqrt{-23})&\simeq \bigl(\mathbf Z[X]/(X^2+23)\bigr)/\bigl((3, 1-X)/(X^2+23)\bigr)\\ &\simeq\mathbf Z[X]/(3,1-X)\simeq \mathbf Z/3\mathbf Z, \end{align*} by the third isomorphism theorem.

Bernard
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  • Could you please explain why $(3,1-X) \equiv 3Z$? – Sahiba Arora May 25 '17 at 18:58
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    @SahibaArora $\mathbb{Z}[X]/(3,1-X) \cong (\mathbb{Z}/3\mathbb{Z})[X]/(1-X) \cong \mathbb{Z}/3\mathbb{Z}$ – reuns May 25 '17 at 18:59
  • @Sahiba Arora: Consider, for any field $K$, the homomorphism $K[X]\longrightarrow K$ which maps $X$ to $1$, i.e. $P\in K[X]\longmapsto P(1)$. It is surjective, and its kernel is the ideal $(X-1)$. – Bernard May 25 '17 at 19:03