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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?

user26857
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  • Do you see that $xy,yz,zx$ all belong to the required ideal $I$? If so, can you prove that $I$ is generated by these three elements? Finally, can you show that you can not get 2 generators out of these three? – Mohan Oct 12 '15 at 04:02
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    What is that bullet supposed to mean? – user26857 Oct 12 '15 at 08:18
  • @user26857 Sorry, just product of ideal. –  Oct 12 '15 at 17:39
  • And what ideal is $V(y,z)$? – user26857 Oct 12 '15 at 18:37
  • @user26857 Oh, really sorry. $V(y,z)$ is just the algebraic set generated by $y=0$ and $z=0$. So it should be x-axis. –  Oct 12 '15 at 20:22
  • @Mohan Do you mean $xy,yz,zx$ all belong to the radical of I? How should I show this? I am not able to handle radical of ideal in polynomial ring...... So could you please write an answer? –  Oct 13 '15 at 00:58

2 Answers2

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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.

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user26857
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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.

Mohan
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  • I don't understand that last step. For example, $xy, yz, xz, xyz$ are also linearly independent over the base field and generate $I$, but are not a minimal generating set. More is needed to argue that $I$ "cannot be generated by two elements", since maybe there's some other generating set that is smaller than $xy,yz,xz$. I imagine local cohomology is needed here. – Jake Levinson Oct 13 '15 at 14:44
  • No, just simple algebra. If possible, let $f_1, f_2$ be a set of generators. Write $f_i=a_ixy+b_iyz+c_izx$. Since $xy,yz,zx$ can be written as a linear combination (again, polynomial) of $f_1,f_2$, read all these modulo all polynomials of order greater than 2. Then you get a vector space (of dimension 6) of homogeneous polynomials of degree 2, where the images of $f_i$ and $xy, yz,zx$ generate the same vector subspace, but one is 3 dimensional, while the other is at most 2 dimensional. – Mohan Oct 13 '15 at 16:22
  • Ah, I see -- I overlooked the importance of the fact that the defining equations are homogeneous. Thanks. – Jake Levinson Oct 13 '15 at 18:55