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The exercise is 1.3(3) from HP Kraft, "Appendix A: Basics from Algebraic Geometry."

If a closed subset $X\subseteq \mathbb C^n$ is a cone, show that $I(X)$ is generated by homogeneous functions.

Closed means $X=\cap g_i^{-1}(0)$ the zero set for some $g_i\in\mathbb C[x_1,\dots,x_n]$; cone means $tv\in X$ for all $v\in X,t\in \mathbb C$; the ideal $I(X)$ means polynomials that vanish on $X$; the exercise defines homogeneous to mean $f(tv)=t^df(v)$ for all $v\in \mathbb C^n, t\in \mathbb C$ for some $d\in\mathbb N.$

The exercise appears before Nullstellensatz and affine variety are introduced, but I can't find an elementary proof anywhere. Wolfram and Wikipedia (citing "Chow's theorem") give the statement without proof. The converse is trivial of course.


Solution (paraphrasing Georges below). Every polynomial can be written as the sum of homogeneous polynomials $p(x) = p_0 (x)+\dots +p_d(x).$ These homogeneous components generate "at least" as much, $$\sum_{p\in I(X)} (p_1)+\dots+(p_d) \supset I(X).$$

To show the reverse inclusion, we fix $p\in I(X)$ and show that each component $p_i\in I(X).$ Let $v\in X$ be given. Since $X$ is a cone, $p$ vanishes on the line $\mathbb C v,$ and $$p(tv) = \sum p_i(tv) = \sum p_i(v) \, t^i=0$$ for all $t\in \mathbb C.$ Thus $p(tv)\in \mathbb C[t]$ must be the zero polynomial with every "coefficient" $p_i(v)=0.$

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It is enough to show that if a polynomial $p(T_1,... ,T_n)=\sum p_d(T_1,... ,T_n)$ vanishes on the line $l=\mathbb C\cdot a $ where $a=(a_1,...,a_n)\neq 0\in X\subset \mathbb C^n$, each of the homogeneous components $p_d(T_1,... ,T_n)$ of degree $d$ of the polynomial $p(T_1,... ,T_n)$ vanishes on $l$.

But since $p(z\cdot a)=\sum p_d(z\cdot a_1,... ,z\cdot a_n)=\sum z^d\cdot p_d( a_1,... , a_n)=0$ is a polynomial in $z$ vanishing for all $z\in \mathbb C$, that polynomial is zero and we necessarily must have $p_d( a_1,... , a_n)=0$ for all $d $, i.e. all homogeneous components $p_d$ of $p$ vanish on $l$.

We have thus proved that $I(X)$ is indeed a homogeneous ideal.

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