Why the ideal of the union of coordinate axes in $\mathbb A^3$ can not be generated by two elements, i.e., is not a complete intersection? My plan is to find the ideal first and then prove it can not be generated by two elements. 
x-axis = V(y,z), y-axis = V(x,z), z-axis = V(x,y). Then the ideal should be the radical of I, where I=(V(y,z)$\bullet$V(x,z)$\bullet$V(x,y)). Then things becomes too complicated for me and I do not know how to continue. I even can not judge whether this ideal is radical. So, in general, how should I judge whether the ideal generated by some polynomials is radical or not?
 A: If $X,Y,Z$ are affine algebraic sets in $\mathbb A^3$, then $I(X\cup Y\cup Z)=I(X)\cap I(Y)\cap I(Z)$. In your case, $I(X)=(y,z)$, $I(Y)=(x,z)$, and $I(Z)=(x,y)$, so the ideal you are looking for is $$(y,z)\cap(x,z)\cap(x,y).$$ But these three ideals are monomial ideals, so their intersection can be computed easily. (You can look here in order to understand what follows.) We have $$(y,z)\cap(x,z)=(xy,yz,xz,z)=(xy,yz,z).$$ Then $$(xy,yz,z)\cap(x,y)=(xy,yz,zx).$$
In order to prove that this ideal can not have lesser generators you can use this answer.
A: If you have an algebraic set $X\subset \mathbb{A}^n$, the set of polynomials vanishing on $X$ (thought of as a set) is always radical, since if $f^n$ vanishes on $X$, so does $f$. So, the polynomials $xy, yz, zx$ vanish on your set $X$ and thus is in the radical of $I$. Now, show that  any polynomial vanishing on your $X$ must be a linear (with polynomial coefficients) combination of these three polynomials by showing that $V(xy,yz,zx)=X$. Finally show that this ideal can not be generated by two elements since these three degree 2 homogeneous polynomials are linearly independent over the base field.
