Product and intersection of ideals of polynomial ring

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The discussion focuses on proving that the intersection of two ideals, I and J, in the polynomial ring k[x,y,z,t], satisfies the inclusion I ∩ J ⊆ IJ, given that I = <x,y> and J = <z, x-t>. The user is attempting to show this inclusion but is struggling with the implications of a polynomial g in the intersection being expressible as a linear combination of generators from both ideals. They note that since g must be in I, its terms must be multiples of x or y, while being in J implies it can be expressed in terms of z and (x-t). The challenge lies in demonstrating how these conditions lead to the conclusion that the components g' and g" must also belong to I, which remains unresolved in the discussion. The conversation highlights the complexities of working within polynomial rings and the specific properties of ideals in this context.
camilus
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Let k[x,y,z,t] be the polynomial ring in four variables and let I=&lt;x,y&gt;, J=&lt;z, x-t&gt; be ideals of the ring.

I want to show that IJ=I \cap J and one direction is trivial. But proving I \cap J \subset IJ has stumped me so far. Anyone have any ideas?
 
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K is an algebraically closed field, of course.
 
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 IJ=I\cap J.
 
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
 
OK, well, let's take a polynomial in g(x,y,z,t) in I\cap J. 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 g(x,y,z,t)=xa(x,y,z,t)+yb(x,y,z,t). Now, g must also lie in J, what does that imply?
 
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..
 
So, let us look at zg^\prime+(x-t)g^{\prime\prime}. We know that each individual term of the polynomial must be divisble by x or y. So we can write g^\prime=xh+yh^\prime, can we not? And the same for g^{\prime\prime}.
 
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