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Use the toric scheme over a base ring construction over $\mathbb{Z}$. Take the cocharacter lattice of an algebraic torus $N={\mathbb{Z}}^2$ and dual character lattice of an algebraic torus $M=\mathrm{Hom}(N,\mathbb{Z})$. For every $i\in \mathbb{Z}$, setThese cones in toric geometry, together with their rays and the zero cone, form an infinite fan in toric geometry. Adjacent two-dimensional cones meet in their common ray; all other pairs meet only at zero. Thus the face and intersection axioms hold even though there are infinitely many charts.

The consecutive vectors $v_i,v_{i+1}$ are a basis of $N$, since their determinant is $-1$. The dual cone has integral generatorswith $⟨a_i,v_i⟩=0$, $⟨a_i,v_{i+1}⟩=1$, $⟨b_i,v_i⟩=1$, $⟨b_i,v_{i+1}⟩=0$. Its integral points form the free commutative monoid $\mathbb{N}{a}_i+\mathbb{N}{b}_i$. Hence the associated semigroup algebra chart is simplyOn the dense algebraic torus, write $t={\chi }^{(1,0)}$ and $q={\chi }^{(0,1)}$. The chart coordinates areAlthough these expressions use inverses on the torus, $x_i,y_i,q$ are regular on the indicated charts.

The common-ray chart of $U_i$ and $U_{i+1}$ is $D(y_i)\subset U_i$, identified with $D(x_{i+1})\subset U_{i+1}$ byThe inverse is $y_i=x_{i+1}^{-1}$, $x_i=y_{i+1}{x}_{i+1}^2$. Nonadjacent charts meet in the common dense torus, and all transition maps agree there with the same characters of an algebraic torus $t,q$, so their cocycle identities follow. By gluing of schemes along open subschemes, these charts construct a scheme $X$.

The lattice projection $(a,b)\mapsto b$ carries every ${\sigma }_i$ into the nonnegative ray of the fan of the affine line. It therefore defines a toric morphismExplicitly its pullback on each chart sends $q$ to $x_i{y}_i$. The adjacent-chart formula preserves that product, so the local morphisms glue.

Compute its scheme-theoretic fibre over $q=0$:This is the union of the two coordinate affine lines, each with multiplicity one. Let $D_i$ denote the toric divisor corresponding to ${\rho }_i$. Its part in $U_i$ is $y_i=0$, with coordinate $x_i$; its part in $U_{i-1}$ is $x_{i-1}=0$, with coordinate $y_{i-1}$. On the overlap these coordinates satisfy $x_i=y_{i-1}^{-1}$. Consequently these two affine lines glue toThe neighbouring components $D_i,D_{i+1}$ meet at the section $p_i=(x_i,y_i)=(0,0)$, isomorphic to $\mathrm{Spec}\mathbb{Z}$. No other components meet. Every special-fibre chart has exactly these two branches and no embedded or extra components. ThusThe union is an infinite chain, not a finite-type curve disguised as one: its infinitely many components account for the infinitely many toric charts.

Over the principal open subset $D(q)$, both $x_i$ and $y_i$ are units because their product is $q$. Every chart then becomes the same torus chart:The transition maps identify the coordinate $t$ exactly, soThis is the generic region described in the question; over the actual generic point of $\mathrm{Spec}\mathbb{Z}[q]$ its fibre is ${\mathbb{G}}_{m,\mathbb{Q}(q)}$.

The construction is also flat. Indeed, each node chart has a free $\mathbb{Z}[q]$-module basisEvery monomial $x_i^u{y}_i^v$ reduces uniquely by taking out $q^{\min (u,v)}$, and the resulting monomials are independent. Thus the chart morphisms are flat morphisms. This verifies directly that the semistable infinite chain from a toric fan is a degeneration of the multiplicative group scheme rather than an unrelated special fibre.

Write $\hat{X}=X_{/Y}$. The formal completion of a scheme restricts on each chart toThus $\hat{X}$ is a formal scheme over $\mathrm{Spf}\mathbb{Z}[[q]]$. The ordinary torus $D(q)$ has disappeared from its underlying space: every point of $\hat{X}$ lies on $Y$. In particular, only adjacent completed charts intersect, and their two possible branch overlaps inside any one chart are disjoint.

Construct the action using the integral lattice automorphismIt carries $v_i$ to $v_{i+3}$ and ${\sigma }_i$ to ${\sigma }_{i+3}$, so induces a toric morphism $g:X\to X$ that is an automorphism. It preserves the lattice projection to the second coordinate, hence preserves $q$ and $Y$. On the dense algebraic torus it isIn the local coordinates this sends $U_i$ to $U_{i+3}$ withThese identities show regularity on every chart without requiring $q$ to be invertible there. They also make the action continuous in the $q$-adic topology, so it extends to $\hat{X}$. Its powers shift $D_i$ to $D_{i+3k}$; no nonzero power is the identity or fixes a point of the formal support. Therefore $G=⟨g⟩\cong \mathbb{Z}$ is a nontrivial infinite cyclic group acting freely.

Construct the quotient explicitly, using three representative chartswith $q=X_i{Y}_i$. Glue their branch opens cyclically, taking subscripts modulo three, byThese are genuine isomorphisms of formal schemes. For example, the completed localization of an adic ring on the first open iswhere $X_i=q/Y_i$. On the second it is $\mathbb{Z}[X_{i+1},X_{i+1}^{-1}][[q]]$, and the gluing sends $X_{i+1}$ to $Y_i^{-1}$ and $Y_{i+1}$ to $q{Y}_i$. Thus it preserves $q$ and is continuous, with a continuous inverse.

Inside ${\mathfrak{V}}_i$ we haveConsequently there are no triple-overlap conditions left to check: the two branch identifications in a chart have disjoint domains. By gluing of schemes along open subschemes in its formal version, the three charts produce a formal scheme $\mathfrak{C}$ over $\mathrm{Spf}\mathbb{Z}[[q]]$. Equivalently, reduce the same gluing modulo $q^{n+1}$ for every $n$; this constructs compatible schemes $C_n$, all on the same underlying space, with $C_n=C_{n+1}\times \mathrm{Spec}\mathbb{Z}[q]/(q^{n+1})$. Their inverse system has precisely the displayed formal charts.

For every integer $i$, map ${\hat{U}}_i$ isomorphically to ${\mathfrak{V}}_{\bar{i}}$ by identifying its coordinates with the corresponding capital-letter coordinates. The adjacent-chart formulas agree with the cyclic identifications, so these maps glue toThey are invariant under $g$, and ${\pi }^{-1}({\mathfrak{V}}_i)$ is the disjoint union of the translated opens ${\hat{U}}_{i+3k}$, each mapping isomorphically onto ${\mathfrak{V}}_i$. The group permutes these copies transitively. Thus the fibres of $\pi$ are exactly the $G$-orbits, and invariant sections on this inverse image are exactly the sections on ${\mathfrak{V}}_i$.

To verify the quotient property also for morphisms, take a $G$-invariant morphism $h:\hat{X}\to \mathfrak{T}$. Its restrictions to ${\hat{U}}_0,{\hat{U}}_1,{\hat{U}}_2$ agree on the adjacent branch opens. On the closing overlap they agree because the translated chart ${\hat{U}}_3$ has the same restriction as ${\hat{U}}_0$ under $g$. Hence they descend to a unique morphism $\mathfrak{C}\to \mathfrak{T}$. This proves that the gluing is the cyclic formal quotient of the infinite toric chain:Its fibre modulo $q$ is a triangle of three projective lines over $\mathbb{Z}$, each meeting its two neighbours at its two boundary sections. More generally, a shift by any $m\ge 3$ gives an $m$-gon. Completion is important here: translates of a formal chart are disjoint, whereas all ordinary toric charts still share the generic torus.

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.