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How to show that all prime ideals of $\mathbb Z[\sqrt {-3}]$ are maximal?

My attempt:

$\mathbb Z[\sqrt {-3}]\cong \mathbb Z[x]/(x^2+3)$

Let p be prime ideal of $\mathbb Z[\sqrt {-3}]$

SO $\mathbb Z[\sqrt {-3}]/(p(x))$ is integral domain

I could not prove but I think it will be finite.

So it will be field so p become maximal ideal .

Please Help me show above claim

Any Help will be appreciated

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You need to exclude the zero ideal: it is prime but not maximal.

If $I$ is a nonzero ideal of $R=\Bbb Z[\sqrt{-3}]$, then $I$ has finite index in $R$, so $R/I$ is a finite ring.

An ideal $I$ in a commutative ring $R$ is maximal iff $R/I$ is a field and is prime iff $R/I$ is an integral domain.

A finite integral domain is a field, by a well-known theorem, so in our example, if $I$ is a non-zero prime ideal, then $R/I$ is an integral domain, so $R/I$ is a field, and $I$ must be maximal.

This argument also works when $R$ any order in an algebraic number field.

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  • $\begingroup$ Dear Sir, Can You please tell me Why I has finite index as I had no argument for that? $\endgroup$ – MathLover Mar 30 at 3:28
  • $\begingroup$ @MathLover Show that every nonzero principal ideal has finite index: in fact the ideal generated by $(a+b\sqrt{-3})$ has index $a^2+3b^2$. $\endgroup$ – Lord Shark the Unknown Mar 30 at 3:30
  • $\begingroup$ Dear Sir How can we say given ring's ideal is of form (a+b/sqrt3i). Please can you elaborate? $\endgroup$ – SRJ Apr 19 at 18:44
  • $\begingroup$ Every principal ideal has that form. @SRJ $\endgroup$ – Lord Shark the Unknown Apr 19 at 18:58
  • $\begingroup$ But sir all ideal need not be principal in that ring na? $\endgroup$ – SRJ Apr 20 at 2:38

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