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Let $n={\mathrm{rank}}_{\mathbb{Z}}M$. Work with commutative rings and take a quotient to mean a surjection $p:M{\otimes }_{\mathbb{Z}}R\twoheadrightarrow Q$, considered up to an isomorphism of $Q$ commuting with $p$. Equivalently, retain its kernel as a submodule of $M\otimes R$. Here a rank-$r$ locally free module is finite locally free, or equivalently a finite projective module of constant rank $r$. A ring homomorphism $R\to R^{\prime }$ sends the quotient toThe tensor product of modules is right exact and preserves finite projectivity, so this defines the functorial action. We prove scheme representability in the usual functor of points sense:naturally in $R$. The Grassmannian of locally free quotients will be constructed explicitly.

Assume first $0<r<n$, and choose a basis $e_1,\dots ,e_n$ of $M$. For every ordered subset $I=\{i_1<\cdots <i_r\}\subseteq \{1,\dots ,n\}$, consider those quotients for which $p(e_{i_1}),\dots ,p(e_{i_r})$ are a basis of $Q$. This choice gives a unique identification $Q\cong R^r$, in which the quotient is represented by an $r\times n$ matrix $A_I$ whose columns indexed by $I$ are the identity matrix. Its remaining $r(n-r)$ entries are arbitrary elements of $R$. Thus this subfunctor is represented by the affine schemeThere is no further change-of-basis ambiguity: the selected quotient images already specify its basis uniquely.

For another subset $J$, write $A_I^{(J)}$ for the square submatrix consisting of the $J$ columns. The same quotient lies in the $J$ chart exactly when its determinant is a unit. Consequently the chart overlap is the principal open subsetOn that overlap the matrix normalized in the $J$ basis isThe entries are regular on $D({\delta }_{IJ})$ by the adjugate matrix. Conversely, the $I$ columns of $A_J$ form $(A_I^{(J)}{)}^{-1}$, so normalization back to $I$ recovers $A_I$. The transition is therefore an isomorphism $U_{IJ}\cong U_{JI}$.

The cocycle identity follows from the same matrix computation, not just an assertion about bases. On a triple overlap, normalization first to $J$ and then to $K$ givesThis is direct normalization from $I$ to $K$. By gluing of schemes along open subschemes, the standard affine charts of the quotient Grassmannian therefore glue to a scheme $G$ over $\mathbb{Z}$.

The universal matrices also glue a universal quotient. On $U_I$, use the surjectionOn overlaps, identify its targets by $(A_I^{(J)}{)}^{-1}$, precisely the transition matrix above. The cocycle identity glues these free sheaves into a rank-$r$ locally free sheaf $\mathcal{Q}$ and glues the maps into $M\otimes {\mathcal{O}}_G\twoheadrightarrow \mathcal{Q}$. Its kernel $\mathcal{S}$ is locally free of rank $n-r$: on $U_I$ an explicit basis isHence we have the universal exact sequence

It remains to verify that these charts capture every quotient, including a nonfree projective target. Given $p:R^n\twoheadrightarrow Q$, define $V_I\subseteq \mathrm{Spec}R$ as the locus where the selected images $p(e_i)$, $i\in I$, form a basis. This is open: locally trivialize the locally free sheaf associated with $Q$, and it is the nonvanishing locus of the corresponding maximal minor. The sets $V_I$ cover $\mathrm{Spec}R$. Indeed, at every prime $\mathfrak{p}$, the images of all $n$ vectors span the $r$-dimensional residue-field quotient, so some $r$ of them form a basis there. Their determinant is a unit in $R_{\mathfrak{p}}$, hence they form a basis over that local ring and on an open neighbourhood of $\mathfrak{p}$.

On $V_I$, trivialize $Q$ using exactly those selected images. The remaining columns give regular functions and hence a morphism $V_I\to U_I$. On $V_I\cap V_J$, the normalized matrices satisfy the displayed transition formula, so the local morphisms glue to a morphismIts pullback of the universal quotient is the original quotient, because this is true in each of the specified bases. Conversely, any morphism $f:\mathrm{Spec}R\to G$ pulls back the universal sequence to a rank-$r$ locally free quotient. On each inverse-image chart its matrix is exactly the normalized matrix used to reconstruct $f$, so the two constructions are inverse. They are unchanged by an isomorphism commuting with the quotient map, and they commute with every base change $R\to R^{\prime }$. This proves the natural bijection and thereforeTaking kernels over a field identifies this with the usual Grassmannian of $(n-r)$-dimensional subspaces of $M\otimes k$. The construction itself does not assume the target module is globally free.

For $r=0$, the only quotient is zero; for $r=n$, a surjection to a locally free module of the same rank is an isomorphism, as can be checked locally by its determinant. Both endpoint functors are represented by $\mathrm{Spec}\mathbb{Z}$. If $r>n$, no such quotient exists over a nonempty test scheme, and the representing scheme is empty. Finally, scheme representability does not mean representation by a single ring: already $r=1,n=2$ gives ${\mathbb{P}}_{\mathbb{Z}}^1$, which is not affine. This is the standard algebraic-geometric meaning of representability for the functor on rings here.