On the Central Geometry of Nonnoetherian Dimer Algebras

ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS

CHARLIE BEIL

Abstract. Let Z be the center of a nonnoetherian dimer algebra on a torus. Although Z itself is also nonnoetherian, we show that it has Krull dimension 3, and is locally noetherian on an open dense set of Max Z. Furthermore, we show that the reduced center Z/ nil Z is depicted by a Gorenstein singularity, and contains precisely one closed point of positive geometric dimension.

1. Introduction In this article, all dimer quivers are nondegenerate and embed in a two-torus. A dimer algebra A is noetherian if and only if its center Z is noetherian, if and only if A is a noncommutative crepant resolution of its 3-dimensional toric Gorenstein center [Br, D, B4]. We show that the center Z of a nonnoetherian dimer algebra is also 3-dimensional, and may be viewed as the coordinate ring for a toric Gorenstein singularity that has precisely one ‘smeared-out’ point of positive geometric dimension. This is made precise using the notion of a depiction, which is a finitely generated overring that is as close as possible to Z, in a suitable geometric sense (Definition 2.5). Denote by nil Z the nilradical of Z, and by Zˆ := Z/ nil Z the reduced ring of Z. Our main theorem is the following. Theorem 1.1. (Theorems 3.19, 3.20.) Let A be a nonnoetherian dimer algebra with center Z, and let Λ := A/ hp − q | p, q a non-cancellative pairi be its ghor algebra with center R. (1) The nonnoetherian rings Z, Zˆ, and R each have Krull dimension 3, and the arXiv:1903.10460v2 [math.RA] 10 Sep 2021 integral domains Zˆ and R are depicted by the cycle algebras of A and Λ. (2) The reduced scheme of Spec Z and the scheme Spec R are birational to a noetherian affine scheme, and each contain precisely one closed point of positive geometric dimension. Dimer models were introduced in string theory in 2005 in the context of brane tilings [HK,FHVWK]. The dimer algebra description of the combinatorial data of a

2010 Mathematics Subject Classification. 16S38,14A20,13C15. Key words and phrases. Non-noetherian ring, non-noetherian geometry, dimer algebra. 1 2 CHARLIE BEIL brane tiling arose from the notion of a superpotential algebra (or quiver with poten- tial), which was introduced a few years earlier in [BD]. Stable (i.e., ‘superconformal’) brane tilings quickly made their way to the mathematics side, but the more difficult study of unstable brane tilings was largely left open, in regards to both their mathematical and physical properties. There were two main difficulties: in contrast to the stable case, the ‘mesonic chiral ring’ (closely related to what we call the cycle algebra1) did not coincide with the center of the dimer algebra, and (ii) although the mesonic chiral ring still appeared to be a nice ring, the center certainly was not. The center is supposed to be the coordinate ring for an affine patch on the extra six compact dimensions of spacetime, the so-called (classical) vacuum geometry. But examples quickly showed that the center of an unstable brane tiling could be infin- itely generated. To say that the vacuum geometry was a nonnoetherian scheme – something believed to have no visual representation or concrete geometric interpre- tation – was not quite satisfactory from a physics perspective. However, unstable brane tilings are physically allowable theories. To make matters worse, almost all brane tilings are unstable, and it is only in the case of a certain uniform symme- try (an ‘isoradial embedding’) that they become stable. Moreover, in the context of 11-dimensional M-theory, stable and unstable brane tilings are equally ‘good’. The question thus remained: What does the vacuum geometry of an unstable brane tiling look like? The aim of this article is to provide an answer. In short, the vacuum geometry of an unstable brane tiling looks just like the vacuum geometry of a stable brane tiling, namely a 3-dimensional complex cone, except that there is precisely one curve or surface passing through the apex of the cone that is identified as a single ‘smeared- out’ point.

2. Preliminary definitions Throughout, k is an uncountable algebraically closed ﬁeld. Given a quiver Q, we denote by kQ the path algebra of Q; and by Q0 and Q1 the sets of vertices and arrows of Q respectively. The vertex idempotent at vertex i ∈ Q0 is denoted ei, and the head and tail maps are denoted h, t : Q1 → Q0. By monomial, we mean a nonconstant monomial.

2.1. Dimer algebras, ghor algebras, and cyclic contractions. Deﬁnition 2.1. • Let Q be a ﬁnite quiver whose underlying graph Q embeds into a real two-torus T 2, such that each connected component of T 2 \ Q is simply connected and bounded

1The mesonic chiral ring is the ring of gauge invariant operators; these are elements of the dimer algebra that are invariant under isomorphic representations, and thus are cycles in the quiver. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 3 by an oriented cycle of length at least 2, called a unit cycle.2 The dimer algebra of Q is the quiver algebra A := kQ/I with relations

I := hp − q | ∃ a ∈ Q1 such that pa and qa are unit cyclesi ⊂ kQ, where p and q are paths. Since I is generated by certain diﬀerences of paths, we may refer to a path modulo I as a path in the dimer algebra A. • Two paths p, q ∈ A form a non-cancellative pair if p 6= q, and there is a path r ∈ kQ/I such that rp = rq 6= 0 or pr = qr 6= 0. A and Q are called non-cancellative if there is a non-cancellative pair; otherwise they are called cancellative. • The ghor algebra of Q is the quotient Λ := A/ hp − q | p, q is a non-cancellative pairi . A dimer algebra A coincides with its ghor algebra if and only if A is cancellative, if and only if A is noetherian, if and only if Λ is noetherian [B4, Theorem 1.1]. • Let A be a dimer algebra with quiver Q.

–A perfect matching D ⊂ Q1 is a set of arrows such that each unit cycle contains precisely one arrow in D. –A simple matching D ⊂ Q1 is a perfect matching such that Q \ D supports a simple A-module of dimension 1Q0 (that is, Q \ D contains a cycle that passes through each vertex of Q). Denote by S the set of simple matchings of A. – A is said to be nondegenerate if each arrow of Q belongs to a perfect matching.3 Each perfect matching D deﬁnes a map

nD : Q≥0 → Z≥0 that sends path p to the number of arrow subpaths of p that are contained in D. nD is 0 0 additive on concatenated paths, and if p, p ∈ Q≥0 are paths satisfying p + I = p + I, 0 then nD(p) = nD(p ). In particular, nD induces a well-deﬁned map on the paths of A. Now consider dimer algebras A = kQ/I and A0 = kQ0/I0, and suppose Q0 is ∗ obtained from Q by contracting a set of arrows Q1 ⊂ Q1 to vertices. This contraction deﬁnes a k-linear map of path algebras ψ : kQ → kQ0.

2In forthcoming work, we consider the nonnoetherian central geometry of ghor algebras on higher genus surfaces. Dimer quivers on other surfaces arise in contexts such as Belyi maps, cluster categories, and bipartite ﬁeld theories; see e.g., [BGH, BKM, FGU]. 3For our purposes, it suﬃces to assume that each cycle contains an arrow that belongs to a perfect matching; see [B1]. 4 CHARLIE BEIL

If ψ(I) ⊆ I0, then ψ induces a k-linear map of dimer algebras, called a contraction, ψ : A → A0. Denote by 0 B := k [xD : D ∈ S ] the polynomial ring generated by the simple matchings S0 of A0. To each path p ∈ A0, associate the monomial Y nD(p) τ¯(p) := xD ∈ B. D∈S0 0 For each i, j ∈ Q0, this association extends to a k-linear map 0 τ¯ : ejA ei → B. This map is an algebra homomorphism if i = j, and injective if A0 is cancellative [B2, Proposition 4.29]. For each i, j ∈ Q0, we also consider the k-linear map given by the composition of the k-linear maps ψ andτ ¯,

ψ 0 τ¯ τ¯ψ : ejAei −→ eψ(j)A eψ(i) −→ B. 0 Given p ∈ ejAei and q ∈ e`A ek, we will write

p :=τ ¯ψ(p) :=τ ¯(ψ(p)) and q :=τ ¯(q). ψ is called a cyclic contraction if A0 is cancellative and 0 0 S := k [∪ τ¯ (e Ae )] = k ∪ 0 τ¯(e A e ) =: S . i∈Q0 ψ i i i∈Q0 i i In this case, we call S the cycle algebra of A. The cycle algebra is independent of the choice of cyclic contraction ψ [B3, Theorem 3.14], and is isomorphic to the center of A0 [B2, Theorem 1.1.3]. Moreover, every nondegenerate dimer algebra admits a cyclic contraction [B1, Theorem 1.1]. In addition to the cycle algebra, the ghor center of A,

R := k [∩i∈Q0 τ¯ψ(eiAei)] , also plays an important role. This algebra is isomorphic to the center of the ghor algebra Λ of Q [B2, Theorem 1.1.3].

2 2 Notation 2.2. Let π : R → T be a covering map such that for some i ∈ Q0, 2 π Z = i. Denote by Q+ := π−1(Q) ⊂ R2 the covering quiver of Q. For each path p in Q, denote by p+ the unique path in Q+ with tail in the unit square [0, 1) × [0, 1) ⊂ R2 satisfying π(p+) = p. For u ∈ Z2, denote by Cu the set of cycles p in A such that + + + h(p ) = t(p ) + u ∈ Q0 . ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 5

Notation/Lemma 2.3. We denote by σi ∈ A the unique unit cycle (modulo I) at i ∈ Q0, and by σ the monomial Y σ := σi = xD. D∈S0 The sum P σ is a central element of A. i∈Q0 i Lemma 2.4. (1) If p ∈ C(0,0) is nontrivial, then p = σn for some n ≥ 1. (2) Let u ∈ Z2 and p, q ∈ Cu. Then p = qσn for some n ∈ Z. Proof. (1) is [B2, Lemma 5.2.1], and (2) is [B2, Lemma 4.18].

2.2. Nonnoetherian geometry: depictions and geometric dimension. Let S be an integral domain and a finitely generated k-algebra, and let R be a (possibly nonnoetherian) subalgebra of S. Denote by Max S, Spec S, and dim S the maximal spectrum (or variety), prime spectrum (or affine scheme), and Krull dimension of S respectively; similarly for R. For a subset I ⊂ S, set ZS(I) := {n ∈ Max S | n ⊇ I}. Definition 2.5. [B5, Definition 3.1] • We say S is a depiction of R if the morphism4

ιS/R : Spec S → Spec R, q 7→ q ∩ R, is surjective, and

(1) {n ∈ Max S | Rn∩R = Sn} = {n ∈ Max S | Rn∩R is noetherian}= 6 ∅. • The geometric height of p ∈ Spec R is the minimum

n −1 o ght(p) := min htS(q) | q ∈ ιS/R(p),S a depiction of R . The geometric dimension of p is gdim p := dim R − ght(p). We will denote the subsets (1) of the algebraic variety Max S by

US/R := {n ∈ Max S | Rn∩R = Sn} , (2) ∗ US/R := {n ∈ Max S | Rn∩R is noetherian} .

The subset US/R is open in Max S [B5, Proposition 2.4.2].

4 The morphism ιS/R is well-deﬁned: Suppose q ∈ Spec S. Then S/q is an integral domain. Whence the subalgebra R/(q ∩ R) ⊆ S/q is an integral domain. Thus q ∩ R is a prime ideal of R. 6 CHARLIE BEIL

Example 2.6. Let S = k[x, y], and consider its nonnoetherian subalgebra R = k[x, xy, xy2,...] = k + xS. R is then depicted by S, and the closed point xS ∈ Max R has geometric dimension 1 [B5, Proposition 2.8]. Furthermore, US/R is the complement of the line Z(x) = {x = 0} ⊂ Max S. 2 In particular, Max R may be viewed as 2-dimensional aﬃne space Ak = Max S with the line Z(x) identiﬁed as a single ‘smeared-out’ point. From this perspective, xS is a positive dimensional point of Max R. In the next section, we will show that the reduced center and ghor center of a nonnoetherian dimer algebra are both depicted by its cycle algebra, and both contain precisely one point of positive geometric dimension.

3. Proof of main theorem Throughout, A is a nonnoetherian dimer algebra with center Z and reduced center Zˆ := Z/ nil Z. By assumption A is nondegenerate, and thus there is a cyclic contraction ψ : A → A0 to a noetherian dimer algebra A0 [B1, Theorem 1.1]. The center Z0 of A0 is isomorphic to the cycle algebra S [B2, Theorem 1.1.3], and the reduced center Zˆ of A is isomorphic to a subalgebra of R [B6, Theorem 4.1].5 We may therefore write (3) Zˆ ⊆ R. The following structural results will be useful. Lemma 3.1. Let g ∈ B be a monomial. (1) If g ∈ R, h ∈ S are monomials and g 6= σn for each n ≥ 0, then gh ∈ R. (2) If g ∈ R and σ - g, then g ∈ Zˆ. (3) If g ∈ R, then there is some m ≥ 1 such that gm ∈ Zˆ. (4) If g ∈ S, then there is some m ≥ 0 such that for each n ≥ 1, gnσm ∈ Zˆ. (5) If gσ ∈ S, then g ∈ S. (6) If a monomial h ∈ S \ R satisﬁes σ - h, then hn 6∈ R for each n ≥ 1. Furthermore, a monomial h ∈ S \ R exists for which σ - h. Proof. (1) is [B6, Lemma 6.1]; (2) - (4) is [B6, Lemma 5.3]; and (5) is [B2, Lemma 4.18] for u 6= (0, 0), and Lemma 2.4.1 for u = (0, 0); and (6) is [B4, Proposition 3.14]. Lemma 3.2. The cycle algebra S is a ﬁnite type integral domain.

5It is often the case that Zˆ is isomorphic to R; an example where Zˆ =6∼ R is given in [B6, Example 4.3]. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 7

Proof. The cycle algebra S is generated by theτ ¯ψ-images of cycles in Q with no nontrivial cyclic proper subpaths. Since Q is finite, there is only a finite number of such cycles. Therefore S is a finitely generated k-algebra. S is also an integral domain since it is a subalgebra of the polynomial ring B. It is well-known that the Krull dimension of the center of any cancellative dimer algebra (on a torus) is 3 (e.g. [Br]). The isomorphism S ∼= Z0 therefore implies that the Krull dimension of the cycle algebra S is 3. In the following we give a new and independent proof of this result. Lemma 3.3. The cycle algebra S has Krull dimension 3.

0 0 Proof. Fix j ∈ Q0 and cycles in ejA ej, 0(1,0) 0(−1,0) 0(0,1) 0(0,−1) (4) s1 ∈ C , t1 ∈ C , s2 ∈ C , t2 ∈ C . Consider the algebra 0 (i) T := k[σ, s1, s2, t1, t2] ⊆ S = S, where (i) holds since the contraction ψ is cyclic. Since A0 is cancellative, if p ∈ C0u and q ∈ C0v are cycles in A0 satisfying p = q, then u = v [B4, Lemma 3.9]. Thus there are no relations among the monomials s1, s2, t1, t2, by our choice of cycles (4). However, by Lemma 2.4.1, there are integers n1, n2 ≥ 1 such that

n1 n2 (5) s1t1 = σ and s2t2 = σ . (i) We claim that dim T = 3. Since T is a ﬁnite type integral domain and k is algebraically closed, the variety Max T is equidimensional [E, Ch. 13, Theorem A]. It thus suﬃces to show that the chain of ideals of T ,

(6) 0 ⊂ (σ, s1, s2) ⊆ (σ, s1, s2, t1) ⊆ (σ, s1, s2, t1, t2), is a maximal chain of distinct primes. The inclusions in (6) are strict since the relations among the monomial generators are generated by the two relations (5). Moreover, (6) is a maximal chain of primes of T : Suppose si is in a prime p of T . Then σ is in p, by (5). Whence si+1 or ti+1 is also in p, again by (5). Thus (σ, s1, s2) is a minimal prime of T . (ii) We now claim that dim S = dim T . By Lemma 2.4.2, we have S[σ−1] = T [σ−1]. Furthermore, S and T are ﬁnite type integral domains, by Lemma 3.2. Thus Max S and Max T are irreducible algebraic varieties that are isomorphic on their open dense sets {σ 6= 0}. Therefore dim S = dim T . 8 CHARLIE BEIL

Corollary 3.4. The Krull dimension of the center of a noetherian dimer algebra is 3. 0 ∼ Proof. Follows from Lemma 3.3 and the isomorphism Z = S. Lemma 3.5. The morphisms κ : Max S → Max Z,ˆ n 7→ n ∩ Z,ˆ (7) S/Zˆ κS/R : Max S → Max R, n 7→ n ∩ R, and ˆ ˆ ιS/Zˆ : Spec S → Spec Z, q 7→ q ∩ Z,

ιS/R : Spec S → Spec R, q 7→ q ∩ R, are well-deﬁned and surjective.

Proof. (i) We first claim that κS/Zˆ and κS/R are well-defined maps. Indeed, let n be in Max S. By Lemma 3.2, S is of finite type, and by assumption k is algebraically closed. Therefore the intersections n ∩ Zˆ and n ∩ R are maximal ideals of Zˆ and R respectively (e.g., [B5, Lemma 2.1]). ˆ (ii) We claim that κS/Zˆ and κS/R are surjective. Fix m ∈ max Z. Then Sm is a proper ideal of S since S is a subalgebra of the polynomial ring B. Thus, since S is noetherian, there is a maximal ideal n ∈ Max S containing Sm. Whence, m ⊆ Sm ∩ Zˆ ⊆ n ∩ Z.ˆ ˆ ˆ ˆ But n ∩ Z is a maximal ideal of Z by Claim (i). Therefore m = n ∩ Z. Similarly, κS/R is surjective.

(iii) It is clear that ιS/Zˆ and ιS/R are well-defined maps (see footnote 4). Finally, we claim that ιS/Zˆ and ιS/R are surjective. By [B5, Lemma 3.6], if D is a finitely generated algebra over an uncountable field k, and C ⊆ D is a subalgebra, then ιD/C : Spec D → Spec C is surjective if and only if κD/C : Max D → Max C is surjective. Therefore, ιS/Zˆ and ιS/R are surjective by Claim (ii). Lemma 3.6. If p ∈ Spec Zˆ contains a monomial, then p contains σ. Proof. Suppose p contains a monomial g. Then there is a nontrivial cycle p such that p = g. Let q+ be a path from h(p+) to t(p+). Then the concatenated path (pq)+ is a cycle in Q+. Thus, there is some n ≥ 1 such that pq = σn, by Lemma 2.4.1. By Lemma 3.5, there is a prime ideal q ∈ Max S such that q∩R = p. Furthermore, pq = σn is in q since p ∈ p and q ∈ S. Whence σ is also in q since q is prime. But ˆ ˆ σ ∈ Z. Therefore σ ∈ q ∩ Z = p. Denote the origin of Max S by

n0 := (s ∈ S | s a nontrivial cycle) S ∈ Max S. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 9

Consider the maximal ideals of Zˆ and R respectively, ˆ z0 := n0 ∩ Z and m0 := n0 ∩ R. ˆ Proposition 3.7. The localizations Zz0 and Rm0 are nonnoetherian. Proof. There is a monomial g ∈ S \ R such that gn 6∈ R for each n ≥ 1, by Lemma 3.1.6. n (i) Fix n ≥ 1. We claim that g is not in the localization Rm0 . Assume otherwise; × then there is an a ∈ R, b ∈ m0, β ∈ k , such that a gn = ∈ R , β + b m0 since m0 is a maximal ideal of R. Whence, (8) bgn − a = −βgn 6∈ R. Thus bgn 6∈ R since a ∈ R and βgn 6∈ R. Since m0 is generated by monomials in R, we may write the polynomial b ∈ m0 as a sum of monomials in m0, X X 0 b = bj + bk, j k where for each j and k, n 0 n bjg 6∈ R and bkg ∈ R. Consider the polynomial X n n X 0 n h := bjg + βg = a − bkg . j k P n n P If h = 0, then j bjg = −βg . Whence j bj = −β since B is an integral P × domain. But j bj is in m0 whereas β ∈ k is not, a contradiction. Therefore h 6= 0. View R ⊂ S as k-vector subspaces of the polynomial ring B. These subspaces have bases given by all monomials with scalar coeﬃcient 1 in R and S respectively. Deﬁne a symmetric bilinear form on S: for monomials m1, m2 ∈ S in the basis, set ( 1 if m1 = m2 (m1, m2) := , 0 otherwise and extend k-bilinearly to S. P n n Now h is nonzero and equals j bjg + βg , and thus may be written entirely in terms of basis vectors that lie in the orthogonal complement to R (with respect to P 0 n (·, ·)). Thus h itself does not lie in R. But h also equals a − k bkg , and thus lies n in R, a contradiction. Therefore g is not in the localization Rm0 .

(ii) We now claim that Rm0 is nonnoetherian. By Lemma 3.1.4, there is an m ≥ 1 such that for each n ≥ 1, n m hn := g σ ∈ R. 10 CHARLIE BEIL

Consider the chain of ideals of Rm0 ,

0 ⊂ (h1) ⊆ (h1, h2) ⊆ (h1, h2, h3) ⊆ .... Assume to the contrary that the chain stabilizes. Then there is an N ≥ 1 such that N−1 X hN = cnhn, n=1 with cn ∈ Rm0 . Thus, since R is an integral domain, N−1 N X n g = cng . n=1 Furthermore, since R is a subalgebra of the polynomial ring B and g is a monomial, there is some 1 ≤ n ≤ N − 1 such that N−n cn = g + b, N−n with b ∈ Rm0 . But then g = cn − b ∈ Rm0 , contrary to Claim (i). ˆ (iii) Similarly, Zz0 is nonnoetherian. Recall that by monomial, we mean a nonconstant monomial. Lemma 3.8. Suppose that each monomial in Zˆ is divisible (in B) by σ. If p ∈ Spec Zˆ contains a monomial, then p = z0. Proof. Suppose p ∈ Spec Zˆ contains a monomial. Then σ is in p by Lemma 3.6. Furthermore, there is some q ∈ Spec S such that q ∩ Zˆ = p, by Lemma 3.5. Suppose g is a monomial in Zˆ. By assumption, there is a monomial h in B such that g = σh. By Lemma 3.1.5, h is also in S. Whence g = σh ∈ q since σ ∈ p ⊆ q. But g ∈ Zˆ. Therefore g ∈ q ∩ Zˆ = p. Since g was arbitrary, p contains all monomials ˆ in Z. Remark 3.9. In Lemma 3.8, we assumed that σ divides all monomials in Zˆ. An example of a dimer algebra with this property is given in Figure 1, where the center is the nonnoetherian ring Z ∼= Zˆ = R = k + σS, and the cycle algebra is the quadric cone, S = k[xz, xw, yz, yw]. The contraction ψ : A → A0 is cyclic.

Lemma 3.10. Suppose that there is a monomial in Zˆ which is not divisible (in B) ˆ ˆ by σ. Let m ∈ Max Z \{z0}. Then there is a monomial g ∈ Z \ m such that σ - g. ˆ Proof. Let m ∈ Max Z \{z0}. (i) We ﬁrst claim that there is a monomial in Zˆ \ m. Assume otherwise. Then ˆ n0 ∩ Z ⊆ m. ˆ But z0 := n0 ∩ Z is a maximal ideal by Lemma 3.5. Thus z0 = m, contrary to assumption. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 11

y 1 1 x y x 2o ··oo 1/ / 2 2o 1/ 2 O O 1 z 1 ψ −→ ···OO w z w w 1 w 1/··/ 2o oo 1 1/ 2o 1 1 x y 1 x y QQ0

Figure 1. An example where σ divides all monomials in R. The quivers are drawn on a torus, the contracted arrows are drawn in dotted green, and the arrows are labeled by theirτ ¯ψ andτ ¯-images.

··········X / X / X / X / X X / X / X / X / X

··········Ö / Ö / Ö / Ö / Ö / Ö / Ö / Ö / a column subquiver

/·····X / X X / X / U ·m ·I ·····- Ö / Ö / Ö / Ö a pillar subquiver

Figure 2. A column and pillar of a dimer quiver Q. The black interior arrows are arrows of Q; the blue and red boundary arrows are paths of length at least one in Q; and each interior cycle is a unit cycle of Q. The leftmost and rightmost brown arrows of the column are identiﬁed. Note that the blue and red bounding paths of the pillar are equal modulo I.

(ii) We now claim that there is a monomial in Zˆ \ m which is not divisible by σ. Indeed, assume to the contrary that every monomial in Zˆ, which is not divisible by σ, is in m. By assumption, there is a monomial in Zˆ that is not divisible by σ. Thus there is at least one monomial in m. Therefore σ is in m, by Lemma 3.6. There is an n ∈ Max S such that n ∩ Zˆ = m, by Lemma 3.5. Furthermore, σ ∈ n since σ ∈ m. Suppose g ∈ Zˆ is a monomial for which σ | g; say g = σh for some monomial h ∈ B. Then h ∈ S by Lemma 3.1.5. Whence, g = σh ∈ n. Thus g ∈ n ∩ Zˆ = m. It follows that every monomial in Zˆ, which is divisible by σ, is also in m. Therefore ˆ every monomial in Z is in m. But this contradicts our choice of m by Claim (i). 12 CHARLIE BEIL

Deﬁnition 3.11. A column and pillar of a dimer quiver Q are subquivers as shown in Figure 2. Lemma 3.12. Let z = P q be a central element of A such that for each i ∈ Q , i∈Q0 i 0 qi is a cycle in eiAei. (In particular, qi = qj for each i, j ∈ Q0.) If σ - qi, then the set of representatives q˜i ∈ eikQei of the qi partition Q into columns and pillars. Proof. See [B2, Lemmas 4.8.3 and 4.12]. Lemma 3.13. Let p ∈ Cu, q ∈ Cv be cycles in Q such that u, v ∈ Z2 are linearly independent over R. If p ∈ S \ Z,ˆ q ∈ Z,ˆ and σ - q, then pq ∈ Z.ˆ ˆ Proof. Since q is a monomial in Z, for each j ∈ Q0 there is a cycle qj ∈ ejAej such that qj = q, qt(q) = q, and the sum X z := qj

j∈Q0 is in the center Z of A. Furthermore, since σ - q, the set of representatives of the qj partition Q into columns and pillars, by Lemma 3.12. Fix a representativep ˜ of p (that is,p ˜ + I = p), and for each j ∈ Q0, choose a representativeq ˜j of qj. By assumption, u and v are linearly independent over R. Thus, for each j ∈ Q0, the cyclesq ˜j andp ˜ intersect at some vertex i(j) ∈ Q0. Factor q˜j andp ˜ into paths

q˜j = qj2ei(j)qj1

p˜ = pj2ei(j)pj1, and consider the cycles

(9) rj := qj2pj1pj2qj1 + I ∈ ejAej. Note that rj = pqj = pq. Thus, to prove the lemma, it suﬃces to show that the element X r := rj

j∈Q0 is in the center Z of A. (a) We ﬁrst claim that r is independent of the choice of representativesq ˜j of qj. 0 Fix j ∈ Q0, and supposeq ˜j andq ˜j are two representatives of qj. By Lemma 3.12, it 0 suﬃces to suppose thatq ˜j andq ˜j bound a pillar, andp ˜ intersects this pillar as shown 0 in Figure 3. Factorp ˜,q ˜j,q ˜j into paths 0 0 0 0 p˜ = p2bp1, q˜j = q5q3q2q1, q˜j = q5q4q3q2q1. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 13

Then

(i) 0 0 q5q3bp1p2q2q1 ≡ q5q4q3cbp1p2q2q1 0 ≡ q5q4σh(p1)p1p2q2q1

(ii) 0 ≡ q5q4p1p2σt(p2)q2q1 0 0 ≡ q5q4p1p2bq3cq2q1

(iii) 0 0 ≡ q5q4p1p2bq2q1, where (i) and (iii) hold by the dimer relations since each arrow in the interior of the column is an arrow in the quiver, and (ii) holds by Lemma 2.3. Therefore the two 0 representativesq ˜j,q ˜j of qj deﬁne the same cycle rj ∈ ejAej in (9). (b) We now claim that r is in Z. Since r is a sum of cycles, r trivially commutes with the vertex idempotents. Thus, to show that r is in Z, it suﬃces to show that ra = ar for each arrow a ∈ Q1. Fix an arrow a. Set j := t(a) and k := h(a). If a is a leftmost arrow subpath of a representativer ˜k of rk, then there is a representativer ˜j of rj that is a cyclic permutation ofr ˜k, by Claim (a). Whence

ra = rka = arj = ar.

So suppose a is not a leftmost arrow subpath of any representative of rk. There are three cases to consider, shown in Figure 4. Case (i): Supposeq ˜j andq ˜k bound a column, andp ˜ intersects this column as shown in Figure 4.i. Factorp ˜,q ˜j,q ˜k into paths

0 0 p˜ = p2bp1, q˜j = q3q2q1, q˜k = q2q1.

Then

0 0 (i) 0 (q2(bp1p2)q1)a ≡ q2bp1p2bq2q1 (i) ≡ a(q3(p1p2b)q2q1), where (i) and (ii) hold by the dimer relations since each arrow in the interior of the column is an arrow in the quiver. Case (ii): Supposeq ˜j andq ˜k bound a column, andp ˜ intersects this column as shown in Figure 4.ii. Factorp ˜,q ˜j,q ˜k into paths

0 0 p˜ = p2cp1, q˜j = q3q2q1, q˜k = q2q1. 14 CHARLIE BEIL

·O

q2 q3 /······/ / / q1 X X X X U j c b j /·m ·I q / ······Ö Ö Ö Ö Ö 5 - / 0 0 /K 0 / q2 q3 q4

0 0 0 0 Figure 3. The pathsp ˜ = p2bp1,q ˜j = q5q3q2q1, andq ˜j = q5q4q3q2q1 for Claim (a) in the proof of Lemma 3.13.

Thus,

0 0 (i) 0 (q2(p1p2c)q1)a ≡ q2p1p2cbq2q1 0 ≡ q2p1p2σt(q2)q1

(ii) 0 ≡ q2σh(p1)p1p2q1 0 ≡ q2bq2cp1p2q1 (iii) ≡ a(q3q2(cp1p2)q1), where (i) and (iii) hold by the dimer relations since each arrow in the interior of the column is an arrow in the quiver, and (ii) holds by Lemma 2.3. Case (iii): Supposeq ˜j andq ˜k bound a pillar, andp ˜ intersectsq ˜j as shown in Figure 4.iii. Factorp ˜,q ˜j,q ˜k into paths 0 0 p˜ = p2p1, q˜j = q4q3q2q1, q˜k = q4q3q2q1. Whence,

0 0 (i) 0 (q4q3(p1p2)q2q1)a ≡ q4q3p1p2q2σt(q2)q1

(ii) 0 ≡ q4σh(q3)q3p1p2q2q1 (iii) ≡ a(q4q3(p1p2)q2q1) where (i) and (iii) hold by the dimer relations since each arrow in the interior of the pillar is an arrow in the quiver, and (ii) holds by Lemma 2.3. Therefore ra = ar holds in each case. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 15

·O

0 0 q1 q2 ········X / X / X X / X / X / X a (i): a c b ········Ö / Ö / Ö / Ö Ö / Ö / q1 q2 K q3

q0 q0 1 2 ········X / X / X X / X / X / X a (ii): a c b ········Ö / Ö / Ö / Ö Ö / Ö / q1 q2 q3

·O

p2 0 0 q1 q4 ···X / X / U q q /·X /·X (iii): 2 3 a a ···I / O / m ····Ö / Ö - Ö / Ö /· q1 q4

Figure 4. The three cases for Claim (b) in the proof of Lemma 3.13.

Lemma 3.14. Let p ∈ Cu, q ∈ Cv be cycles in Q such that u, v ∈ Z2 are linearly dependent over R. If p ∈ S \ Z,ˆ q ∈ Z,ˆ and σ does not divide p or q, then there is a cycle r and integers m, n ≥ 1 such that rn = q, r ∈ Z,ˆ and prm ∈ Z.ˆ 16 CHARLIE BEIL

Proof. (i) First suppose u = (0, 0). Since p 6∈ Zˆ, p is not a vertex. Thus, there is some ` ≥ 1 such that p = σ`, by Lemma 2.4.1. Whence σ | p, contrary to assumption. (ii) Assume to the contrary that there are positive integers n1, n2 ≥ 1 such that n1u = n2v. By assumption, σ - p. Thus there is a simple matching D such that n1 n2 xD - p. Whence σ - p . Similarly, σ - q . Consequently,

(i) (ii) (10) pn1 = qn2 ∈ Zˆ ⊆ R, where (i) holds by Lemma 2.4.2, and (ii) holds by [B6, Theorem 1.1.3]. Now if a monomial g ∈ S is not in R and σ - g, then g` is also not in R for each ` ≥ 1, by Lemma 3.1.6. Thus (10) implies that p is in R. Therefore p is in Zˆ since σ - p, by Lemma 3.1.2. But this contradicts our choice of p. (iii) Finally, suppose there are positive integers n1, n2 ≥ 1 such that n1u = −n2v. Letv ˆ ∈ Z2 be the vector of minimal length (with respect to the standard R2 metric) vˆ such that v = n3vˆ for some n3 ≥ 1. Then there is a cycle r ∈ C such that σ - r, by [B1, Corollary 3.9]. Thus r ∈ Zˆ since q ∈ Zˆ, by Lemma 3.1.6. Furthermore, setting m := n2n3 we have n1 u = −mv.ˆ Therefore there is some ` ≥ 1 such that prm (i)= σ` ∈ Z,ˆ where (i) holds by Lemma 2.4.1. Recall the subsets (2) of Max S and the morphisms (7). Proposition 3.15. Let n ∈ Max S. Then ˆ ˆ n ∩ Z 6= z0 if and only if Zn∩Zˆ = Sn, and n ∩ R 6= m0 if and only if Rn∩R = Sn. Consequently, ˆ κS/Zˆ(US/Zˆ) = Max Z \{z0} and κS/R(US/R) = Max R \{m0}. Proof. ˆ (i) Set m := n ∩ Z, and suppose m 6= z0. We ﬁrst claim that ˆ (11) S ⊂ Zm. Consider g ∈ S \ Zˆ. By [B2, Proposition 5.14], S is generated by σ and a set of monomials in B not divisible by σ. Furthermore, σ is in Zˆ. It thus suﬃces to suppose that g is a monomial which is not divisible by σ; let p be a cycle for which p = g. (i.a) First suppose σ does not divide all monomials in Zˆ. By Lemma 3.10, there is a nontrivial cycle q ∈ A such that q ∈ Zˆ \ m and σ - q. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 17

Let u, v ∈ Z2 be such that p ∈ Cu and q ∈ Cv. If u, v are linearly independent over R, then pq is in Zˆ, by Lemma 3.13. Whence −1 ˆ g = p = (pq)q ∈ Zm. If instead u, v are linearly dependent over R, then there is a cycle r and integers m, n ≥ 1 such that rn = q, r ∈ Z,ˆ prm ∈ Z,ˆ by Lemma 3.14. Furthermore, r 6∈ m since rn = q and q 6∈ m. Thus, m −m ˆ g = p = (pr )r ∈ Zm. ˆ Therefore, in either case, g is in the localization Zm. Consequently, (11) holds if σ does not divide all monomials in Zˆ. (i.b) Now suppose σ divides all monomials in Zˆ. Then m does not contain any monomials since m 6= z0, by Lemma 3.8. In particular, σ 6∈ m. By Lemma 3.1.4, there is an n ≥ 0 such that gσn ∈ Zˆ. Thus n −n ˆ g = (gσ )σ ∈ Zm. Therefore (11) also holds if σ divides all monomials in Zˆ. ˆ ˆ ˆ (ii) Denote by m˜ := mZm the maximal ideal of Zm. Then, since Z ⊂ S, we have

(i) Zˆ = Zˆ ⊆ S ⊆ (Zˆ ) = (Zˆ ) = Zˆ , m m˜∩Zˆ m˜∩S m m˜∩Zˆm m mZˆm m where (i) holds by (11). Therefore ˆ Sn = Zm.

(iii) Now suppose n ∩ R 6= m0. We claim that Rn∩R = Sn. Since n ∩ R 6= m0, there is a monomial g ∈ R \ n. Thus there is some n ≥ 1 such that gn ∈ Zˆ, by Lemma 3.1.3. Furthermore, gn 6∈ n since n is a prime ideal. Consequently, gn ∈ Zˆ \ (n ∩ Zˆ). Whence ˆ (12) n ∩ Z 6= z0. Therefore (ii) (i) ˆ Sn = Zn∩Zˆ ⊆ Rn∩R ⊆ Sn, where (i) holds by (12) and Claim (i), and (ii) holds by (3). It follows that Rn∩R = Sn. (iv) Finally, we claim that ˆ Zz0 6= Sn0 and Rm0 6= Sn0 . 18 CHARLIE BEIL ˆ These inequalities hold since the local algebras Zz0 and Rm0 are nonnoetherian by Proposition 3.7, whereas Sn is noetherian by Lemma 3.2. Lemma 3.16. Let q and q0 be prime ideals of S. Then q ∩ Zˆ = q0 ∩ Zˆ if and only if q ∩ R = q0 ∩ R. Proof. (i) Suppose q ∩ Zˆ = q0 ∩ Zˆ, and let s ∈ q ∩ R. Then s ∈ R. Whence there is some n ≥ 1 such that sn ∈ Zˆ, by Lemma 3.1.3. Thus sn ∈ q ∩ Zˆ = q0 ∩ Z.ˆ Therefore sn ∈ q0. Thus s ∈ q0 since q0 is prime. Consequently, s ∈ q0 ∩ R. Therefore q ∩ R ⊆ q0 ∩ R. Similarly, q ∩ R ⊇ q0 ∩ R. (ii) Now suppose q ∩ R = q0 ∩ R, and let s ∈ q ∩ Zˆ. Then s ∈ Zˆ ⊆ R. Thus s ∈ q ∩ R = q0 ∩ R. 0 ˆ ˆ 0 ˆ ˆ 0 ˆ Whence s ∈ q ∩ Z. Therefore q ∩ Z ⊆ q ∩ Z. Similarly, q ∩ Z ⊇ q ∩ Z.

Proposition 3.17. The subsets US/Zˆ and US/R of Max S are equal. Proof. (i) We ﬁrst claim that

US/Zˆ ⊆ US/R. ˆ Indeed, suppose n ∈ US/Zˆ. Then since Z ⊆ R ⊆ S, we have ˆ Sn = Zn∩Zˆ ⊆ Rn∩R ⊆ Sn. Thus

Rn∩R = Sn.

Therefore n ∈ US/R, proving our claim. (ii) We now claim that

US/R ⊆ US/Zˆ.

Let n ∈ US/R. Then Rn∩R = Sn. Thus by Proposition 3.15,

n ∩ R 6= n0 ∩ R. Therefore by Lemma 3.16, ˆ ˆ n ∩ Z 6= n0 ∩ Z. But then again by Proposition 3.15, ˆ Zn∩Zˆ = Sn.

Whence n ∈ US/Zˆ, proving our claim. We denote the complement of a set W ⊆ Max S by W c. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 19

Theorem 3.18. The following subsets of Max S are open, dense, and coincide: U ∗ = U = U ∗ = U S/Zˆ S/Zˆ S/R S/R (13) = κ−1 (Max Zˆ \{z }) = κ−1 (Max R \{m }) S/Zˆ 0 S/R 0 c c = ZS(z0S) = ZS(m0S) . In particular, Zˆ and R are locally noetherian at all points of Max Zˆ and Max R except at z0 and m0.

Proof. For brevity, set Z(I) := ZS(I). (i) We ﬁrst show the equalities of the top two lines of (13).

By Proposition 3.17, US/R = US/Zˆ. By Lemma 3.2, S is noetherian. Thus for each n ∈ Max S, the localization Sn is noetherian. Therefore, by Proposition 3.15, ∗ ∗ US/Zˆ = US/Zˆ and US/R = US/R. Moreover, again by Proposition 3.15, U = κ−1 (Max Zˆ \{z }) and U = κ−1 (Max R \{m }) . S/Zˆ S/Zˆ 0 S/R S/R 0

(ii) We now claim that the complement of US/R ⊂ Max S is the zero locus Z(m0S). Suppose n ∈ Z(m0S); then m0S ⊆ n. Whence,

m0 ⊆ m0S ∩ R ⊆ n ∩ R.

Thus n ∩ R = m0 since m0 is a maximal ideal of R. But then n 6∈ US/R by Claim (i). c Therefore US/R ⊇ Z(m0S). c Conversely, suppose n ∈ US/R. Then κS/R(n) 6∈ κS/R(US/R), by the deﬁnition of US/R. Thus n ∩ R = κS/R(n) = m0, by Claim (i). Whence,

m0S = (n ∩ R)S ⊆ n, c and so n ∈ Z(m0S). Therefore US/R ⊆ Z(m0S). (iii) Finally, we claim that the subsets (13) are open dense. Indeed, there is a maximal ideal of R distinct from m0, and κS/R is surjective by Lemma 3.5. Thus −1 the set κS/R (Max R \{m0}) is nonempty. Moreover, this set equals the open set c c Z(m0S) , by Claim (ii). But S is an integral domain. Therefore Z(m0S) is dense since it is nonempty and open. Theorem 3.19. The center Z, reduced center Zˆ, and ghor center R of A each have Krull dimension 3, dim Z = dim Zˆ = dim R = dim S = 3. Furthermore, the fraction ﬁelds of Zˆ, R, and S coincide, (14) Frac Zˆ = Frac R = Frac S. 20 CHARLIE BEIL

Proof. Recall that S is of ﬁnite type by Lemma 3.2, and Zˆ ⊆ R ⊆ S are integral domains since they are subalgebras of the polynomial ring B. ˆ The sets US/Zˆ and US/R are nonempty, by Theorem 3.18. Thus Z, R, and S have equal fraction ﬁelds [B5, Lemma 2.4]; and equal Krull dimensions [B5, Theorem 2.5.4]. In particular, dim Zˆ = dim R = dim S = 3, by Lemma 3.3. Finally, each ˆ prime p ∈ Spec Z contains the nilradical nil Z, and thus dim Z = dim Z.

Recall that the reduction Xred of a scheme X, that is, its reduced induced scheme structure, is the closed subspace of X associated to the sheaf of ideals I, where for each open set U ⊂ X,

I(U) := {f ∈ OX (U) | f(p) = 0 for all p ∈ U} .

Xred is the unique reduced scheme whose underlying topological space equals that of

X. If R := OX (X), then OXred (Xred) = R/ nil R. Theorem 3.20. Let A be a nonnoetherian dimer algebra, and let ψ : A → A0 a cyclic contraction. (1) The reduced center Zˆ and ghor center R of A are both depicted by the center Z0 ∼= S of A0. (2) The aﬃne scheme Spec R and the reduced scheme of Spec Z are birational to the noetherian scheme Spec S, and each contain precisely one closed point of positive geometric dimension, namely m0 and z0. Proof. (1) We ﬁrst claim that Zˆ and R are depicted by S. By Theorem 3.18, ∗ ∗ US/Zˆ = US/Zˆ 6= ∅ and US/R = US/R 6= ∅. 6 Furthermore, by Lemma 3.5, the morphisms ιS/Zˆ and ιS/R are surjective. (2.i) As schemes, Spec S is isomorphic to Spec Zˆ and Spec R on the open dense subset US/Zˆ = US/R, by Theorem 3.18. Thus all three schemes are birationally equivalent. Furthermore, Spec Zˆ and Spec R each contain precisely one closed point ˆ where Z and R are locally nonnoetherian, namely z0 and m0, again by Theorem 3.18. (2.ii.a) We claim that the closed point m0 ∈ Spec R has positive geometric dimension. Indeed, since A is nonnoetherian, there is a cycle p such that σ - p and pn ∈ S \ R for each n ≥ 1, by Lemma 3.1.6. If p is in m0S, then there are monomials g ∈ R, h ∈ S, such that gh = p. Furthermore, g 6= σn for all n ≥ 1 since σ - p. But then p = gh is in R by Lemma 3.1.1, a contradiction. Therefore

p 6∈ m0S.

6The fact that S is a depiction of R also follows from [B5, Theorem D.1], since the algebra homomorphism τψ :Λ → M|Q0|(B) is an impression [B2, Theorem 5.9.1]. ON THE CENTRAL GEOMETRY OF NONNOETHERIAN DIMER ALGEBRAS 21

Consequently, for each c ∈ k, there is a maximal ideal nc ∈ Max S such that

(p − c, m0)S ⊆ nc. Thus, m0 ⊆ (p − c, m0)S ∩ R ⊆ nc ∩ R.

Whence nc ∩ R = m0 since m0 is maximal. Therefore by Theorem 3.18, c nc ∈ US/R. Set \ (15) q := nc. c∈k The intersection of radical ideals is radical, and so q is a radical ideal. Thus, since S is noetherian, the Lasker-Noether theorem implies that there are minimal primes q1,..., q` ∈ Spec S over q such that

(16) q = q1 ∩ · · · ∩ q`.

We claim that at least one qi is non-maximal in S. Setting Z(I) := ZS(I), we have

` ` (i) (ii) ∪i=1Z(qi) = Z(∩i=1qi) = Z(q) = Z(∩c∈knc) = ∪c∈kZ(nc), where (i) holds by (16), and (ii) holds by (15). Thus, if each qi were maximal, then

{Z(q1),..., Z(q`)} = {Z(nc): c ∈ k}. In particular, there there would be an inﬁnite set of distinct zero-dimensional points that is equal to a ﬁnite set of zero-dimensional points, which is not possible. Therefore at least one qi is a non-maximal prime, say q1. Whence, \ \ m0 = (nc ∩ R) = nc ∩ R = q ∩ R ⊆ q1 ∩ R. c∈k c∈k

Consequently, q1 ∩ R = m0 since m0 is maximal in R. Since q1 is a non-maximal prime ideal of S,

ht(q1) < dim S. Furthermore, S is a depiction of R by Claim (1). Thus (i) ght(m0) ≤ ht(q1) < dim S = dim R, where (i) holds by Theorem 3.19. Therefore

gdim m0 = dim R − ght(m0) ≥ 1, proving our claim. ˆ (2.ii.b) Finally, we claim that the closed point z0 ∈ Spec Z has positive geometric dimension. Again, since A is nonnoetherian, there is a cycle p such that σ - p and p ∈ S \ R, by Lemma 3.1.6. 22 CHARLIE BEIL ˆ If p is in z0S, then there are monomials g ∈ Z, h ∈ S, such that gh = p. Further- more, g ∈ R since Zˆ ⊆ R by (3); and g 6= σn for all n ≥ 1 since σ - p. But then p = gh is in R by Lemma 3.1.1, a contradiction. Therefore

p 6∈ z0S. The proof then follows as in Claim (2.ii.a). Remark 3.21. Although Zˆ and R determine the same variety using depictions, their associated aﬃne schemes ˆ (Spec Z, OZˆ) and (Spec R, OR) will not be isomorphic if their rings of global sections, Zˆ and R, are not isomorphic.

Acknowledgments. The author would like to thank an anonymous referee for their helpful comments which have improved the article. The author was supported by the Austrian Science Fund (FWF) grant P 30549-N26. Part of this article is based on work supported by the Heilbronn Institute for Mathematical Research.

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Institut fur¨ Mathematik und Wissenschaftliches Rechnen, Universitat¨ Graz, Hein- richstrasse 36, 8010 Graz, Austria. Email address: [email protected]