Homogeneous ideal vs. cone Let $k$ be a field and $I$ an ideal of $k[X_1, ..., X_n]$. By definition,
$V(I)$ is a cone if $V(I) = V(I') \ne \emptyset$ for some homogeneous ideal
$I'$ (generated by homogeneous polynomials).
If $k$ is infinite and $V(I)$ is a cone, is then $I$ necessarily homogeneous ?
Is the additional condition that $I$ is prime necessary ? Or is there a
counter-example, even if $I$ is prime ?
 A: I assume you are working in a classical setting, not a scheme theoretic one, so that $V(I)$ is a subset of $k^n$. Following @KReiser's question, let me know if this is not right.

Consider the ideal $I=(x^2+y^2, z-x^2)$ in $\mathbb{Q}[x,y,z]$. Clearly $V(I)=\{(0,0,0)\}$; indeed, $x^2+y^2=0$ forces $x=y=0$, and hence forces $z=x^2=0$ as well. Thus $V(I)=V((x,y,z))$. However, $I$ is not a homogeneous ideal; since $I$ contains elements with terms of degree $1$, any homogeneous generators of $I$ would have to include a non-zero polynomial of degree $\leqslant 1$, and no such polynomials lie in $I$.
To see that $I$ is prime, first consider the map $f:\mathbb{Q}[x,y,z]\rightarrow\mathbb{Q}[x,y]$ given by $z\mapsto x^2$. We claim that $\ker f=(z-x^2)$; the $\supseteq$ direction is clear, and for the $\subseteq$ direction suppose $f(p)=0$. Then $p(x,y,x^2)=0$, and so $p$ is divisible by $z-x^2$ in $\mathbb{Q}(x,y)[z]$. By Gauss' lemma, this means that $p$ is divisible by $z-x^2$ in $\mathbb{Q}[x,y][z]$, as desired.
So ${\mathbb{Q}[x,y,z]}\big/{(z-x^2)}\cong\mathbb{Q}[x,y]$, with isomorphism induced by $f$, and $I\supseteq\ker f$. Thus $$\frac{\mathbb{Q}[x,y,z]}{I}\cong\frac{\mathbb{Q}[x,y]}{f(I)}=\frac{\mathbb{Q}[x,y]}{(x^2+y^2)}.$$ Since $x^2+y^2$ is irreducible in the UFD $\mathbb{Q}[x,y]$, the ideal $(x^2+y^2)<\mathbb{Q}[x,y]$ is prime, so the ring above is a domain and hence $I<\mathbb{Q}[x,y,z]$ is prime, as desired.
