# Product of non principal ideals is principal.

• June 10th 2011, 08:31 PM
orbit
Product of non principal ideals is principal.
Hello, Im trying to prove that the product of two non principal ideals is a principal ideal.

I consider the non principals ideals $\left( {3,1 - \sqrt { - 5} } \right),\left( {3,1 + \sqrt { - 5} } \right) \in \mathbb{Z}\left[ {\sqrt { - 5} } \right]$

The book saids the result is (3). But I dont know how to do the product.

Regards...
• June 11th 2011, 02:33 AM
tonio
Quote:

Originally Posted by orbit
Hello, Im trying to prove that the product of two non principal ideals is a principal ideal.

I consider the non principals ideals $\left( {3,1 - \sqrt { - 5} } \right),\left( {3,1 + \sqrt { - 5} } \right) \in \mathbb{Z}\left[ {\sqrt { - 5} } \right]$

The book saids the result is (3). But I dont know how to do the product.

Regards...

Well, SOMETIMES the prod. of two non principal ideals is principal, not always...surely you meant this.

Now, $\left( {3,1 - \sqrt { - 5} } \right)\left( {3,1 + \sqrt { - 5} } \right)=(9,6,3(1\pm \sqrt{-5})) \in (3)$

Tonio
• June 11th 2011, 06:43 AM
orbit
Quote:

Originally Posted by tonio
Well, SOMETIMES the prod. of two non principal ideals is principal, not always...surely you meant this.

Now, $\left( {3,1 - \sqrt { - 5} } \right)\left( {3,1 + \sqrt { - 5} } \right)=(9,6,3(1\pm \sqrt{-5})) \in (3)$

Tonio

Thanks for aswering. But I don´t know how to do the product... wich are the steps to get the result. Thats my question.

Regards
• June 11th 2011, 07:38 PM
tonio
Quote:

Originally Posted by orbit
Thanks for aswering. But I don´t know how to do the product... wich are the steps to get the result. Thats my question.

Regards

Just multiply the ideals's generators with each other.

Tonio
• June 11th 2011, 07:51 PM
Deveno
the elements of IJ are sums of the form Σij, where i is in I, and j is in J. what are the possible products ij?

(3)(3) = 9

(3)(1 + √(-5))

(1 - √(-5))(3)

(1 - √(-5)(1 + √(-5)) = 1 - (-5) = 6.

3 divides all of these products, so certainly IJ is contained in (3).

on the other hand 3 = 9 - 6 which is an element of IJ, so (3) is contained in IJ.
• June 12th 2011, 01:17 PM
orbit
Thanks
Quote:

Originally Posted by Deveno
the elements of IJ are sums of the form Σij, where i is in I, and j is in J. what are the possible products ij?

(3)(3) = 9

(3)(1 + √(-5))

(1 - √(-5))(3)

(1 - √(-5)(1 + √(-5)) = 1 - (-5) = 6.

3 divides all of these products, so certainly IJ is contained in (3).

on the other hand 3 = 9 - 6 which is an element of IJ, so (3) is contained in IJ.

Thank you very much!