1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Product and intersection of ideals of polynomial ring

  1. Nov 29, 2012 #1
    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?
  2. jcsd
  3. Nov 29, 2012 #2
    K is an algebraically closed field, of course.
  4. Nov 29, 2012 #3
    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].
  5. Nov 29, 2012 #4
    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
  6. Nov 29, 2012 #5
    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?
  7. Nov 30, 2012 #6
    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..
  8. Nov 30, 2012 #7
    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].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook