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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 ?

Loic
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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.

Atticus Stonestrom
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  • Thank you for this counter-example, and for the care you take in explaining the details, I find it very instructive. And this answers the question, indeed. But do you think that there is a counter-example for the case that the cone does not consist of the origin only, that is, that the corresponding projective variety is not empty? And what happens if $k$ is algebraically closed but $I$ is not prime? – Loic Dec 18 '20 at 23:28
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    @Loic See here. Either scheme-theoretically or over an algebraically closed field, "being a cone" means that the radical of your ideal is homogeneous, which is implied by but does not imply your ideal being homogeneous, for general ideals. If $I$ is prime, it is its own radical and is therefore homogeneous. – KReiser Dec 18 '20 at 23:41
  • @Loic happy it helped! :) – Atticus Stonestrom Dec 19 '20 at 05:01
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    I defer to @KReiser's comment for your questions, although I will add a small comment on your first question; it's certainly possible to have examples where the vanishing set is not just the origin. for instance, consider the example above but in the ring $\mathbb{Q}[x,y,z,w]$. if that seems too trivial, replace $z-x^2$ in $I$ by $(z-w)-x^2$. the result is the image of $I$ under the automorphism $\mathbb{Q}[x,y,z,w]\rightarrow\mathbb{Q}[x,y,z,w]$ taking $z\mapsto z-w$, which remains prime, but perhaps feels like a more "organic" example. (the homogeneous ideal in question becomes $(x,y,z-w)$) – Atticus Stonestrom Dec 19 '20 at 05:01
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    Atticus Stonestrom @KReiser: Thanks for your explanations, which, in my opinion, completely clarify the definition of a cone. Other not-so-experienced readers (like me) may find the following summary helpful. For any ideal $I$, $I \subset \sqrt{I} \subset I(V(I))$, where $I(V(I))$ is radical (and if $I$ is prime, then $I = \sqrt{I}$); if $k$ is infinite and $V(I)$ is a cone, then $I(V(I))$ homogeneous, and if, additionally, $k$ is algebraically closed, then $\sqrt{I} = I(V(I))$ (Nullstellensatz). – Loic Dec 20 '20 at 22:04