intersection of ideals...

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December 18th 2008, 12:21 PM
aurora

intersection of ideals...

Hi... Thank you.
I'm confused regarding the problem of finding an intersection of two specific ideals. How do I approach such a problem?
I'm given R = Q[x] - the polynoms over Q.
and these two ideals:
I = I(x)R = (2x^3 + x^2 -6x -3)R; J= J(x)R = (2x^3-x^2-x)R.
(I marked the specific polynoms as I(x) and J(x) for convinience).
I unerstand what this means, and I understand that if p(x) is in I and J, then there exists r1(x) and r2(x) in R so that:
p(x) = I(x)r1(x) = J(x)r2(x)
However I have no idea how to simplify that and I feel I'm off the track...

:-\
Help?
Thank you.
Tomer.
December 18th 2008, 12:57 PM
NonCommAlg

Quote:

Originally Posted by aurora

Hi... Thank you.
I'm confused regarding the problem of finding an intersection of two specific ideals. How do I approach such a problem?
I'm given R = Q[x] - the polynoms over Q.
and these two ideals:
I = I(x)R = (2x^3 + x^2 -6x -3)R; J= J(x)R = (2x^3-x^2-x)R.
(I marked the specific polynoms as I(x) and J(x) for convinience).
I unerstand what this means, and I understand that if p(x) is in I and J, then there exists r1(x) and r2(x) in R so that:
p(x) = I(x)r1(x) = J(x)r2(x)
However I have no idea how to simplify that and I feel I'm off the track...

:-\
Help?
Thank you.
Tomer.

first find the prime factorization of the generators of your ideal: $I(x)=(x^2-3)(2x+1), \ \ J(x)=x(x-1)(2x+1).$ thus: $I \cap J = \text{lcm}(I(x),J(x))\mathbb{Q}[x]=x(x-1)(2x+1)(x^2 - 3)\mathbb{Q}[x]. \ \Box$

Note: recall that $\mathbb{Q}[x]$ is a UFD (a PID atually) and things in a UFD are very easy!
December 20th 2008, 05:25 AM
aurora

Thank you.
A friend of mine had offered an explenation to why the intersection is what you offered.... but I would be glad to hear your explenation, becuase I'm not sure I understood it.
I mean, you said the intersection is the lcm of both polynoms. I haven't studied anything like that. Are you basing this claim on a more general one? Could you quote it perhaps?
Oh, and what's PID? I just started studying "fraction rings" (I think that's the translation... I''m not studying in English), and I think that's what you mean by UFD (but I could be wrong).
If you could give a short explenation I would, as usual, be very glad (Smirk)

Thank you!
Tomer.
December 20th 2008, 09:47 AM
NonCommAlg

Quote:

Originally Posted by aurora

Thank you.
A friend of mine had offered an explenation to why the intersection is what you offered.... but I would be glad to hear your explenation, becuase I'm not sure I understood it.
I mean, you said the intersection is the lcm of both polynoms. I haven't studied anything like that. Are you basing this claim on a more general one? Could you quote it perhaps?
Oh, and what's PID? I just started studying "fraction rings" (I think that's the translation... I''m not studying in English), and I think that's what you mean by UFD (but I could be wrong).
If you could give a short explenation I would, as usual, be very glad (Smirk)

Thank you!
Tomer.

see here for the definition of UFD. a PID (principal ideal domain) is an integral domain in which every ideal can be generated by one element only. ok, here's a simple explanation of what i did:

following your notation, let $K(x)=x(x-1)(2x+1)(x^2 - 3),$ and $K=K(x)R.$ obviously $K \subseteq I \cap J.$ for the other side, suppose $f(x) \in I \cap J.$ then $f(x)=I(x)g(x)=J(x)h(x),$ for some

$g(x), h(x)$ in $R.$ thus: $(x^2 - 3)g(x)=x(x-1)h(x).$ hence $x^2 - 3 \mid x(x-1)h(x).$ but clearly $\gcd(x^2 - 3, x(x-1))=1,$ thus $x^2 - 3 \mid h(x).$ so let $h(x)=(x^2 - 3)u(x).$ then we will have:

$f(x)=J(x)h(x)=x(x-1)(2x+1)(x^2 - 3)u(x) \in K.$ therefore $I \cap J \subseteq K,$ and we're done!
December 20th 2008, 10:39 AM
aurora

Very nice.
I'm tired of thanking you. (Wink)