# Question about proof that the Grassmannian is a parameter space

Edit: If it is easier to give a reference where this is written down in detail, I would gladly accept that as an answer.

Fix a base scheme $B$, and fix $n$ and $k$ with $k<n$.

In section 28.3 of Vakil's notes, there is a sketch of the proof that the Grassmannian -- defined as the scheme that represents the functor classifying surjections $\mathcal O_B^n \rightarrow \mathcal G$ onto rank $k$ locally free sheaves -- is the parameter space for closed subschemes of $\mathbb P^{n-1}_B$ (finitely presented and flat over $B$) whose fibers over points in the base are linearly embedded copies of $\mathbb P^k_{k(b)}$. In it, he shows how to get from rank $k$ surjections to closed subschemes, and from closed subschemes to rank $k$ surjections. I'm having trouble showing these are actually inverse operations.

Fix a closed subscheme $X\subset \mathbb P^{n-1}_B$ of the appropriate form over $B$. Twisting the subscheme exact sequence gives $$0\rightarrow I_X(1)\rightarrow \mathcal O_{\mathbb P^{n-1}}(1) \rightarrow \mathcal O_X(1) \rightarrow 0.$$

Now, $\mathcal O_X(1)$ restricted to each fiber of the projection $\pi$ to $B$ is $\mathcal O(1)$ on the linearly embedded projective subspace, and this sheaf has no higher cohomology. So, by the theorem on base change and cohomology (see Vakil for details), the higher pushforwards of $\mathcal O_X(1)$ vanish, and there is an exact sequence

$$0\rightarrow \pi_* I_X(1)\rightarrow \pi_* \mathcal O_{\mathbb P^{n-1}}(1) \rightarrow \pi_*\mathcal O_X(1) \rightarrow 0.$$

The middle term is $\mathcal O_B^n$, and we obtain a surjection $$\mathcal O_B^n\rightarrow \pi_*\mathcal O_X(1)$$ onto a locally free sheaf of rank $k$.

Now we go the other way, applying $\operatorname{Proj}_B (\operatorname{Sym}^\bullet)$ to both sides to yield a closed subscheme $$\operatorname{Proj}_B (\operatorname{Sym}^\bullet \pi_*\mathcal O_X(1)) \rightarrow \operatorname{Proj}_B (\operatorname{Sym}^\bullet \mathcal O_B^n)=\mathbb P^n_B.$$

Why does this recover the original subscheme?

• Here's a hint, since right now I don't have the time to do it with all the dirty details: consider the correspondence between closed sub-schemes of $\mathrm{Proj}_B(\mathop{S}\mathcal{O}_B^n)$ and homogenous ideal sheaves in $\mathop{S}\mathcal{O}_B^n$. If $X\subset \mathbb{P}^{n-1}_B$ is linear over $B$, then its homogenous ideal in $\mathop{S}\mathcal{O}_B^n$ should be generated in degree one, i.e., by $\pi_*I_X(1)$... – Ben Mar 30 '16 at 12:01
• @Ben Hi. It's not clear to me that $X$ is linear if we only know its fibers are. This is the part that seems hard. – Potato Mar 30 '16 at 13:34

From the quotient $\mathcal{O}_B^n\to\pi_*\mathcal{O}_X(1)$, whose kernel is $\pi_* \mathrm{I}_X(1)$, we get an exact sequence $$\operatorname{S}\pi_* \mathrm{I}_X(1)\to \operatorname{S}\mathcal{O}_B^n\to\operatorname{S}\pi_*\mathcal{O}_X(1)\to0$$

That is, $\operatorname{S}\pi_* \mathrm{I}_X(1)$ surjects onto the homogenous ideal in $\operatorname{S}\mathcal{O}_B^n$ corresponding to the closed sub-scheme $\mathrm{Proj}(\operatorname{S}\pi_*\mathcal{O}_X(1))\subset\mathrm{Proj}(\operatorname{S}\mathcal{O}_B^n)$. What we have to prove is that this homogenous ideal equals the ideal corresponding to $X$ (eventually in all degrees, but in this case they are actually equal in all degrees).

Since such functors are completely understood on affine schemes we may suppose that $B = \mathrm{Spec}(A)$ is affine. Then we can identify $\operatorname{S}\mathcal{O}_B^n$ with the quasi-coherent $\mathcal{O}_B$-module associated with the free symmetric $A$-algebra $\operatorname{S}A^n$ (alias the polynomial ring). If $I\subset \operatorname{S}A^n$ is the homogenous ideal corresponding to $X$, i.e., such that $I^\sim = I_X\subset \operatorname{S}\mathcal{O}_B^n$, then the image of $\operatorname{S}\pi_*I_X(1)$ corresponds to the homogenous ideal in $\operatorname{S}A^n$ generated by the degree one part $I^1 = I\cap\operatorname{S}^1A^n$. Therefore, to finally show that these ideals agree, we have to show that $I$ is generated by $I^1$.

The assumptions (finitely presented and linearly embedded) imply that

1. we may suppose that $I$ is a finitely generated ideal. In particular, it can be finitely generated by homogenous elements; inductively, it's easy to see that from this, and finiteness of $\operatorname{S}A^n$ over $A$, it follows that each $I^d$, $d\geq 0$, is a finitely generated $A$-module;
2. for all $b\in B$, the map $\operatorname{S}I^1\otimes_B k(b)\to I\otimes_Bk(b)$ is surjective. In particular, each of the maps $\operatorname{S}^dI^1\otimes_B k(b)\to I^d\otimes_Bk(b)$, $d\geq 0$, is surjective.

According to 1. we can apply Nakayama to the surjective maps in 2. and get that for each prime ideal $p\subset A$, all of the localised maps $(\operatorname{S}^dI^1)_p\cong\operatorname{S}^dI_p^1\to I_p^d$, $d\geq 0$, are surjective. Finally, since surjectivity is local, the claim follows.

[Note: There probably is some general statement like "A morphism between $B$-smooth schemes which is an isomorphism over all points of $B$ is an isomorphism", perhaps with further assumptions, but I couldn't find a reference.]