Special case: rational points over a field. Let $S = \mathrm{Spec}(k)$ for some field $k.$ Then the elements of $\mathrm{Gr}(r, n)(k) := \mathrm{Gr}_{\mathrm{Spec}(k)}(r, n)(\mathrm{Spec}(k))$ are given by the $k$-linear surjections $k^{n} \twoheadrightarrow k^{r},$ which can be viewed as $r \times n$ matrices with at least one $r \times r$ submatrix having nonzero determinant. (In other words, the matrices have rank $r,$ which is the full rank.) We are these maps up to isomorphisms, which amounts to the action of $\mathrm{GL}_{r}(k)$ on the left. Thus, we have established: $$\mathrm{Gr}(r, n)(k) \leftrightarrow \mathrm{GL}_{r}(k) \backslash \mathrm{Mat}_{r \times n}(k).$$ The elements of the right-hand side can be seen as picking $r$-linearly independent vectors up to $k$-linear isomorphisms of the vector spaces they generate in $k^{n}.$ Thus, we see that $\mathrm{Gr}(r, n)(k)$ parametrizes $r$-dimensional subspaces of $k^{n}.$ The following lemma explains this a bit more clearly:
Lemma. Let $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ two sets of $r$ $k$-linearly independent vectors in $k^{n}.$ Denote by $A$ the $r \times n$ matrix whose $i$-th row is given by $v_{i}$ and define $B$ similarly using $w_{i}.$ Then the following are equivalent:
- $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ generate the same subspace in $k^{n}$;
- $B = gA$ for some $g \in \mathrm{GL}_{r}(k).$
Proof. Write $v_{i} = (v_{i,1}, \dots, v_{i,n})$ and $w_{i} = (w_{i,1}, \dots, w_{i,n})$ with $v_{i,j}, w_{i,j} \in k.$ Then for any $g = (g_{i,j}) \in \mathrm{Mat}_{r}(k),$ saying that $B = gA$ is equivalent to saying $$w_{ij} = g_{i,1}v_{1,j} + \cdots + g_{i,r}v_{r,j}$$ for $1 \leq i \leq r$ and $1 \leq j \leq n.$ This condition can be rewritten as $$w_{i} = g_{i,1}v_{1} + \cdots + g_{i,r}v_{r}$$ for $1 \leq i \leq r.$ Now, note that $\{v_{1}, \dots, v_{r}\}$ and $\{w_{1}, \dots, w_{r}\}$ generate the same subspace in $k^{n}$ if and only if the last condition we mentioned is true for some $g_{i,j} \in k.$ When this happens, it is immediate that $(g_{ij})$ is necessarily invertible, so this finishes the proof. $\Box$
Back to functorial nonsense. We are going to show that $$\mathrm{Gr}_{S}(r, n) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ is representable. Since all representable functor is a Zariski sheaf (i.e., a sheaf on the big Zariski site of $S$), the strategy is to show that $\mathrm{Gr}_{S}(r, n)$ is a Zarski sheaf and then check a certain property on top of that so that we can guarantee that it is representable by some $S$-scheme (which will be necessarily a unique one). That is, after showing $\mathrm{Gr}_{S}(r, n)$ is a Zarski sheaf, we will use the following theorem we studied in a previous posting:
Theorem. A functor $F : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ is representable if and only if the following two conditions are satisfied:
- $F$ is a Zariski sheaf;
- $F$ can be covered by representable open subfunctors.
Checking Grassmanianns are Zariski sheaves. Saying that $\mathrm{Gr}_{S}(r, n)$ is a sheaf means that for any $S$-scheme $U$ and an open cover $U = \bigcup_{i \in I}U_{i},$ the following two conditions are satisfied:
- if $\phi, \psi \in \mathrm{Gr}_{S}(r, n)(U)$ such that $\phi|_{U_{i}} = \psi|_{U_{i}}$ for all $i \in I,$ then $\phi = \psi$;
- if $(\phi_{i})_{i \in I} \in \prod_{i \in I}\mathrm{Gr}_{S}(r, n)(U_{i})$ satisfies that $\phi_{i}|_{U_{i} \cap U_{j}} = \phi_{j}|_{U_{i} \cap U_{j}}$ for all $i, j \in I,$ then there is $\phi \in \mathrm{Gr}_{S}(r, n)(U)$ such that $\phi|_{U_{i}} = \phi_{i}$ for all $i \in I.$
Proof for the first condition. To check the first among the two conditions above, consider $\phi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ and $\psi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}'.$ Since we are working with representatives of isomorphism classes, instead of checking needed equalities, we aim to establish corresponding isomorphisms. Denote by $s_{1}, \dots, s_{n} \in \mathscr{Q}(U)$ and $t_{1}, \dots, t_{n} \in \mathscr{Q}'(U)$ the images of $e_{1}, \dots, e_{n} \in \mathscr{O}(U)^{\oplus n},$ writing $\mathscr{O} = \mathscr{O}_{U}$ for convenience.
Suppose that $\phi|_{U_{i}} \simeq \psi|_{U_{i}}$ for all $i \in I.$ Our goal is to show that $\phi \simeq \psi.$ For this, we may assume that $U_{i}$'s are refined enough so that both $\mathscr{Q}|_{U_{i}}$ and $\mathscr{Q}'|_{U_{i}}$ are free of rank $r$ over $U_{i}.$ The restrictions $\phi|_{U_{i}} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}|_{U_{i}}$ and $\psi|_{U_{i}} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}'|_{U_{i}}$ are isomorphic so that we have $\eta_{i} : \mathscr{Q}|_{U_{i}} \overset{\sim}{\longrightarrow} \mathscr{Q}'|_{U_{i}}$ with $\psi|_{U_{i}} = \eta_{i} \circ \phi|_{U_{i}}$ for $i \in I.$ Given $i \in I,$ the commutativity condition can be written as $$t_{l}|_{U_{i}} = \eta_{i, U_{i}}(s_{l}|_{U_{i}})$$ for $1 \leq l \leq n.$ An important observation is the following:
Claim. We have $\eta_{i, U_{i} \cap U_{j}} = \eta_{j, U_{i} \cap U_{j}}.$
Proof of Claim. Note that $$\eta_{i, U_{i} \cap U_{j}}(s_{l}|_{U_{i} \cap U_{j}}) = t_{l}|_{U_{i} \cap U_{j}} = \eta_{j, U_{i} \cap U_{j}}(s_{l}|_{U_{i} \cap U_{j}})$$ for all $1 \leq l \leq n.$ Since $s_{1}|_{U_{i} \cap U_{j}}, \dots, s_{l}|_{U_{i} \cap U_{j}}$ generate $\mathscr{Q}(U_{i} \cap U_{j}),$ this establishes the identity. $\Box$ (Claim)
From now on let us write $U_{ij} = U_{i} \cap U_{j},$ as the notations get messier. Let $V \subset U$ be any open subset. We write $V_{i} = V \cap U_{i}$ and $V_{ij} = V \cap U_{ij}.$ Given any $s \in \mathscr{Q}(V),$ we have $$\eta_{i, V_{i}}(s|_{V_{ij}})|_{V_{ij}} = \eta_{i, V_{ij}}(s|_{V_{ij}}) = \eta_{j, V_{ij}}(s|_{V_{ij}}) = \eta_{j, V_{j}}(s|_{V_{j}})|_{V_{ij}}$$ for $i, j \in I.$ Since $\mathscr{Q}'$ is a sheaf, there is unique $\eta_{V}(s) \in \mathscr{Q}'(V)$ such that $\eta_{V}(s)|_{V_{i}} = \eta_{i, V_{i}}(s|_{V_{i}})$ for $i \in I.$ This defines a map $\eta_{V} : \mathscr{Q}(V) \rightarrow \mathscr{Q}'(V),$ which can be immediately recognized as a sheaf map $\eta : \mathscr{Q} \rightarrow \mathscr{Q}'$ by intersection with $U_{i}$'s to check some commutative diagrams. $\Box$
Proof for the second condition. Consider an $\mathscr{O}_{U_{i}}$-linear epimorphism $$\phi_{i} : \mathscr{O}_{U_{i}}^{\oplus n} \twoheadrightarrow \mathscr{Q}_{i}$$ for each $i \in I$ such that $\phi_{i}|_{U_{ij}} \simeq \phi_{j}|_{U_{ij}}$ for $i, j \in I.$ This means that, for $i, j \in I,$ we have $\eta_{ij} : \mathscr{Q}_{i}|_{U_{ij}} \overset{\sim}{\longrightarrow} \mathscr{Q}_{j}|_{U_{ij}}$ such that $\phi_{j}|_{U_{ij}} = \eta_{ij} \circ \phi_{i}|_{U_{ij}}.$ Our goal is to find $$\phi : \mathscr{O}_{U}^{\oplus n} \twoheadrightarrow \mathscr{Q}$$ such that $\phi|_{U_{i}} \simeq \phi_{i}$ for $i \in I.$ Writing $U_{ijk} = U_{i} \cap U_{j} \cap U_{k},$ we note that we have $$\eta_{jk}|_{U_{ijk}} \circ \eta_{ij}|_{U_{ijk}} \circ \phi_{i}|_{U_{ijk}} = \eta_{jk}|_{U_{ijk}} \circ \phi_{j}|_{U_{ijk}} = \phi_{k}|_{U_{ijk}} = \eta_{kj}|_{U_{ijk}} \circ \phi_{i}|_{U_{ijk}}.$$ This implies that we have the cocycle condition $$\eta_{jk}|_{U_{ijk}} \circ \eta_{ij}|_{U_{ijk}} = \eta_{kj}|_{U_{ijk}},$$ so there must be a unique sheaf (that is necessarily locally free of rank $r$) $\mathscr{Q}$ gluing $\mathscr{Q}_{i}$'s. (For example, see SP:00AK.) Now, we are given $\phi_{i} : \mathscr{O}_{U_{i}}^{\oplus n} \rightarrow \mathscr{Q}|_{U_{i}}$ with $\phi_{i}|_{U_{ij}} \simeq \phi_{j}|_{U_{ij}}$ for $i, j \in I.$ Working similarly to the last paragraph of the last proof, we are done. $\Box$
Hence, we have shown that $\mathrm{Gr}_{S}(r, n)$ is a Zariski sheaf.
What remains to show Grassmannians are representable. Now that we know $\mathrm{Gr}_{S}(r, n)$ is a Zariski sheaf, to show it is representable, it remains to show that $\mathrm{Gr}_{S}(r, n)$ is covered by reprsentable open subfunctors.
Given any subset $I \subset \{1, \dots, n\}$ of size $r,$ and an $S$-scheme $Y,$ denote by $\mathrm{Gr}_{S}(r, n)_{I}(Y)$ the subset of $\mathrm{Gr}_{S}(r, n)$ consisting of $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ such that on every trivializing open subset $U \subset Y$ of $\mathscr{Q}$ (which in particular gives $\mathscr{Q}|_{U} \simeq \mathscr{O}_{U}^{\oplus r}$), the $I$-th minor of the $r \times n$ matrix given by $\mathscr{O}_{U}^{\oplus r} \twoheadrightarrow \mathscr{Q}|_{U}$ is nonzero at every point of $U.$
More concretely, given $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q},$ let $s_{1}, \dots, s_{n} \in \mathscr{Q}(Y)$ be the images of $$e_{1}, \dots, e_{n} \in \mathscr{O}_{Y}(Y)^{\oplus n}.$$ Then $s_{1}|_{U}, \dots, s_{n}|_{U}$ are the columns of the $r \times n$ matrix in discussion. Fix any $S$-scheme map $\pi : X \rightarrow Y.$ We know this induces $$\mathrm{Gr}_{S}(r, n)(Y) \rightarrow \mathrm{Gr}_{S}(r, n)(X)$$ by the pullback $\pi^{*}.$ Moreover, we note that $\pi^{*}s_{1}, \dots, \pi^{*}s_{n} \in \mathscr{O}_{X}(X)^{\oplus n}$ are the images of $e_{1}, \dots, e_{n} \in \mathscr{O}_{X}(X)^{\oplus n},$ so considering trivializing open subsets, we may observe that the above map restricts to $$\mathrm{Gr}_{S}(r, n)_{I}(Y) \rightarrow \mathrm{Gr}_{S}(r, n)_{I}(X)$$ for each $I \subset \{1, \dots, n\}$ of size $r.$ Thus, we have seen that $\mathrm{Gr}_{S}(r, n)_{I}$ is a functor $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$ We have a map $\mathrm{Gr}_{S}(r, n)_{I} \rightarrow \mathrm{Gr}_{S}(r, n)$ of functors given by the relevant inclusions of sets.
Claim. The functor $\mathrm{Gr}_{S}(r, n)_{I} \rightarrow \mathrm{Gr}_{S}(r, n)$ is an open subfunctor. That is, given any $S$-scheme $Y$ and a map $h_{Y} \rightarrow \mathrm{Gr}_{S}(r, n)$ of functors, the base chage $$\mathrm{Gr}_{S}(r, n)_{I} \times_{\mathrm{Gr}_{S}(r, n)} h_{Y} \rightarrow h_{Y}$$ is given by an open embedding $V_{I} \hookrightarrow Y$ (i.e., $h_{V_{I}} \rightarrow h_{Y}$).
Construction of open subsets $V_{I}$. We are given be an $S$-scheme $Y$ and an element of $$\mathrm{Hom}(h_{Y}, \mathrm{Gr}_{S}(r, n)) \simeq \mathrm{Gr}_{S}(r, n)(Y),$$ where the hom-set is taken over the functors $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}.$ The correspondence is given by the Yoneda lemma, and explicitly, it is given as $\xi \mapsto \xi_{Y}(\mathrm{id}_{Y}).$ We denote by $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ (a reperesentative of) the given element in $\mathrm{Gr}_{S}(r, n)(Y).$ It is convenient to denote by $s_{1}, \dots, s_{n} \in \mathscr{Q}(Y)$ the images of $e_{1}, \dots, e_{n} \in \mathscr{O}_{Y}(Y)^{\oplus n}$ under $\phi_{Y}.$
It necessarily follows that $h_{Y}(X) \rightarrow \mathrm{Gr}_{S}(r, n)(X)$ is given by $$[\pi : X \rightarrow Y] \mapsto [\mathscr{O}_{X}^{\oplus n} \simeq \pi^{*}\mathscr{O}_{Y}^{\oplus n} \xrightarrow{\pi^{*}\phi} \pi^{*}\mathscr{Q}.]$$ Thus, the set $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X)$$ consists of the elements of the form $(\pi^{*}\phi, \pi)$ such that $\pi^{*}\phi \in \mathrm{Gr}_{S}(r, n)(X)_{I}$ (i.e., the $I$-th minor of $[(\pi^{*}s_{1})|_{W} | \cdots | (\pi^{*}s_{1})|_{W}]$ is nowhere zero in any trivializing open $W \subset X$ for $\pi^{*}\mathscr{Q}$). The map $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X) \rightarrow h_{Y}(X)$$ is simply given by $(\pi^{*}\phi, \pi) \mapsto \pi.$ Of course, we note that $\phi$ is only given up to an isomorphism, but the above discussion still makes sense even if we just consider isomorphism classes.
Let's construct $V_{I}$'s now. Again, note that $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ is fixed. Let $U \subset Y$ be any trivializing open subset for $\mathscr{Q}.$ Denote by $f_{I} \in \mathscr{O}_{Y}(U)$ the $I$-th minor (i.e., the determinant of the $I$-th $r \times r$ submatrix of $[s_{1}|_{U} | \cdots | s_{n}|_{U}]$). We define $V_{I}$ to be the union of $D_{U}(f_{I})$ where we vary trivializing open subsets $U \subset Y$ for $\mathscr{Q}.$
Proof of Claim. For any $\pi : X \rightarrow Y$ in $h_{Y}(X) = \mathrm{Hom}_{S}(X, Y),$ we have $\pi^{*}\phi : \mathscr{O}_{X}^{\oplus n} \twoheadrightarrow \pi^{*}\mathscr{Q}$ in $\mathrm{Gr}_{S}(r, n)(X).$ The upshot is that the following are equivalent:
- we have $\pi^{*}\phi \in \mathrm{Gr}_{S}(r, n)(X)_{I}$;
- $\pi(X) \subset V_{I}.$
This immediately gives an explicit bijection $$\mathrm{Gr}_{S}(r, n)_{I}(X) \times_{\mathrm{Gr}_{S}(r, n)(X)} h_{Y}(X) \simeq h_{V_{I}}(X) = \mathrm{Hom}_{S}(V_{I}, X),$$ which does the job. $\Box$
Our $V_{I}$'s form an open cover of $Y$. For each $y \in U,$ we have a surjective $\kappa(y)$-linear map $\mathscr{O}_{Y}^{n}|_{y} \twoheadrightarrow \mathscr{Q}|_{y},$ so there exists at least one $I \subset \{1, \dots, n\}$ with $|I| = r$ such that $y \in D_{U}(f_{I}).$ This implies that we have an open cover $$U = \bigcup_{I \subset \{1, \dots, n\}}D_{U}(f_{I}).$$ Denote by $V_{I}$ the union of all $D_{U}(f_{I}) \subset Y$ where $U$ varies among all the trivializing open subsets of $Y$ for $\mathscr{Q}.$ This gives an open cover $$Y = \bigcup_{I \subset [n] \text{ with } |I| = r} V_{I},$$ where we started to write $[n] = \{1, \dots, n\}$ since the notations are getting messy.
Each $\mathrm{Gr}_{S}(n, k)_{I}$ is represented by $\mathbb{A}^{r(n-r)}_{S}$. Let $Y$ be an $S$-scheme. Given $[\phi] \in \mathrm{Gr}_{S}(r, n),$ the map $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ can be described by $n$ sections $\phi_{j} : \mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q}.$ That is, we have $\phi = \phi_{1} \oplus \cdots \oplus \phi_{n}.$
Fix $I = \{i_{1}, \dots, i_{r}\}$ with $1 \leq i_{1} < \cdots < i_{r} \leq n.$ If $[\phi] \in \mathrm{Gr}_{S}(r, n)_{I},$ then $\phi_{i_{1}} \oplus \cdots \oplus \phi_{i_{r}} : \mathscr{O}_{Y}^{\oplus r} \rightarrow \mathscr{Q}$ is an isomorphism.
Proof. To see this, it is enough to show that its localization at a point $y \in Y$ is an isomorphism. Surjectivity is immediate from Nakayama. To show injectivity, consider the map $\mathscr{O}_{Y,y}^{\oplus r} \rightarrow \mathscr{Q}_{y} \simeq \mathscr{O}_{Y,y}^{\oplus r}$ in question as an $r \times r$ matrix whose columns are given by $$\phi_{i_{1},y}(1), \dots, \phi_{i_{r},y}(1) \in \mathscr{Q}_{y} \simeq \mathscr{O}_{Y,y}^{\oplus r}.$$ The determinant of this matrix is not in the maximal ideal $\mathfrak{m}_{Y,y}$ of $\mathscr{O}_{Y,y},$ so it must be invertible in the local ring. By Cramer's rule, this implies that the matrix is invertible over $\mathscr{O}_{Y,y},$ which implies the injectivity we wanted. $\Box$
Hence, we now may write $$\mathscr{Q}(Y) = \mathscr{O}_{Y}(Y)\phi_{i_{1},Y}(1) \oplus \cdots \oplus \mathscr{O}_{Y}(Y)\phi_{i_{r},Y}(1).$$ This implies that given $j \in [n] \setminus I,$ we may write $$\phi_{j,Y}(1) = a_{1,j}(\phi)\phi_{i_{1},Y}(1) + \cdots +a_{r,j}(\phi)\phi_{i_{r},Y}(1) \in \mathscr{Q}(Y),$$ where $a_{i,j}(\phi) \in \mathscr{O}_{Y}(Y) = \Gamma(Y, \mathscr{O}_{Y}),$ which can be viewed as an $S$-scheme map $a_{i,j}(\phi) : Y \rightarrow \mathbb{A}^{1}_{S},$ by gluing $R$-algebra maps $R[t] \rightarrow \Gamma(Y, \mathscr{O}_{Y})$ given by $t \mapsto a_{ij}(\phi)$ for affine open $\mathrm{Spec}(R) \subset S.$
Thus, we have $$(a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} : Y \rightarrow \mathbb{A}^{r(n-r)}_{S}.$$ This defines a map $$\mathrm{Gr}_{S}(r, n)(Y)_{I} \rightarrow \mathrm{Hom}_{S}(Y, \mathbb{A}^{r(n-r)}_{S}),$$ as we may note that $a_{i,j}(\phi)$ only depends on the isomorphism class of $\phi.$
Given a map $\pi : X \rightarrow Y$ of $S$-schemes, we have $$(\pi^{*}\mathscr{Q})(X) = \mathscr{O}_{X}(X)(\pi^{*}\phi_{i_{1},Y})(1) \oplus \cdots \oplus \mathscr{O}_{X}(X)(\pi^{*}\phi_{i_{r},Y})(1),$$ so we may see that $a_{i,j}(\pi^{*}\phi) = \pi^{*}a_{i,j}(\phi) \in \mathscr{O}_{X}(X)$ for $j \in [n] \setminus I.$ Since $\pi : X \rightarrow Y$ comes with the map $\Gamma(Y, \mathscr{O}_{Y}) \rightarrow \Gamma(X, \mathscr{O}_{X})$ such that $a_{i,j}(\phi) \mapsto \pi^{*}a_{i,j}(\phi),$ we see that $$(a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} \circ \pi = (\pi^{*}a_{i,j}(\phi))_{1 \leq i \leq r, j \in [n] \setminus I} = (a_{i,j}(\pi^{*}\phi))_{1 \leq i \leq r, j \in [n] \setminus I}.$$ This shows that our map is actually a map $$\mathrm{Gr}_{S}(r, n)(-)_{I} \rightarrow \mathrm{Hom}_{S}(-, \mathbb{A}^{r(n-r)}_{S})$$ of functors.
Conversely, given $$(a_{i,j})_{1 \leq i \leq r, j \in [n] \setminus I} : Y \rightarrow \mathbb{A}^{r(n-r)}_{S},$$ which we interpret as $a_{ij} \in \Gamma(Y, \mathscr{O}_{Y}) = \mathscr{O}_{Y}(Y).$ We define $\phi : \mathscr{O}_{Y}^{n} \rightarrow \mathscr{O}_{Y}^{\oplus r}$ as follows. We declare $\phi_{i_{1},Y}(1) := e_{1}, \dots, \phi_{i_{r},Y}(1) := e_{r} \in \mathscr{O}_{Y}(Y)^{\oplus r}$ so that $$\phi_{i_{1},Y} \oplus \cdots \oplus \phi_{i_{r},Y} : \mathscr{O}_{Y}^{\oplus r} \rightarrow \mathscr{O}_{Y}^{\oplus r}$$ is the identity map. For $j \in [n] \setminus I,$ we may define $$\phi_{j,Y}(1) := a_{1,j}\phi_{i_{1},Y}(1) + \cdots + a_{r,j}\phi_{i_{r},Y}(1) \in \mathscr{O}_{Y}(Y)^{\oplus r}.$$ One may check this gives a desired inverse so that we establish $$\mathrm{Gr}_{S}(r, n)(-)_{I} \simeq \mathrm{Hom}_{S}(-, \mathbb{A}^{r(n-r)}_{S}).$$ That is, the $S$-scheme $\mathbb{A}^{r(n-r)}_{S}$ respresents the functor $\mathrm{Gr}_{S}(r, n)(-)_{I}.$
Upshot. Since these are open subfunctors that cover $\mathrm{Gr}_{S}(r, n),$ we see that $\mathrm{Gr}_{S}(r, n)$ is representable by an $S$-scheme.
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