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

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 .

Any Help will be appreciated

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.

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