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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

    micromass

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    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

    micromass

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    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

    micromass

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    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].
     
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