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## Homework Statement

Show that the elements 3 and 2+[tex]\sqrt{-5}[/tex] in [tex]\textbf{Z}[/tex][[tex]\sqrt{-5}[/tex]] have a greatest common divisor of 1, but the ideal I = (3, 2+[tex]\sqrt{-5}[/tex]) is not the total. Conclude that I is not principle.

## Homework Equations

## The Attempt at a Solution

I have got as far as attempting to show the first part. If I show that both 3 and 2+[tex]\sqrt{-5}[/tex] are both irreducible in [tex]\textbf{Z}[/tex][[tex]\sqrt{-5}[/tex]] then they are relatively prime and hence have gcd of 1?

Also, the norms of both the numbers are 9, so does this do anything at all?

The way I have argued that both 3 and 2+[tex]\sqrt{-5}[/tex] are irreducible is:

Let 3=ab with a and b in [tex]\textbf{Z}[/tex][[tex]\sqrt{-5}[/tex]].

Now |a|[tex]^{2}[/tex]*|b|[tex]^{2}[/tex]=9 which has only 1,3,9 as positive factors.

If |a|[tex]^{2}[/tex]=3 then a=x+y[tex]\sqrt{-5}[/tex] with x,y satisfying x[tex]^{2}[/tex]+5y[tex]^{2}[/tex]=3.

There are no such integers, so |a|[tex]^{2}[/tex]=/=3.

So either |a|=1 or |b| = 1. so either a or b is a unit, so 3 is irreducible.

Exactly the same for 2+[tex]\sqrt{-5}[/tex] since 4+5=9, then argue the same way.

But I don't really know what to do from here,

And how do I argue that the Ideal is not the total?

As for the last statement, is it true that if an ideal of two coprime numbers is not the whole ring, then the ideal is not principal?

Many thanks in advance.