An exercise in Shafarevich's Basic Algebraic Geometry I is to show that the ideal of the projective embedding of the $(2,n)$-Grassmannian is generated by the Plücker equations (this is true for the $(k,n)$-Grassmannian generally, but I guess it's harder to show).

Let us write the Plücker equations as $F_1,...,F_m$, and the ideal of all polynomials vanishing on the Grassmannian let us denote as $\mathfrak{U}_{Gr}$. The suggestion in Shafarevich is to show that the "affine" versions of $\mathfrak{U}_{Gr}$ are generated by the "de-homogenized" Plücker equations (i.e. $\frac{F_1}{x_i^{\deg F_1}},\dots,\frac{F_m}{x_i^{\deg F_m}}$) on the usual affine patches of projective space given by non-vanishing of $x_i$, and then somehow use this to show that the actual ideal $\mathfrak{U}_{Gr}$ is generated by the homogeneous polynomials $F_1,...,F_m$.

I can show the statement for the affine ideals, but I am struggling to show that this implies what is desired for the actual ideal. I think a concise way to state the difficulty I am having is maybe how to show the following, if it is true:

(*) Let $\mathbb{P}^n$ be projective space, and $X$ be an irreducible projective variety defined set-theoretically by the vanishing of homogeneous polynomials $F_1=\cdots=F_m=0$. If for $i=0,...,n$ we have that the ideal of all functions vanishing on the affine piece of $X$ given by $X\cap\{x_i\ne 0\}$ is $(\frac{F_1}{x_i^{\deg F_1}},\dots,\frac{F_m}{x_i^{\deg F_m}})$, then the homogeneous ideal given by all homogeneous polynomials vanishing on $X$ is equal to $(F_1,...,F_m)$.

If anyone can give me a hint towards seeing why this is true, I would appreciate it.


1 Answer 1


After discussing this a bit with some people in real life, it seems that the reduction (*) of the question that I asked is not in fact true. Here is a counter-example: In $\mathbb{P}^2$, consider the line given by $x_0=0$. The ideal of this variety is given by $(x_0)$, but $x^2_0=x_0x_1=x_0x_2$ defines it set-theoretically. For any $i$, the dehomogenized version of the ideal will contain the generator, but $(x^2_0,x_0x_1,x_0x_2)$ is clearly properly contained in $(x_0)$.

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    $\begingroup$ One thing you can read about is the saturation of a homogeneous ideal $I$. I think these are elements $a$ such that $(x_0, \dots, x_n)^ka \in I$ for some $k$. $\endgroup$
    – Hoot
    Sep 7, 2016 at 4:13
  • $\begingroup$ I have actually been referred to that term by the same people I discussed this problem with, but thank you for your suggestion. I am still stuck on my original problem to show that the Plücker equations generate the ideal of the Grassmannian, though, so if you have any ideas in that direction I'd appreciate it! $\endgroup$
    – A. S.
    Sep 7, 2016 at 4:21
  • $\begingroup$ Well I think it relates to the original problem in that if you can find polynomials that dehomogenize to the right thing on all affine patches and the ideal they generate is saturated then you're done. I will try to think about how feasible this is. $\endgroup$
    – Hoot
    Sep 7, 2016 at 4:32
  • $\begingroup$ Yeah I was wondering about that, but it seems quite messy. $\endgroup$
    – A. S.
    Sep 7, 2016 at 18:53

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