The Stacks project

Proof. Set $F = G(k, n)$. To prove the lemma we will use the criterion of Schemes, Lemma 26.15.4. The reason $F$ satisfies the sheaf property for the Zariski topology is that we can glue sheaves, see Sheaves, Section 6.33 (some details omitted).

The family of subfunctors $F_ i$. Let $I$ be the set of subsets of $\{ 1, \ldots , n\} $ of cardinality $n - k$. Given a scheme $S$ and $j \in \{ 1, \ldots , n\} $ we denote $e_ j$ the global section

of $\mathcal{O}_ S^{\oplus n}$. Of course these sections freely generate $\mathcal{O}_ S^{\oplus n}$. Similarly, for $j \in \{ 1, \ldots , n - k\} $ we denote $f_ j$ the global section of $\mathcal{O}_ S^{\oplus n - k}$ which is zero in all summands except the $j$th where we put a $1$. For $i \in I$ we let

which is the direct sum of the coprojections $\mathcal{O}_ S \to \mathcal{O}_ S^{\oplus n}$ corresponding to elements of $i$. More precisely, if $i = \{ i_1, \ldots , i_{n - k}\} $ with $i_1 < i_2 < \ldots < i_{n - k}$ then $s_ i$ maps $f_ j$ to $e_{i_ j}$ for $j \in \{ 1, \ldots , n - k\} $. With this notation we can set

Given a morphism $f : T \to S$ of schemes the pullback $f^*s_ i$ is the corresponding map over $T$. Since $f^*$ is right exact (Modules, Lemma 17.3.3) we conclude that $F_ i$ is a subfunctor of $F$.

Representability of $F_ i$. To prove this we may assume (after renumbering) that $i = \{ 1, \ldots , n - k\} $. This means $s_ i$ is the inclusion of the first $n - k$ summands. Observe that if $q \circ s_ i$ is surjective, then $q \circ s_ i$ is an isomorphism as a surjective map between finite locally free modules of the same rank (Modules, Lemma 17.14.5). Thus if $q : \mathcal{O}_ S^{\oplus n} \to \mathcal{Q}$ is an element of $F_ i(S)$, then we can use $q \circ s_ i$ to identify $\mathcal{Q}$ with $\mathcal{O}_ S^{\oplus n - k}$. After doing so we obtain

mapping $e_ j$ to $f_ j$ (notation as above) for $j = 1, \ldots , n - k$. To determine $q$ completely we have to fix the images $q(e_{n - k + 1}), \ldots , q(e_ n)$ in $\Gamma (S, \mathcal{O}_ S^{\oplus n - k})$. It follows that $F_ i$ is isomorphic to the functor

This functor is isomorphic to the $k(n - k)$-fold self product of the functor $S \mapsto \Gamma (S, \mathcal{O}_ S)$. By Schemes, Example 26.15.2 the latter is representable by $\mathbf{A}^1_\mathbf {Z}$. It follows $F_ i$ is representable by $\mathbf{A}^{k(n - k)}_\mathbf {Z}$ since fibred product over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ is the product in the category of schemes.

The inclusion $F_ i \subset F$ is representable by open immersions. Let $S$ be a scheme and let $q : \mathcal{O}_ S^{\oplus n} \to \mathcal{Q}$ be an element of $F(S)$. By Modules, Lemma 17.9.4. the set $U_ i = \{ s \in S \mid (q \circ s_ i)_ s\text{ surjective}\} $ is open in $S$. Since $\mathcal{O}_{S, s}$ is a local ring and $\mathcal{Q}_ s$ a finite $\mathcal{O}_{S, s}$-module by Nakayama's lemma (Algebra, Lemma 10.20.1) we have

Let $f : T \to S$ be a morphism of schemes and let $t \in T$ be a point mapping to $s \in S$. We have $(f^*\mathcal{Q})_ t = \mathcal{Q}_ s \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{T, t}$ (Sheaves, Lemma 6.26.4) and so on. Thus the map

induced by $(f^*q \circ f^*s_ i)_ t$ is the base change of the map $\kappa (s)^{\oplus n - k} \to \mathcal{Q}_ s/\mathfrak m_ s\mathcal{Q}_ s$ above by the field extension $\kappa (t)/\kappa (s)$. It follows that $s \in U_ i$ if and only if $t$ is in the corresponding open for $f^*q$. In particular $T \to S$ factors through $U_ i$ if and only if $f^*q \in F_ i(T)$ as desired.

The collection $F_ i$, $i \in I$ covers $F$. Let $q : \mathcal{O}_ S^{\oplus n} \to \mathcal{Q}$ be an element of $F(S)$. We have to show that for every point $s$ of $S$ there exists an $i \in I$ such that $s_ i$ is surjective in a neighbourhood of $s$. Thus we have to show that one of the compositions

is surjective (see previous paragraph). As $\mathcal{Q}_ s/\mathfrak m_ s\mathcal{Q}_ s$ is a vector space of dimension $n - k$ this follows from the theory of vector spaces. $\square$