# Product and intersection of ideals of polynomial ring

Let $k[x,y,z,t]$ be the polynomial ring in four variables and let $I=<x,y>, J=<z, x-t>$ be ideals of the ring.

I wanna 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?

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 $g^{\prime\prime}$.