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Let $U$ be the union of the $x,y,z$ axes in complex affine 3-space. The set of functions that vanish on $U$ is an ideal. Can we neatly express their generators?

There are a lot of similar such problems and I'm trying to get the flavor for how this is done.

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    What is the simplest possible function you can think of that vanishes on this space? Will this function generate the ideal or do you need more? Here's a hint: the ideal of functions vanishing on $U$ is the intersection of the ideal of functions vanishing on each of the axes.2012-11-30

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Let $U,V$ be two algebraic sets in $\mathbb{A}_\mathbb{C}^3$, with ideals $I(U)$ and $I(V)$. Then the ideal $I(U \cup V)$ is the set of functions vanishing on either $U$ or $V$ (or both). Then we see (show this!) that $I(U \cup V) = I(U) \cap I(V)$.

So in your case, the respective ideals are $(y,z), (x,z)$ and $(x,y)$, the ideals of the x,y and z-axis, respectively. The ideal of the union is the interesection of the ideals, so $I(U)=(yx,xz,yz)$. Make sure you can do the computation yourself.

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    @WouterZeldenthuis: No. Take the function $x$, for example. It does not vanish on (all of the) the x-axis. The x-axis are precisely the points with y- and z-coordinate zero.2012-11-30