# Thread: Product of non principal ideals is principal.

1. ## 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 $\displaystyle$\left( {3,1 - \sqrt { - 5} } \right),\left( {3,1 + \sqrt { - 5} } \right) \in \mathbb{Z}\left[ {\sqrt { - 5} } \right]$$The book saids the result is (3). But I dont know how to do the product. Regards... 2. Originally Posted by orbit Hello, Im trying to prove that the product of two non principal ideals is a principal ideal. I consider the non principals ideals \displaystyle \left( {3,1 - \sqrt { - 5} } \right),\left( {3,1 + \sqrt { - 5} } \right) \in \mathbb{Z}\left[ {\sqrt { - 5} } \right]$$

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

Tonio

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

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

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

Regards

4. Originally Posted by orbit
Thanks for aswering. But I donīt 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

5. 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.

6. ## Thanks

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!