T. Hibi Binomial ideals arising from combinatorics

T. Hibi Binomial ideals arising from combinatorics lecture notes written by Filip Rupniewski (email: frupniewski@impan.pl) Binomial Ideals conference 3-9 September 207, Łukęcin, Poland Lecture Let S = k[x,..., x n ] be the polynomial ring in n variables over a field k and let be the set of monomials of S. a) Dickson s Lemma Mon(S) = {x a = x a...xa n n : a = (a,..., a n ) Z n 0} Let x a = x a...xan n and x b = x b...xbn n. We say that x a divides x b if a i b i for all i n. Let M Mon(S). We say that x a M is minimal if for x b M such that x b x a we have b = a. Let M min be the set of minimal monomials in M. Theorem. (Dickson s Lemma). Let = M Mon(S). Then M min is a finite set. Proof. We use induction on n. For n = the proof is easy. Let n 2 and let y = x n. Write S = k[x,..., x n, y] and B = k[x,..., x n ]. Set N = {x a Mon(B) x a y b M for some b 0}. By induction we have that N min is a finite set. Let N min = {u,..., u s } and u y b,..., u s y bs M. Let b = max{b,..., b s }. For each 0 c < b define Again, we know N min c is a finite set, say N min c N c = {x a N : x a y c M} N = {u (c),..., u(c) s c }. Consider the following monomials: u y b, u 2 y b 2,... u s y b s u (0),... u(0) s 0 u (b ) y b,... u (b ) s b It then follows easily that every monomial in M is divisible by one of the monomials on the above list b) Monomial order A monomial order on S is a total order < on Mon(S) such that i) < u for u Mon(S), ii) if u, v Mon(S) and u < v, then uw < vw for all w Mon(S). Example.2. Let x a = x a...xan n and x b = x b...xb n n. a) (Lexicographic order) We say that x a < lex x b if either n i= a i < n i= b i or n i= a i = n i= b i and the leftmost non-zero component of the vector a b is negative. We call < lex the lex order on S induced by x >... > x n.

b) (Reverse lexicographic order) We say that x a < rev x b if either n i= a i < n i= b i or n i= a i = n i= b i and the rightmost non-zero component of the vector a b is positive. We call < rev the reverse lex order on S induced by x >... > x n. c) (Purely lexicographic order) We say that x a < purelex x b if the leftmost non-zero component of the vector a b is negative. The reverse purely lexicographic order is not a monomial order since > x. We have: x 2 x 3 < lex x x 4 Lemma.3. If u v and u v then u < v. x x 4 < rev x 2 x 3 x 5 2 < purelex x 3 Proof. We have v = uw for some w Mon(S). From the first property of the monomial order we have that < w. Hence, from the second property of the monomial order we have u < uw = v. Lemma.4. There exists no infinite descending sequence of monomials of the form...u 2 < u < u 0. Proof. Suppose that there exists such a sequence M. Let M min = {u i0, u i,..., u is } with i 0 < i <... < i s. Then we have u ij u is + for some 0 j s. Hence from the previous lemma we get u ij < u is +. Thus i j > i s +, which is a contradiction. c) Gröbner bases Fix a monomial order < on S. Given a polynomial 0 f = u Mon(S) c uu (c u k). We define the support of f to be supp(f) = {u Mon(S) c u 0}. Define also the initial monomial of f to be in < f = the biggest monomial w.r.t < belonging to supp(f). Given an ideal 0 I S we define the initial ideal of I: in < (I) = ({in < f : 0 f I}). Lemma.5. There exists polynomials g,..., g s I s.t in < (I) = (in < g,..., in < g s ). Proof. From. we have {in < f : 0 f I} = {in < g,..., in < g s } for some polynomials g,.., g s. It follows that in < (I) = (in < g,..., in < g s ). Let 0 I S be an ideal. A Gröbner basis of I w.r.t. the monomial order < is a finite set G = {g,..., g s } of polynomials where each 0 g i I, such that in < (I) = (in < g,..., in < g s ). A Gröbner basis always exists but cannot be unique. d) Hilbert s basis theorem Fix a monomial order < on S. Theorem.6. If G = {g,..., g s } is a Gröbner basis of an ideal 0 I S, then I is generated by g,..., g s. In other words, every Gröbner basis of I is a system of generators of I. Proof by Gordan. Let 0 f I. Since in < f in < I one has in < g i0 in < f for some i 0 s. Let in < f = w 0 in < g i0 for w 0 Mon(S). Set h 0 = f c i o c 0 w 0 g i0 I where LT (f) = c 0 in < f and c i0 in < g i0 = LT (g i0 ). If h 0 = 0, then f (g,..., g s ). If h 0 0, then in < h 0 < in < f. Continue this procedure and use Lemma.4 to finish the proof. Corrolary.7 (Hilbert s basis theorem). Every ideal of the polynomial ring is finitely generated. 2

e) Macaulay theorem Notation.8. S = K[x,..., x n ], 0 I S ideal, < monomial order Definition.9. A monomial u Mon(S) is called standard with respect to in < (I) if u in < (I) Theorem.0 (.8 Macaulay). The set of standard monomials with respect to in < (I) is a K-basis of S/I. Proof. Let B = {ū = u + I S/I : u Mon(S) is standard with respect to in < (I)} We show that B is a K - basis of S/I. B is linearly independent : let c ū +...c n ū n = 0 in S/I where c i K and u < u 2 <...u n are standard. Then 0 f = c u +...c n u n I and in < (f) = u n in < (I). This is impossible since u n is standard S/I is spanned by B: Let B denote the subspace of S/I spanned by B. Let 0 f S. We show f B by using induction (lemma.4) on in < (f): Suppose ū = in < (f) B. By assumption of induction we know f cu B (coefficient of min f). Since u B, one has f B Suppose ū = in < (f) B. Then u is not standard, i.e. u in < (I). Hence 0 g I u = in < (g). Then (by induction) c f cg B. However in S/I c f = c f cg B. Thus c f B and f B. Corrolary. (.9). 0 I S ideal, < monoid order, h,..., h s I with each h i 0. Let H = {u Mon(S) : i s in < (h i ) u} Suppose H is linearly independent over K in S/I. Then {h,..., h s } is a GB of I w.r.t. <. In particular {h,...h s } is a system of generators of I Example.2 (.0). Consider the semigroup ring A = K[t, xt, yt, xyt, yzt, xyzt] k[x, y, z, t]. Define the surjective ring homomorphism u : S = k[x, x 2,..., x 6 ] A by setting u : x t, x 2 xt,..., x 6 xyzt, I = ker(u). We know T = x 2 x 3 x x 4, T 2 = x 2 x 5 x x 6, T 3 = x 4 x 5 x 3 x 6 I = (T, T 2, T 3 ) (this equality is not obvious). By using (.9) we can show that {T, T 2, T 3 } a GB w.r.t. rev. lex. order induced by x > x 2 >... (Problem ) Problem.3. In (.0) show that {T, T 2, T 3 } is a GB, w.r.t < rev H = {x a xa2 2 xa4 4 xa6 6, xb xb3 3 x b4 4 xb6 6, xc xc3 3 xc5 u, v H, u v π(u) π(v) 5 xc6 6 } is linearly independent if: Problem.4. (chsugi) S = K[x,..., x 0 ], I = (T,..., T 5 ), where T = 8 26, T 2 = 29 37, T 3 = 30 48, T 4 = 46 59, T 5 = 57 0. Show that monomial order < on S for which {T,..., T 5 } is a GB of I w.r.t. < Suppose, on the contrary, that there exists a monomial order < on S such that G = f,..., f 5 is a Grobner basis of I with respect to <. First, note that each of the five polynomials: x x 8 x 9 x 3 x 6 x 7, x 2 x 9 x 0 x 4 x 7 x 8, x 2 x 6 x 0 x 5 x 7 x 8, x 3 x 6 x 0 x 5 x 8 x 9, x x 9 x 0 x 4 x 6 x 7 belongs to I. Let, say, xx8x9 > x3x6x7. Since xx8x9 in < (I), there is g G such that in < (g) divides xx8x9. Such g G must be f. Hence xx8 > x2x6. Thus x2x6 in < (I). Hence there exists no g G such that in < (g) divides x2x6x0. Hence x2x6x0 < x5x7x8. Thus x5x7 > xx0. Continuing these arguments yields xx8x9 > x3x6x7, x2x9x0 > x4x7x8, x2x6x0 < x5x7x8, x3x6x0 > x5x8x9, xx9x0 < x4x6x7 and xx8 > x2x6, x2x9 > x3x7, x3x0 > x4x8, x4x6 > x5x9, x5x7 > xx0. Hence (xx8)(x2x9)(x3x0)(x4x6)(x5x7) > (x2x6)(x3x7)(x4x8)(x5x9)(xx0). However, both sides of the above inequality coincide with xx2 x0. This is a contradiction. 3

2 Toric rings and toric ideals a) Configuration matrix A = (a ij ) i d, j n Z d n, column vector a j = [a,j,..., a d,j ] T, j n Definition 2.. We call A a configuration matrix if 0 c R d for which j n a j c = usual inner product in R d ca = 0 Example 2.2. Given A Z (d ) n, define A # Z d n = A matrix with A on top, ones below. Then A # is a configuration matrix with c = [0,..., 0, ] Example 2.3. If a j +...a dj = h 0 j n then A is a configuration matrix with c = [/h,.../h] b) toric ideal Definition 2.4. A binomial is a polynomial of the form u v where u and v are monomials with deg u = deg v A binomial ideal of S = k[z,..., z n ] is an ideal of S generated by binomials Given a configuration matrix A Z d n define Ker Z A = {b Z n : Ab = 0} Lemma 2.5 (2.2). If b = [b,..., b n ] Ker Z A, then b +..., b n = 0. Proof. Since A is a configuration matrix, one has 0 c R d with j n a j c = 0. Since Ab = 0, one has n j= b ja j = 0. Hence 0 = ( n j=0 b ja j )c = n j=0 b j(a j c) = b j. Definition 2.6. Now, for each b = [b,..., b n ] ker Z A, define the binomial f b = b n >0 xbi i b i <0 xbi i = f + b f b S. By (2.2) one has deg f + b = deg f b. Let us define I A := ({f b : b ker Z A}) 0 0 Example 2.7. A = 0 0 0 0 Z4 5 configuration matrix. One has A [,,,, 2] T =: Ab = 0, b ker Z A, f b = x 2 x 3 x 4 x x 2 5. One can show that I A = (f b ) c) Toric ring Definition 2.8. A = (a ij ) Z d n is a configuration matrix if 0 c R d such that for all horizontal vector a j, a j c = ca = [,..., ] t a j := t aj t a2j 2...t a dj d K[t, t,..., t d, t d ] Definition 2.9. The toric ring of A is the subring K[A] k[t, t...t d, t d ] generated by ta, t a 2,..., t a n K[A] = K[t a, t a 2,..., t a n ]( k[t, t,..., t d, t d ]) Example 2.0. A= 0 0 K[A] = K[t, t 3, t 2 t 3, t t 2 t 3 ] Now define the surjective ring homomorphism π : S = k[x,..., x n ] k[a] = k[t a,..., t a n ] x i t a i k[t, t,..., t d, t d ] Theorem 2.. I A = ker(π). Corrolary 2.2. I A is a prime ideal. 4

Proof of Theorem 2.. (First Step) We will show that, for u, v Mon(S), if π(u) = π(v), then deg u = deg v. Let u = n j= xc j j, v = n j= xd j j. Then π(u) = n j= (ta j ) c j, π(v) = n j= (ta j ) d j In other words, π(u) = t n j= cjaj, π(v) = t n j= d ja j. If π(u) = π(v), then n j= c ja j = n j= d ja j. Thus ( n j= c ja j ) c = ( n j= d ja j ) c = n j= d j = n j= c j. Hence deg u = deg v. (Second step) We will show that ker(π) is a binomial ideal. Write f S = k[x,..., x n ] as f = f +... + f t where each f i S and for monomials u supp(f i ) and v supp(f j ), one has π(u) = π(v) if and only if i = j. Let f i = s i k= c iku ik where 0 c ik k, u ik Mon(S). Since π(u i ) = π(u ik ) for k = 2,..., s i it follows that π(f i ) = s i k= c ikπ(u ik ) = ( s i k= c ik)π(u i ). Hence π(f) = π(f )+...+π(f t ) = t i= ( s i k= c ik)π(u i ). If i j, then π(u i ) π(u j ). Thus, if f ker(π) then s i k= c ik = 0 for all i t. Hence c i = s i k=2 c ik. We have f i = s i k= c iku ik = s i k=2 c ik(u ik u i ). Thus f = t i= ( s i k=2 c ik(u ik u i )). We have u ik u i ker(π). Therefore, first step shows that u ik u i is a binomial. Hence ker(π) is generated by those binomials u v with π(u) = π(v). (Third step) We will show that I A = ker(π). Let f = n j= xc j j n j= xd j j be a binomial. Then π(f) = n j= (ta j ) c j n j= (ta j ) d j = t a j c j t a j d j. Hence π(f) = 0 if and only if n j= c ja j = n j= d ja j if and only if f b = f, b = c d Ker Z A. Thus binomials belonging to ker(π) must belong to I A. The converse is clear. Hence I A = ker(π). d) Toric ideals arising from finite graphs Let G be a finite connected simple graph on the vertex set [d] = {, 2,..., d} with the set of edges E(G) = {e,..., e n }. For each edge e i connecting vertices p i and q i [d] define t ei = t pi t qi k[t,..., t d ]. The toric ring (or edge ring)of G is k[g] = k[t e,..., t e n ]. Define π : S = k[x,..., x n ] k[g], by x i t ei. We call ker(π) the toric ideal of G and we denote it by I G. Graph terminology: (even, odd) cycle, chord, (closed) walk. Problem 2.3. Find a configuration matrix A with I A = I G The matrix M Z n n where n is a number of vertices. In columns there are 0 and, every column corresponds to a one edge, ones are in vertices of the edge. Proof follows from the definition of surjections in I G and I A. Problem 2.4. If Γ is an even closed walk, then show that f Γ I G going through the graph and taking even edges we will take edges with all vertices possible. the same with even edges. Closed walk was even so there is the same number of even and odd edges. When we multiplicate we obtain the same monomials. Hence f γ I G. Problem 2.5. Show that I G is generated by those binomials f Γ, where Γ is an even closed walk Let I G denote the binomial ideal generated by these binomials f Γ, where Γ is an even closed walk of G. Choose a binomial f = q k= x i k q k= x j k I G. We prove f I G by induction on q = degf. One can assume that i k j k for all k and k. Let say, π(x il ) = t t 2. Since π( q k= x i k ) = π( q k= x j k ) one has π(x jm ) = t 2 t r for some m with r. Say m =, r = 3. Thus π(x j ) = t 2 t 3. Then π(x il ) = t 3 t s for some l with s 2. Say l = 2, s = 4. Repeated application of these procedure yields an even closed walk Γ = (e i, e j,..., e il, e jl ) with f Γ = p k= x i k p k= x j k I g. This one has π( q k=p+ x i k ) = π( q k=p+ x j k ). Hence g = q k=p+ x i k q k=p+ x j k I G. By induction one has g I G. Now one has: f = ( q k=f+ x i k )f Γ + ( p k= x j k )g. f Γ, g I G. Thus f I G as desired 5

Problem 2.6. We say that an even closed walk Γ is primitive if there is no even closed walk Γ with Γ Γ such that f + Γ f + Γ and f Γ f Γ. Show that I G is generated by those binomials f Γ, where Γ is a primitive even closed walk. Problem 2.7. Find a minimal system of binomial generators of I G for the following graphs G =<><>, G 2 = hexagon with the diameter. for <><>: we take binomials representing <> and <> - it s enough for hexagon: we don t need to take the whole hexagon, we only need both halfs as a cycles. x x 3 x 2 x 7, x 5 x 7 x 4 x 6 Problem 2.8. Let G be a finite connected simple bipartite graph. Show that I G is generated by those binomials f C, where C is an even cycle without chord (cięciwa) 3 Regular triangulation of lattice polytopes a) Triangulation of lattice polytopes (integral polytopes Definition 3.. A convex polytope is a convex hull of a finite set A convex polytope P R d of dimension d is called a lattice (or integral) polytope if each vertex Z d. Let P R d be a lattice polytope of dim = d and P Z d = {a i, a 2,..., a n } [ ] Write A(P ) Z (d+) n a a for the configuration matrix A(P ) = 2... a n Z... (d+) n since dimp = d one have rank A(P ) = d + 0 0 Example 3.2. for two tetrahedrons with common base 0 0 0 0 Definition[ 3.3. ] [ A ] simplex [ ] belonging to P of a dimension s is a subset F = {a i,..., a is } P Z d ai ai2 ais for which... are linearly independent over Q In particular is a simplex belonging to P of dimension A maximal simplex = {a i,..., a id+ } belonging to P is a simplex of dimension d. A maximal simplex is called fundamental if [ ] [ ] [ ] [ ] [ ] ai ai2 aid+ a a2 ZA(P ) = ZA(F ), where ZA(F ) := Z +Z +...+Z and ZA(P ) := Z +Z + [ ] an... + Z Z d+ Definition 3.4. A collection of simplices belonging to P is called a triangulation of P if the following conditions are satisfied::. If F and F F, then F 2. If F, G, then conv(f ) conv(g) = conv(f G) 3. P = F conv(f ) (convex hull of F in Rd ) Definition 3.5. A triangulation of P is called unimodular if every maximal simplex F is fundamental. 6

b) Regular triangulations [ ] P R d lattice polytope of dimension d, P Z d a a = {a,...,, a n }, A(P ) = 2... a n Z... (d+) n K[A(P )] = K[t a s,..., t an s] K[t, t,..., t d, t d, s] toric ring, π : S = K[x,..., x n ] K[A(P )] π(x i ) = t a i s, I A(P ) = ker(π) toric ideal Fix monomial order on S and let in < (I A(P ) ) denote initial ideal Recall that the radical of in < (I A(P ) ) is the ideal of S generated by those polynomials f S with f N in < (I A(P ) ) for some N = N f > 0 Example 3.6. if in < (I A(P ) ) = (x 3 x 2 x 5 3x 4, x 3 2x 5 x 2 6), then in < (I A(P ) ) = (x x 2 x 3 x 4, x 2 x 5 x 6 ) generated by square free monomials Lemma 3.7 (3.). A subset F P Z d is a simplex belonging to P if a j F x j in < (I A(P ) ) [ ] [ ] [ ] ai ai2 aid+ Sketch of proof. Let F = {a i,..., a is }. Suppose that F satisfies the inclusion. We show that... [ ] [ ] are linear independent. If not, then (0,..., 0) (a i,..., a is ) Z s ai ai2 such that: a i + a i2... + [ ] ais a is = 0. Then one can easily show that 0 q k >0 xq n jk q n <0 x q n j k =: u v I A(P ) Thus u or v in < (I A(P ) ). Hence q k >0 x j k in < (I A(P ) ) or q k <0 x j k in < (I A(P ) ). This contradicts a j F x j in < (I A(P ) ). Definition 3.8. Let (in < (I A(P ) )) := {F P Z d : a j F x j in < (I A(P ) )} Theorem 3.9 (3.2 Strumfels). (in < (I A(P ) )) is a triangulation of P. We omit the proof Example 3.0. in the example of tetrahedrons with common base: I A(P ) = (x 2 x 3 x 4 x x 2 5), in < (I A(P ) ) = (x x 2 5), (in < (I A(P ) )) = (x x 5 ) Definition 3.. A triangulation of P is called regular if = (in < (I A(P ) )) for some monomial order < Theorem 3.2. (in < (I A(P ) )) is unimodular in < (I A(P ) ) = in < (I A(P ) ) ( in < (I A(P ) ) is generated by square free monoids) Commutative Algebra unimodular triangulation toric ring K[A(P )] is normal and Cohen - Macaulay. 4 The join-meet ideals of finite lattice a) Review on classical lattice theory Definition 4.. A lattice is a poset L in which any two elements a and b of L has a meet a b and a join a b. In particular, a finite lattice has both the minimal element ˆ0 and the maximal element ˆ A finite lattice L is called distributive if a (b c) = (a b) (a c) and a (b c) = (a b) (a c) A finite L is called modular if a c a (b c) = (a b) c Every distributive lattice L is modular. In fact if L is distributive lattice and a c, then a (b c) = (a b) (a c) = (a b) c Problem 4.2. Let G be a finite group and L(G) the set of normal subgroups of G. We can regard L(G) as a poset ordered by inclusion. Show that:. L(G) is a lattice 2. L(G) is a modular lattice 7

3. For G a finite abelian group: L(G) is a distributive lattice G is a cyclic group [for.c) ] If G Z/nZ. Subgroups are Z/kZ for k n Z/k Z Z/k 2 Z = Z/lcm(k, k 2 )Z, Z/k Z + Z/k 2 Z = Z/gcd(k, k 2 )Z. Enough to check gcd, lcm satisfies distributive laws. G not cycle F = Z/n Z Z/n 2 Z... such that n n2... Let s take 3 subgroups isomorphic to Z/n Z H := (, 0, 0,...), H := (0, n2 n, 0,...), H := (, n2 n, 0,...) so we obtain a sublattice of diamond type, so it s not distributive. Definition 4.3. N 5 pentagon lattice M 5 the diamond lattice (quadrangle with a diagonal and a vertex on a diagonal) Fact 4.4. N 5 is not modular (in fact, even though a < c, one has a (b c) = a 0 = a, (a b) c = c = c) M 5 modular, but not distributive (In fact a (b c) = a = a, (a ) (a c) = 0 0 = 0) Theorem 4.5 ((4..) Dedekind).. a finite lattice L is modular no sublattice of L is N 5 2. a modular lattice L is distributive no sublattice of L is M 5 3. a finite lattice L is distributive neither N 5 nor M 5 is a sublattice of L. Definition 4.6. Let P = {p,.., p n } be a finite poset with a partial order <. A poset ideal of P is a subset α P such that if p i α, p j p i,then p j α. Let J(P ) denote the set of poset ideals of P. Fact 4.7. If α and β are poset ideals, α β and α β are also poset ideals of P. Hence J(P ) can be a finite lattice, ordered by inclusion. It is distributive. Theorem 4.8 (4.2 Birkhoff). Give a finite distributive lattice L, there is a unique poset P such that L = J(P ). Definition 4.9. Join irreducible element of a lattice is an element which has only one arrow going down. b) Join-meet ideals of finite lattices Let L be a finite lattice and K[L] := K[{x a : a L}] the polynomial ring in L - variables over a field K. Given a, b L, define the binomial f a,b by setting f a,b = x a x b x a b x a b In particular f a,b = 0 a and b are comparable (either a b or b a) Definition 4.0. The join-meet ideal of L is finite binomial ideal I L := ({f a,b : a and b are incomparable}) Example 4.. I N5 = (x a x b x 0 x, x b x c x 0 x ) I M5 = (x a x b x 0 x, x a x c x 0 x, x b x c x 0 x, ) Definition 4.2. A monomial order < on K[L] is called compatible if for any a and b of L for which a and b are incomparable, one has in < (f a,b ) = x a x b Example 4.3. L = {x, x 2,..., x n }, x i < x j in L i > j. Then < rev induced by x >... > x n. Then < rev is a compatible monomial order on K[L]. We called it rank reversed lexicographic order. Theorem 4.4 ((4.4)). Let L be a finite lattice and fix a compatible monomial order < on K[L]. Let G L := {f a,b : a, b L are incomparable} Then the following are equivalent:. G L is a Grobner basis with respect to < 2. L is distributive Theorem 4.5 (4.5). Give a finite lattice L. The following conditions are equivalent:. I L is a prime ideal 2. L is distributive In both upper theorems implication from top to bottom is easy - exercise 8

c) Toric ring R K [L] with L = J(P ) distributive lattice Let P = {p, p 2,..., p n } be a finite poset and L = J(P ) the distributive lattice consisting of all poset ideals of P, ordered by the inclusion. Let S = K[x, x 2,..., x n, t] denote the polynomial ring in (n + ) variables over a field K. Give a poset ideal α L = J(P ). We introduce the monomial u α by setting u α = ( p i α x i)t S. In particular u = t, u P = x x n t. Let R K [L] denote the toric ring R K [L] := K[{u α : α L = J(P )}]. Example 4.6. rysunki Define the surjective ring homomorphism π : K[L] = K[{x α : α L = J(P )}] R k [L] by setting π(x α ) = u α for all α L = J(P ). Lemma 4.7 (4.6). I L ker(π) Proof. α, β L = J(P ), α β = α β, α β = α β, π(x α β x α β ) = ( p i α β x i)( p i α β x i)t 2, π(x α x β ) = u α u β = ( p i α x i)( p i β x i)t 2 Hence u α u β = u α β u α β in R K [L]. Thus x α x β x α β x α β ker(π) Theorem 4.8 (4.7). Let L = J(L) and fix compatible monomial order < on K[L]. Then G L := {f α,β : α, β L = J(P ) are incomparable} is a Grobner basis of ker(π) with respect to <. In particular I L = ker(π), so 2 in Theorems 4.4 and 4.5. Proof. the technique of corollary of Macaulay s theorem in paragraph. can be applied. Let In < (G L ) := {in < (f α,β ) : f α,β G L }. In other words in < (G L ) is the set of monomials x α x β K[L] for which α and β are incomparable. Lemma (4.6) says that in < (G L ) in < (ker π). Let B denote the set of those monomials w K[L] such that xα x β in < (G L ) x α x β w and B those monomials w K[L] with ω in < (ker π). Recall that, Macaulay s theorem B is a K basis of R K [L] = K[L]/ ker π. Since B B, in order to show that B = B, our work is to show that B is linearly independent in R K [L] = K[L]/ ker π. Now, we prove, for w, w B with w w one has π(w) π(w ): Let w = x α x α2 x αp, w = x β x β2 x βq and α α 2... α p, β β 2... β q. π(w) = (monomials in x i )t p, π(w ) = (monomials in x i )t q. We may assume that p = q. Induction on deg w (= deg w ) one can assume that i,j α i β j. Thus α β. Take p i0 α \ β. As subsets of P one has α α 2... α p, β β 2... β p. Since i p p i0 α i, π(x i0 ) p appears in π(w) = π(x α )π(x α2 ) π(x αp ). However, since p i0 β, the power r of x i0 for which π(x i0 ) r appears in π(w ) is at most p α = β. Contradiction. Problem 4.9.. Find a configuration matrix A with I L = I A where lattice L = tree squares connected to look like a sign >. 2. Find a finite poset P with L = J(P ) where L = two cubes with common edge. ) x t, x x t, x,2 x x 2 t, x,3 x x 2 t, x 2,3 x 2 x 3 t, x,2,3 x x 2 x 3 t, x,2,4 x x 2 x 4 t, x,2,3,4 0 0 0 0 0 x x 2 x 3 x 4 t so the matrix should be : 0 0 0 0 0 0 0 0 0 lattice of ideals are always distributive, so from theorem (4.7) I L = ker π Problem 4.20. By using Dedekind theorem, prove 2 of theorem (4.4) and 2 of theorem (4.5) 9

5 Order polytopes of finite posets a) Order polytopes P = {p, p 2,..., p n } finite poset, e = [, 0,...0] T, e 2 = [0,, 0,..., 0] T,..., e n = [0,..., 0, ] T R n. α J(P ), P (α) := p i α e i R n. In particular P ( ) = [0,..., 0] T R n, P (P ) = [,..., ] T R n Definition 5.. The order polytope of P is the convex polytope O(P ) R n which is the convex hull of {P (α) : α J(P )} R n Example 5.2. rysunek b) Linear extensions Definition 5.3. A permutation i i 2...i n of [n] = {,..., n} is called a linear extension of poset P if p ik < p il in poset P, then k < l Definition 5.4. e(p ) := the number of linear extension of P Example 5.5. rysunek Lemma 5.6 (5.2).. Suppose that i, i 2,..., i n is a linear extension of P. Then α j = {p i, p i2,..., p ij } P is a poset ideal for all j n. Moreover, = α 0 < α < α 2 <... < α n = P is a maximal chain of L = J(P ) 2. If = α 0 < α < α 2 <... < α n = P is a maximal chain of L = J(P ) then i, i 2,..., i n is a linear extension of P. where p ij α j α j Proof.. If p ik < p il α j, then k < l j. Hence p ik α j 2. Let p ik < p il. Since α l = {p i, p i2,..., p il } and α l is a poset ideal of P, one has p ik α l. Hence k < l. Example 5.7. rysunek Corrolary 5.8 (5.3). {linear extensions of P } : {maximal chains of L = J(P )}. In particular, e(p ) is equal to the number of maximal chains of L = J(P ). Book: R.Stanley, Enumerative combinatorics, vol, chapter 3. Let i, i 2,..., i n be a linear extension of P and α j {p i, p i2,..., p ij } L = J(P ). Since P (α j ) = e i + e i2 +...e ij O(P ) and convex hull conv({p ( ), P (α ),..., P (α n )}) O(P ) is a standard lattice simplex in R n. Standard means volume = n! Example 5.9. rysunek Proposition 5.0 (5.4). dim O(P ) = n Proposition 5. (5.5). The set of vertices of O(P ) is V (O(P )) = {P (α) : α J(P )}. In particular O(P ) is a lattice polytope. Lemma 5.2 (5.6). O(P ) Z n = V (O(P )) extension 0

c) Toric rings of order polytopes Recall that, in general, given a lattice polytope P R n of [ dimension n, the ] toric ring of P is a a the toric ring K[A(P )] of the configuration matrix A(P ) = 2... a n Z... (n+) N where P Z n = {a, a 2,.., a N }. In other words, K[A(P )] = K[x a t,..., x a N t] K[x, x,..., x n, x n ], x a j = x aj x a2j 2 x anj n Now we discuss the toric ring of O(P ). O(P ) Z n = V (O(P )) = {P (α) : α L = J(P )}. One has x P (α) p = x i α ei = p i α x i. Hence toric ring of O(P ), K[{x P (α) t : α J(P )}] = K[{( p i α x it : α J(P )}] K[x,..., x n, t] Recalling section 4: L = J(P ), R K [L] = K[{u α t : α J(P )}] where u α = p i α x i. Example 5.3. rysunek d) Regular triangulation of O(P) π : K[{x P (α) : α J(P )] K[A(O(P ))] toric ring x P (α) x P (α) t = ( p x i α i)t. I A(O(P )) = ker π, toric ideal of O(P ). Since K[A(O(P ))] = R K [L] with L = J(P ) it follows that I A(O(P )) = ({x P (α) x P (β) x P (α β) x P (α β) : α, β L = J(P ) are incomparable in L } =: G L ) Fix a compatible monomial order < on K[{x P (α) : α J(P )}] Then we notice G L is a Grobner base of I A(O(P )) with respect to <. In < (I A(O(P )) ) = ({x P (α) x P (β) : α, β L = J(P ) are compatible }. Now we discuss the regular (unimodular) triangulation = (In < (I A(O(P )) )). F V (O(P )) = O(P ) Z n belongs to (see section 3) (definition) P (δ) F x P (δ) in < (I A (O(P ))) = in < (I A(O(P )) ) x P (α) x P (β) x P (δ) F for all α, β L = J(P ) which are incomparable in L if F = {P (α i ), P (α i2 ),..., P (α is )} then α i < α i2 <..., α is in L = J(P ). Thus in particular Proposition 5.4 (5.8). F = {P (α 0 ), P (α ),.., P (α n )} is a maximal simplex of = α 0 < α,..., α n = P is a maximal chain of L = J(P ). Furthermore conv(f ) is a standard lattice simplex in R n. Theorem 5.5 (5.9). The volume of O(P ) is e(p ) n!, where e(p ) is the number of linear extensions of P. Proof. Since is a triangulation of O(P ), one has volume of O(P ) = F maximal (volume of conv(f )) = ( the number of maximal simplex of )/n! = (the number of maximal chains of L)/n! = e(p ) n! (because of corollary 5.3). Problem 5.6. Let V R n be a set of (0, ) vectors of R n and P R n the convex polytope which is the convex hull of V in R n. Show that:. The set of vertices of P coincides with V ( (5.5) ) 2. P Z n = V ( (5.6) ) Proof. Let w be a (0, ) vector with w V. Then w conv(v ), V = {v,..., v n }. If w conv(v ), w = λ v +... + λ s v s, λ + λ 2 +...λ s =, λ i 0... Definition 5.7. v P vertex (If v = u+w 2, u, w P v = u = w ) Lemma 5.8. If P = conv(v ), then each vertex belongs to V. Moreover, if V (P ) is the set of vertices of P, then P = conv(v (P )).