Product and intersection of ideals of polynomial ring

  • Thread starter camilus
  • Start date
  • #1
146
0
Let [itex]k[x,y,z,t][/itex] be the polynomial ring in four variables and let [itex]I=<x,y>, J=<z, x-t>[/itex] be ideals of the ring.

I wanna show that [itex]IJ=I \cap J[/itex] and one direction is trivial. But proving [itex]I \cap J \subset IJ[/itex] has stumped me so far. Anyone have any ideas?
 

Answers and Replies

  • #2
146
0
K is an algebraically closed field, of course.
 
  • #3
22,089
3,293
Try to prove in general that if R is a unital ring and if I and J are ideals such that I+J=R (we say that I and J are comaximal), then [itex]IJ=I\cap J[/itex].
 
  • #4
146
0
I have already proved that (and thought of that), but the problem is that these are ideals of a polynomial ring, so that if I+J=k[x] then either I or J IS k[x], otherwise you could not generate the scalars in the field.. (since k-field, it has no nontrivial ideals)

So this approach won't work. I want to show just this case, not prove the general statement of when the intersection of two ideals in poly ring is equal to their product.

I just need and argument for I intersect J \subset IJ for this particular case (I already know it is true, I just need to show it).

Thanks anyways micromass
 
  • #5
22,089
3,293
OK, well, let's take a polynomial in [itex]g(x,y,z,t)[/itex] in [itex]I\cap J[/itex]. This polynomial must lie in I. This means that all the individual terms of the polynomial must be a multiple of x or of y. So you can write [itex]g(x,y,z,t)=xa(x,y,z,t)+yb(x,y,z,t)[/itex]. Now, g must also lie in J, what does that imply?
 
  • #6
146
0
That it is a linear combination of z and (x-t), g=zg'+(x-t)g" for g',g" in k[x,y,z,t].

The question is what we do from there.

We know that g(0,0,z,t)=0 (because g in I) hence g(0,0,z,t)=zg'(0,0,z,t)-tg"(0,0,z,t)=0.

But from here can we conclude that g',g" are in I? I don't see how to do it..
 
  • #7
22,089
3,293
So, let us look at [itex]zg^\prime+(x-t)g^{\prime\prime}[/itex]. We know that each individual term of the polynomial must be divisble by x or y. So we can write [itex]g^\prime=xh+yh^\prime[/itex], can we not? And the same for [itex]g^{\prime\prime}[/itex].
 

Related Threads on Product and intersection of ideals of polynomial ring

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
1
Views
2K
Replies
1
Views
3K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
9
Views
41K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
2K
Replies
8
Views
961
Top