Ideal generating the Grassmannian

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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.

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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 $(x_0)$.