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The free commutative monoid on $d$ generators consists of the nonnegative integral sums of those generators, with unique coefficients. Its semigroup algebra over a commutative ring is the polynomial ring in $d$ variables. In a smooth toric chart, the dual-cone lattice points form such a monoid, yielding affine-space coordinates.

An ideal $I$ is idempotent when $I^2=I$. Then $I^j=I$ for every $j\ge 1$, so a nonzero idempotent ideal gives a nonzero intersection of all its powers. A semigroup algebra with arbitrarily divisible positive exponents supplies examples in a non-Noetherian integral domain. If $I$ is finitely generated, the determinant trick applied to $I=I^2$ produces $r\in I$ such that $(1+r)I=0$.

Write $C={\sigma }^{\vee }$ for the dual cone, $S=C\cap M$, and $U_{\sigma }=\mathrm{Spec}\mathbb{C}[S]$ for the affine toric variety. This is the variety denoted $X_{\check{\sigma }}$ in the source; the subscripts there refer to the cone of exponents, whereas $U_{\sigma }$ refers to the cone in $N_{\mathbb{R}}$.

For a face of a polyhedral cone $\tau ⪯\sigma$, define ${\tau }^{\ast }=C\cap {\tau }^{\perp }$. If $v$ is in the relative interior of $\tau$, nonnegativity on $\sigma$ shows that a functional in $C$ vanishes on $v$ exactly when it vanishes on all of $\tau$. Hence ${\tau }^{\ast }=C\cap v^{\perp }$ is a face of $C$. Conversely a face $\gamma$ of $C$ gives ${\gamma }^{\ast }=\sigma \cap {\gamma }^{\perp }$. Every polyhedral face is exposed: the supporting functionals for its containing facets can be added, since they are nonnegative. If $\tau =\sigma \cap {\mu }^{\perp }$, then $\mu \in {\tau }^{\ast }$, soThe same argument applied to $C$ proves $({\gamma }^{\ast }{)}^{\ast }=\gamma$, using the closed-cone identity $C^{\vee }=\sigma$. That identity follows from separation of a point outside a closed convex cone. Finally, enlarging $\tau$ makes its annihilator smaller. Thus these maps are inverse order-reversing bijections, proving face duality for polyhedral cones even when $\sigma$ is not full-dimensional.

For the given integral exposing functional $\mu$, we claimOne inclusion is immediate since $\mu$ vanishes on $\tau$. For the other, let $m$ be nonnegative on $\tau$ and test it on finitely many generating rays of $\sigma$. Where $\mu$ vanishes, those rays belong to $\tau$, so $m$ is already nonnegative. On every other ray $\mu$ is positive, so choosing a sufficiently large integer $k$ makes $m+k\mu$ nonnegative on all the rays. This proves the claim. The corresponding semigroup algebra identity isThus the face chart is an open subvariety.

For a face $\gamma ⪯C$, the span of monomials ${\chi }^m$ with $m\notin \gamma$ is an ideal: if $m+n\in \gamma$ with $m,n\in C$, the face property forces both into $\gamma$. Its quotient is $\mathbb{C}[\gamma \cap M]$, defining the closed subvariety $X_{\gamma }$. If $\mu \notin \gamma$, this subvariety lies in $V({\chi }^{\mu })$.

Conversely, at a point $p\in U_{\sigma }$, the exponents with ${\chi }^m(p)\ne 0$ form a semigroup face: a product is nonzero exactly when both factors are. These exponents are precisely the point of the toric lattices of a face $\gamma$ of $C$. To see the geometric assertion, choose finitely many monoid generators by the Gordan lemma. The nonzero generators span a cone. If that cone were not a face, a rational nonnegative relation would express a sum of its generators using an omitted generator with a positive coefficient. Clearing denominators and evaluating the corresponding monomials would equate a nonzero product to zero. Conversely, any point of the toric lattice in their cone has a positive integral multiple expressible using them, so its monomial must also be nonzero. This proves the monomial support face of a toric point assertion. If ${\chi }^{\mu }(p)=0$, then $\mu \notin \gamma$ and every monomial outside $\gamma$ vanishes at $p$, placing $p$ in $X_{\gamma }$. Therefore, as reduced closed subvarieties,The assertion concerns the closed set with its reduced structure; a monomial's principal zero subscheme can have multiplicities.

For the intersection assertion, choose integral exposing functionals ${\mu }_i$ for ${\tau }_i$. Both are nonnegative on $\sigma$, so ${\tau }_1\cap {\tau }_2=\sigma \cap ({\mu }_1+{\mu }_2{)}^{\perp }$. Localization then givesThis uses the dual of the intersection, as printed in the PDF; the TeX's perpendicular symbol in that position is not the correct transcription.

For a general toric fan, the points with a fixed monomial support face form a torus orbit. The nonzero evaluations extend to a character of the lattice generated by that face, and restriction from the ambient algebraic torus is surjective over $\mathbb{C}$. Thus the affine calculation yields the orbit-cone correspondence: $U_{\sigma }$ contains exactly the orbits $O(\nu )$ for faces $\nu ⪯\sigma$. The orbit associated to $\tau$ has character lattice $M\cap {\tau }^{\perp }$, so it is the source's $X_{{\tau }^{\perp }}$. The same monomial ideal calculation, chart by chart, shows that its closure isIndeed, inside a chart containing $\tau$, the closure has exponent face ${\sigma }^{\vee }\cap {\tau }^{\perp }$, and its smaller support faces correspond exactly to cones containing $\tau$.

Fix $\sigma \in \Sigma$, as required for its affine chart. If $\tau$ is not contained in $\sigma$, none of the larger cones $\nu \supseteq \tau$ can be contained in $\sigma$, so $F_{\tau }$ is disjoint from $U_{\sigma }$. Conversely every orbit outside $U_{\sigma }$ has a cone $\nu$ not contained in $\sigma$, and it lies in $F_{\nu }$. Consequently the global complement is exactly

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 $C=\check{\sigma }={\sigma }^{\vee }$, $S=C\cap M$, and $U_{\sigma }=X_{\check{\sigma }}=\mathrm{Spec}k[S]$. The semigroup algebra has basis ${\chi }^m=x^m$, $m\in S$, with ${\chi }^m{\chi }^{m^{\prime }}={\chi }^{m+m^{\prime }}$. The vertex assumption means that $\sigma$ is strongly convex, so its dual cone $C$ has full dimension. By the Gordan lemma, $S$ is finitely generated: choose integral toric cone generators and subtract integer parts of their nonnegative coefficients; the remainder lies in a bounded parallelepiped with only finitely many lattice points. This argument also applies when $C$ has lineality space, by including generators in both directions there.

Choose an integral interior point $u\in C$. For every $m\in M$, both $tu$ and $tu+m$ belong to $S$ for all sufficiently large integers $t$. Thus $S-S=M$. Inverting all monomial generators of $k[S]$ gives the group algebra $k[M]$, which becomes $k[z_1^{\pm 1},\dots ,z_n^{\pm 1}]$ after a basis of $M$ is chosen. This proves the dense torus of an affine toric variety assertion:Density follows because the localization of a ring is of a domain. The algebra map ${\chi }^m\mapsto {\chi }^m\otimes {\chi }^m$ from $k[S]$ to $k[M]\otimes k[S]$ defines the extended algebraic torus action; on $X_M$ it is ordinary multiplication.

For a face of a polyhedral cone $\nu ⪯C$, set $I_{\nu }=({\chi }^m:m\in S\setminus \nu )$. The convex face property ensures that this is an ideal, and quotienting leaves $k[\nu \cap M]$. Hence $X_{\nu }=\mathrm{Spec}k[\nu \cap M]$ is a closed subvariety of $U_{\sigma }$. Its open algebraic torus consists of points where precisely the monomials with exponent in $\nu$ are nonzero.

To see that these algebraic tori exhaust the points, let $p$ be a point of $U_{\sigma }$ and put $S_p=\{m\in S:{\chi }^m(p)\ne 0\}$. Multiplicativity gives $m+m^{\prime }\in S_p$ if and only if both $m,m^{\prime }\in S_p$. Thus $S_p$ is a semigroup face. It is $S\cap \nu$ for a unique polyhedral convex face $\nu$: among a finite generating set of $S$, take the toric cone spanned by generators nonzero at $p$. A relation expressing a sum with a generator outside this toric cone as a combination of generators inside would, after clearing rational denominators, contradict the displayed multiplicativity property. Therefore the toric cone is a convex face. Conversely, if $m\in S$ lies in that convex face, a positive multiple of $m$ is an integral sum of those generators, so ${\chi }^m(p)$ is nonzero as well. This also proves uniqueness.

The point $p$ lies in $X_M$ exactly when its convex face is all of $C$. ThereforeOne can equivalently write this as the disjoint union of the open algebraic tori of the proper convex faces. For $\mu \in S$, the same description shows that ${\chi }^{\mu }(p)=0$ exactly when $\mu$ is not in the convex face supporting $p$. On every $X_{\nu }$ with $\mu \notin \nu$ the monomial vanishes identically. Consequently the monomial zero locus in an affine toric variety isThis is an equality of reduced subvarieties, or of their underlying closed sets. The principal subscheme itself may have multiplicity, as $x^2=0$ already shows on ${\mathbb{A}}^1$.

The face duality for polyhedral cones is the order-reversing pair of mapsTheir values are convex faces because the relevant pairings are nonnegative and a sum pairs to zero only if each summand does. More explicitly, finitely many nonnegative exposing functionals can be summed to expose the indicated intersection. For a convex face $\tau$, choose $m_0\in C$ exposing it: $\tau =\sigma \cap m_0^{\perp }$. Then $m_0\in C\cap {\tau }^{\perp }$, so $\sigma \cap (C\cap {\tau }^{\perp }{)}^{\perp }\subseteq \tau$, while the reverse inclusion is immediate. Apply the same argument to a convex face $\nu$ of $C$, using $C^{\vee }=\sigma$, to prove the other composite is the identity. Inclusion reversal follows directly from annihilators. This proves the bijection, not just the existence of the maps.

For $\tau ⪯\sigma$, write $\nu =C\cap {\tau }^{\perp }$ and $L=M\cap {\tau }^{\perp }$. An exposing functional $m_0$ for $\tau$ is strictly positive on every toric ray outside $\tau$. Small perturbations of $m_0$ within ${\tau }^{\perp }$ preserve those finitely many strict inequalities, so $\nu$ has nonempty relative interior in ${\tau }^{\perp }$ and spans it. Therefore $\nu \cap M$ generates $L$, by the interior-point argument used above within that subspace. Thus the open algebraic torus of $X_{\nu }$ is $\mathrm{Spec}k[L]=X_{{\tau }^{\perp }}$. Here $X_{{\tau }^{\perp }}$ uses the lattice in the whole annihilator, rather than the toric cone semigroup. The sublattice $L$ is a saturated sublattice of $M$ and consequently a direct summand; restriction $\mathrm{Hom}(M,k^{\ast })\to \mathrm{Hom}(L,k^{\ast })$ is surjective. The extended algebraic torus action is multiplication on this open algebraic torus and hence transitive. All other monomials vanish there. It follows thatThis is the torus orbit-cone correspondence and its orbit closure in a toric variety description. Under convex face duality, $F_{\gamma }={\bigcup }_{\eta ⪰\gamma }O(\eta )$.

Finally, choose integral $m_{\tau }\in C$ exposing $\tau$. Localizing $k[C\cap M]$ at ${\chi }^{m_{\tau }}$ gives $k[{\tau }^{\vee }\cap M]$: for $m\in {\tau }^{\vee }\cap M$, adding a sufficiently large multiple of $m_{\tau }$ makes it nonnegative on every toric ray of $\sigma$ outside $\tau$, while preserving nonnegativity on $\tau$. Hence $X_{\check{\tau }}=U_{\tau }=D({\chi }^{m_{\tau }})$ is the usual affine open chart inside $U_{\sigma }$. An orbit $O(\gamma )$ belongs to it exactly when $\gamma ⪯\tau$. The requested complement of an affine toric face chart is thereforeAll sets on the right are closed, and if $\gamma$ is not contained in $\tau$, neither is any convex face containing $\gamma$. This justifies replacing the excluded orbits by their whole closures.

First suppose $\Delta$ has full dimension in $M_{\mathbb{R}}$. Form the toric cone over it in $M_{\mathbb{R}}\oplus \mathbb{R}$ and the graded semigroup algebrawhere degree zero contains only $k$. The lattice semigroup is finitely generated by the Gordan lemma. The projective toric variety of a lattice polytope is ${\mathbb{P}}_{\Delta }=\mathrm{Proj}{R}_{\Delta }$. Using all lattice points in all dilates avoids any assumption that the degree-one lattice points generate the ring.

For each vertex $v$ of $\Delta$, the degree-zero part after inverting ${\chi }^{v}t$ isIndeed its exponents are $m-lv$ with $m\in l\Delta$. Conversely, every lattice point of the tangent toric cone has this form for sufficiently large integral $l$. These charts cover the Proj construction: each point of a dilate is a rational convex combination of vertices, so some power of its monomial is a product of vertex monomials. Their dual cones are the vertex toric cones of the inward normal fan of a polytope. The normal toric fan is complete because every linear functional on $\Delta$ attains its minimum on a convex face. This describes ${\mathbb{P}}_{\Delta }$ as a complete toric variety. Translating $\Delta$ by $m_0\in M$ gives the graded isomorphism $(m,l)\mapsto (m+l{m}_0,l)$, so containing the origin is unnecessary. If $\Delta$ is not full-dimensional, translate a lattice vertex to zero and use $M\cap {\mathrm{span}}_{\mathbb{R}}(\Delta -\Delta )$; the construction is then complete of dimension $\dim \Delta$.

For the simplex in question put $Q={\prod }_i{q}_i$. Its inequalities are $x_i\ge 0$ and ${\sum }_i{q}_i{x}_i\le Q$, so its inward normal toric rays are ${\mathbb{R}}_{+}{e}_i$ and ${\mathbb{R}}_{+}(-q_1,\dots ,-q_n)$. Here the $e_i$ in the dual space denote the standard dual basis. Apply the real linear map $A(e_i)=e_i/q_i$. It carries these toric rays and their convex face incidences to the standard simplex fan, with toric rays $e_1,\dots ,e_n,-{\sum }_i{e}_i$. The transported lattice isWith $\gcd (q_1,\dots ,q_n)=1$, the primitive vectors of this toric fan in $N^{\prime }$ are $v_i=e_i/q_i$ and $v_0=-{\sum }_i{e}_i$. The latter is primitive because $\lambda {\sum }_i{e}_i\in N^{\prime }$ requires each $\lambda q_i$ to be integral, whose smallest positive solution is $\lambda =1$. Their unique primitive relation isThus mutual coprimality beyond the displayed common gcd condition is not needed.

For a general complete toric variety, introduce one variable $z_{\rho }$ for each toric ray. Its toric homogeneous coordinate ring, or Cox ring, isThe toric divisor class sequence is $0\to M\to {\mathbb{Z}}^{\Sigma (1)}\to \mathrm{Cl}(X_{\Sigma })\to 0$, where the first map is $m\mapsto (⟨m,v_{\rho }⟩{)}_{\rho }$. Thus the graded piece $S_{\alpha }$ has basis the monomials $\prod z_{\rho }^{b_{\rho }}$ with $b_{\rho }\ge 0$ and $\sum b_{\rho }[F_{\rho }]=\alpha$. For an invariant divisor $D=\sum a_{\rho }{F}_{\rho }$, monomials in degree $[D]$ correspond to algebraic torus characters ${\chi }^m$ through $b_{\rho }=a_{\rho }+⟨m,v_{\rho }⟩\ge 0$. This also identifies that graded piece with the global sections of $\mathcal{O}(D)$.

In our example $M^{\prime }=(N^{\prime }{)}^{\ast }={⨁}_i{q}_i\mathbb{Z}{e}_i^{\ast }$, and its map to ${\mathbb{Z}}^{n+1}$ sends ${\sum }_i{q}_i{b}_i{e}_i^{\ast }$ to $(-{\sum }_i{q}_i{b}_i,b_1,\dots ,b_n)$. The quotient is therefore $\mathbb{Z}$, via $(c_0,\dots ,c_n)\mapsto c_0+{\sum }_i{q}_i{c}_i$. In particular $\mathrm{deg}{z}_0=1$ and $\mathrm{deg}{z}_i=q_i$, giving the explicit graded isomorphismThe target has the grading inherited from ordinary total degree in $k[Y_0,\dots ,Y_n]$. Its displayed generators are algebraically independent, which proves injectivity as well as surjectivity onto that subring.

To obtain the quotient, the toric irrelevant ideal is generated by ${\prod }_{\rho \notin \sigma (1)}{z}_{\rho }$. Here each maximal toric cone omits exactly one toric ray, so it is $(z_0,\dots ,z_n)$, and the open set is ${\mathbb{A}}^{n+1}\setminus \{0\}$. The diagonalizable algebraic group with algebraic torus character group $\mathrm{Cl}(X)=\mathbb{Z}$ is $k^{\ast }$. Its action on the Cox coordinates is dictated by their degrees, namelyHere the $y_i$ name the independent Cox coordinates; the graded-subring realization above does not change their weights. More concretely, the quotient chart where $y_i\ne 0$ has coordinate ring $(S[y_i^{-1}]{)}_0$, the invariant ring for this action. These charts glue to $\mathrm{Proj}S$ and to the specified toric fan charts. Positive weights make each orbit closed inside the punctured affine space, and in characteristic zero the finite stabilizers cause no failure of this geometric quotient. This is the weighted projective space construction.

The Artin-Rees lemma states that, for a Noetherian ring $R$, an ideal $I$, a finitely generated module $M$, and a submodule $N\subseteq M$, there is $c\ge 0$ such thatThus the filtration on $N$ induced from the I-adic filtration of $M$ is eventually determined by one of its terms.

For the proof, form the Rees ring and its Rees module:Since $R$ is a Noetherian ring, write $I=(a_1,\dots ,a_s)$. Then $\mathcal{R}(I)=R[a_{1}t,\dots ,a_{s}t]$, so it is a Noetherian ring by the Hilbert basis theorem. Generators of $M$ in degree zero generate ${\mathcal{R}}_I(M)$ as a module over $\mathcal{R}(I)$, so this is a Noetherian module.

The graded submoduleis therefore finitely generated. Choose homogeneous generators $z_{\alpha }{t}^{d_{\alpha }}$ with $d_{\alpha }\le c$ for some $c$. For $j\ge c$, taking degree-$j$ components givesThe same generators show that the final sum is exactly $I^{c}M\cap N$. This proves the Artin-Rees lemma. If $L=0$, one may simply take $c=0$.

Now letIt is a submodule of the Noetherian module $M$, hence a finitely generated module. Apply the Artin-Rees lemma with $N=K$ and $j=c+1$. Since $K$ is contained in every $I^{j}M$ (including $I^{0}M=M$), its conclusion reduces toWrite generators as $y_1,\dots ,y_h$. There is a matrix $A=(a_{ij})$ with entries in $I$ such that $y_i={\sum }_j{a}_{ij}{y}_j$. Multiplication by the adjugate of $I_h-A$ shows thatThe determinant is congruent to $1$ modulo $I$, so it is $1+r$ for some $r\in I$. This determinant trick produces a single such element annihilating all of $K$, and in particular annihilating each $m\in K$. If $K=0$, use $r=0$.

Conversely, if $(1+r)m=0$ with $r\in I$, then $m=-rm$. Iterating gives $m=(-r{)}^{j}m\in I^{j}M$ for every $j\ge 1$. Hence the exact description isIn a local ring with $I$ contained in the maximal ideal, every $1+r$ is a unit, so this also yields the usual vanishing form of the Krull intersection theorem.

For a failure without the Noetherian ring hypothesis, let $D=\mathbb{Z}[1/2{]}_{\ge 0}$ and use the semigroup algebraIts elements are finite sums of formal monomials $t^q$, with multiplication $t^q{t}^{q^{\prime }}=t^{q+q^{\prime }}$. Each ring in the union is a polynomial ring and the inclusions are injective, so $R$ is an integral domain. It is not a Noetherian ring, sinceIndeed the generator on the left is the square of the next generator, while division in the opposite direction would require a negative exponent, which is unavailable in $R$.

Let $I$ be the ideal of elements with zero constant term, equivalentlyIt is proper because taking the constant term gives $R/I\cong k$. Every generator $t^q$ is $(t^{q/2}{)}^2$, with $t^{q/2}\in I$, so $I^2=I$. Thus this is a nonzero idempotent ideal, with $I^j=I$ for every $j\ge 1$. TakeFor every $r\in I$, the constant term of $1+r$ is $1$, so $1+r\ne 0$. Since $R$ is an integral domain and $t\ne 0$, we have $(1+r)t\ne 0$. This gives all three requested objects and proves that no element of the required form annihilates $m$.

Given a dual pair of free lattices $N,M$ and a rational fan, glue $\mathrm{Spec}A[{\sigma }^{\vee }\cap M]$ over common-face charts. The monoid rings are semigroup algebras. A lattice map respecting cones gives a toric morphism. This construction works over a commutative ring $A$, including $\mathbb{Z}$, and also for an infinite fan in toric geometry.

// Target: toric-geometry.bigb