-1

If $J$ is a homogeneous ideal of height $2$ in $\mathbb C[X,Y]$ such that $J\subseteq (X,Y)$, then does there necessarily exist an integer $n\ge 1$ such that $X^n,Y^n \in J$ ?

user521337
  • 3,583
  • 1
  • 7
  • 21
  • 1
    Yes, it does. We have that $\sqrt J=(X,Y)$ since $J$ has a homogeneous primary decomposition, and the only height two homogeneous prime ideal is $(X,Y)$. – user26857 Feb 07 '19 at 09:17

1 Answers1

0

The minimal primes over an ideal $J = (f_1,\ldots,f_n)$ correspond to intersection points $V(f_1) \cap \cdots \cap V(f_n)$. The radical of $J$ is the intersection of the minimal primes, and the condition $x,y \in \sqrt{J}$ thus means the only intersection point is the origin.

If $f_1,\ldots,f_n$ have a common zero at another point $(a,b) \neq 0$, then by homogeneity they have common zeros along the line $\lambda\cdot(a,b)$. But this implies they are divisible by the equation $\ell$ of the line, hence $J \subset (\ell)$ has height 1.

So if $J$ has height two, their only common zero is the origin, and $x,y \in \sqrt{J}$.


First (wrong) attempt: No. The minimal primes over $J = (f,g)$ correspond to common zeros where $f=g=0$. The radical of $J$ is the intersection of the minimal primes, and the condition $x,y \in \sqrt{J}$ thus means the only common zero is at the origin. This is true for monomial ideals but not for more general homogeneous ideals.

For instance take $f=x + y$ and $g=x^3+y^3$, which intersect at $(1,-1)$.

Ben
  • 6,200
  • 11
  • 25
  • If you take $J$ to be the homogeneous ideal generated by $x+y$ and $x+y^3$, then $x,y^3\in J$ ... am I missing something ? – user521337 Feb 07 '19 at 01:20
  • @user521337 Not sure - why do you think $x,y^3 \in J$? I see $(x+y) - (x+y^3) = y(1-y^2)$ is in $J$, but I don't see $x$ or $y^3$. – Ben Feb 07 '19 at 01:57
  • the definition of a homogeneous ideal is if $f \in J$ and $f=f_1+...+f_r$ where $f_i$s are homogeneous and $\deg f_1<\deg f_2<...<\deg f_r$, then each $f_i\in J$ ... – user521337 Feb 07 '19 at 02:11
  • @user521337 Oops. I'll think about it some more. – Ben Feb 07 '19 at 02:13
  • @user521337 If you change $g$ to $x^3 + y^3$ I think it's fixed. – Ben Feb 07 '19 at 02:18
  • $(x+y,x^3+y^3) \subseteq (x+y)$, so that ideal has height $1$ .... – user521337 Feb 07 '19 at 02:32
  • @user521337 Ah right, and clearly any homogeneous polynomial which is zero at $(1,-1)$ is divisible by $x+y$. – Ben Feb 07 '19 at 03:02
  • that $J$ has height $2$, does not necessarily mean $J$ can be generated by two polynomials ... but of course you can choose $f,g \in J$ such that ht$(f,g)=2$ ... – user521337 Feb 07 '19 at 03:13
  • @user521337 True but I think the generalization to more generators works without changing anything. – Ben Feb 07 '19 at 03:21