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Plactic algebras
Journal of Algebra 274 (2004) 97–117 www.elsevier.com/locate/jalgebra
Plactic algebras Ferran Cedó a and Jan Okni´nski b,∗ a Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra (Barcelona), Spain b Institute of Mathematics, Warsaw University, 02-097 Warsaw, Poland
Received 21 March 2002 Communicated by Michel van den Bergh
Abstract The structure of the plactic algebra K[M] of rank 2 or 3 is studied. It is shown that these algebras are semiprimitive. Moreover, the prime spectrum is completely described in the rank 2 case. Also it is proved that the plactic algebra of rank n 4 is not semiprime and the plactic algebra of rank 3 is not prime. 2004 Elsevier Inc. All rights reserved.
The construction of the plactic monoid originated from the papers of Schensted [14] and Knuth [6] on the ground of certain combinatorial problems and operations on Young tableaux. It was systematically studied by Lascoux and Schützenberger in [10] and became a very important tool in the theory of symmetric functions, Young tableaux and various aspects of representation theory [4,9]. The importance of these monoids was also indicated by Schützenberger in [15]. Plactic monoids appeared recently in the theory of modular representations of the symmetric group and in quantum groups, via the representation theory of quantum enveloping algebras [8]. The plactic monoid of rank 2 has also been studied from the point of view of language theory [1]. Our aim in this paper is to study basic ring theoretic properties of the plactic algebra K[M] over a field K. For the general background on semigroup algebras we refer to [12].
* Corresponding author.
E-mail address: [email protected] (J. Okni´nski). 0021-8693/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2003.12.004
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1. Preliminaries In this section, we recall the definition and some of the basic properties of the plactic monoids. These results will be often used in the subsequent sections. Definition 1. The plactic monoid of rank n is the monoid presented with generators: a1 , a2 , . . . , an , subject to the relations: ak ai aj = ai ak aj
if i j < k,
as at ar = as ar at
if r < s t.
We will denote by Mn the plactic monoid of rank n. Definition 2. Since the defining relations of Mn are homogeneous with respect to the degree in every ai , this degree is inherited by Mn . So we denote by di (t) the degree of t ∈ Mn with respect to ai . By deg(t) we mean the total degree of t ∈ Mn . By a row in Mn we mean an element which is a word ai1 · · · air such that r 1 and i1 i2 · · · ir . A column is defined as a word with strictly decreasing indices. We say that a row u = ai1 · · · air dominates a row v = ak1 · · · aks if r s and ij > kj for every j = 1, . . . , r. We write u v in this case. A tableau is a word w = u1 · · · ut , where ui are rows such that u1 u2 · · · ut . One of the main features of Mn is that every element w ∈ Mn , w = 1, is equal to a unique tableau. For example w = a5 a3 a4 a4 a2 a3 a3 a3 a1 a1 a2 a2 a2 a3 is a tableau with the subsequent rows v1 = a5 ,
v2 = a3 a4 a4 ,
v3 = a2 a3 a3 a3 ,
v4 = a1 a1 a2 a2 a2 a3 .
Such tableaux can be interpreted as planar objects. a5 a a a w= 3 4 4 a2 a3 a3 a3 a1 a1 a2 a2 a2 a3 and the subsequent columns of w are w1 = a5 a3 a2 a1 ,
w2 = a4 a3 a1 ,
w3 = a4 a3 a2 ,
w4 = a3 a2 ,
w5 = a2 ,
w6 = a3 .
Here the element is read row by row from left to right. Moreover, if w1 , . . . , wk denote the columns of a tableau, read from the top to the bottom, then the element w1 · · · wk
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called the ‘column reading’ of w satisfies w = w1 · · · wk in Mn . For example, w = a5 a3 a2 a1 a4 a3 a1 a4 a3 a2 a3 a2 a2 a3 for the tableau w displayed above. For a presentation of these and many other features of the plactic monoid we refer to [9]. In the following lemma, we recall some other basic properties of these monoids proved in [10]. Lemma 3. The following conditions are satisfied in Mn (1) the center Z(Mn ) of Mn is equal to the cyclic monoid z, where z = an · · · a1 , (2) if zw = zv for some w, v ∈ Mn then w = v, (3) if w ∈ Mn then w ∈ zM if and only if w = c1 an c2 an−1 · · · a2 cn a1 cn+1 for some ci ∈ Mn .
2. Plactic algebras Throughout K will denote a field. The semigroup algebra K[Mn ] is called the plactic algebra of rank n over K. In this section we begin with a description of the center of K[Mn ]. Then we obtain certain negative results concerning the primeness and semiprimeness of these algebras. As a consequence of the tableau presentation of elements of Mn we get that the growth function of Mn is polynomially bounded, so the Gelfand–Kirillov dimension of the algebra K[Mn ] is finite. We refer to [9, Corollary 6.3.11], for the exact formula for the cardinality of the set of elements of total degree k in Mn , for any given k 1. In particular GKdim K[M2 ] = 3 and GKdim K[M3 ] = 6. Let α be a nonzero element of K[Mn ]. We denote by supp(α) the support of α. We denote by degan (α) the degree of α with respect to ai , that is degai (α) = max di (w) | w ∈ supp(α) . We denote by deg(α) the (total) degree of α deg(α) = max deg(w) | w ∈ supp(α) . Proposition 4. Let z = an · · · a1 ∈ Mn . Then the center of K[Mn ] is Z K[Mn ] = K[z] and the nonzero elements of K[z] are regular elements of the algebra K[Mn ]. Proof. By Lemma 3, we know that Z(Mn ) = z. Hence K[z] ⊆ Z(K[Mn ]). To prove the equality, weproceed by induction on n. The case n = 1 is clear. So assume that n > 1. Let α = m i=1 αi wi ∈ Z(K[Mn ]), with m 1, αi ∈ K \ {0} and distinct elements wi ∈ Mn ; in particular, α = 0. It is clear that the homogeneous components of α also
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belong to Z(K[Mn ]). Hence we may assume that α is homogeneous with respect to every generator ai . Now j an α(an−1 · · · a1 )j = zj α ∈ Z K[Mn ] , for every j 1. Let j = degan (α). Write b = an−1 · · · a1 and β = αbj . Since z is regular in K[Mn ] by Lemma 3, it follows that β = 0. It is easy to see that supp(β) ⊆ a1 , . . . , an−1 , z. Since β is homogeneous with respect to an , we get β = zj γ for some γ ∈ K[a1 , . . . , an−1 ]. Since zj γ = αbj and b ∈ Z(a1 , . . . , an−1 ), it follows that zj γ centralizes K[a1 , . . . , an−1 ]. This implies that γ ∈ Z(K[a1 , . . . , an−1 ]) and γ = 0 because β = 0. Clearly we have supp(α) = supp zj α = supp αbj = supp zj γ = supp(γ ). Then by the induction hypothesis γ ∈ K[b]. Therefore, the homogeneity of γ implies that γ = λbk , for some k 0 and λ ∈ K. Since |supp(α)| = |supp(γ )|, it follows that m = 1 and w1 ∈ Z(Mn ) = z. Therefore α ∈ K[z], as desired. For the second assertion, we need to prove that αβ = 0, with α ∈ K[z], β ∈ K[M] and α = 0, implies that β = 0. Suppose otherwise. Write α = α0 + α1 z + · · · + αt zt , for some t 0 and αj ∈ K. Choose minimal p such that αp = 0. Let γ be the homogeneous component of β of the smallest (total) degree. Clearly, αp zp γ = 0, whence zp γ = 0. From Lemma 3 it follows that γ = 0, a contradiction. This yields the assertion. 2 Now we can construct the central localization of K[Mn ] with respect to K[z] \ {0}. Thus we have −1 K[Mn ] ⊆ K[Mn ] K[z] \ {0} . We denote by Mn the quotient of Mn modulo its center z, and by w the class of w ∈ Mn modulo z. In other words, Mn is obtained from Mn by adding the relation z = 1. By 3.8 of [10], Mn is a simple monoid. For n = 2, it is the bicyclic monoid, see [2]. Lemma 5. Let x be an indeterminate over K. Let ϕ : K[Mn ] → K[x] Mn be defined by ϕ
m
i=1
αi wi
=
m
αi x dn (wi ) wi .
i=1
Then ϕ is an injective homomorphism of K-algebras.
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Proof. It is clear that ϕ is a homomorphism of K-algebras. Furthermore, if
x dn (w) w = x dn (w ) w , for w, w ∈ Mn , then dn (w) = dn w and there exists nonnegative integers m, m such that
zm w = zm w .
Since m + dn (w) = dn (zm w) = dn (zm w ) = m + dn (w ), we have m = m . By Proposition 4, w = w . Thus ϕ is injective. 2 Lemma 6. −1 ∼ K[Mn ] K[z] \ {0} = K(x) Mn . Proof. Let ϕ be the map defined in Lemma 5. It is clear that ϕ(z) = x. Now it is easy to see that the natural extension of ϕ: −1 K[Mn ] K[z] \ {0} → K(x) Mn is an isomorphism. 2 Note that this result allows us to study some properties of K[Mn ] via the properties of K(x)[Mn ]. A first question that we can ask is whether the plactic algebra is a right (or left) primitive ring. The following result gives a partial answer. Theorem 7. Let F be an algebraically closed field. Let A be an F -algebra such that dimF A < |F | and Z(A) = F . Then A is not left primitive. Proof. Suppose that A is left primitive. Let V be a faithful simple left A-module. Let D = EndA (V ). By Density Theorem, D is a division F -algebra which is a homomorphic image of an F -subalgebra of A. So dimF D dimF A < |F |. Since F is algebraically closed, D = F by Lemma 7.1.2(iii) of [13]. Let α ∈ Z(A) \ F . Consider the homomorphism λ : A → EndF (V ),
β → λβ
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such that λβ (v) = βv for all v ∈ V . Since V is a faithful left A-module, Ker(λ) = {0}. We shall see that λα ∈ Z(EndF (V )). Let f ∈ EndF (V ). Suppose that f ◦ λα = λα ◦ f . Then there exists v ∈ V such that f λα (v) = λα f (v) . By Density Theorem, there exists β ∈ A such that f (v) = βv and f λα (v) = βλα (v). Hence λα f (v) = αβv = βαv = βλα (v) = f λα (v) , a contradiction, therefore λα ∈ Z(EndF (V )). Thus there exists k ∈ F such that λα = k IdV . But then α − k ∈ Ker(λ) = {0} and α ∈ / F , a contradiction, therefore A is not left primitive. 2 Since Z(Mn ) is not trivial (for n > 0), we have the following consequence of Theorem 7. Corollary 8. If F is an algebraically closed uncountable field then F [Mn ] is neither left primitive nor right primitive. In fact we have the following result. Theorem 9. Let K[Mn ] be the plactic algebra of rank n. Then (1) if n 3 then K[Mn ] is not prime, (2) if n 4 then K[Mn ] is not semiprime. Proof. Suppose that n 3. Consider the nonzero elements α = (an an−1 · · · a2 a1 )(an−1 · · · a2 ) − (an−1 · · · a2 a1 )(an an−1 · · · a2 ), β = (an a1 ) − (a1 )(an ),
γ = (an a2 ) − (a2 )(an )
in K[Mn ]. Note that the words between parentheses are columns. Let c = (ai1 ai2 · · · air ) ∈ Mn be a column, i.e., i1 > i2 > · · · > ir . By Lemma 3, if i1 < n and 1 < ir or if i1 = n, ir = 1 and r = n then αc = cα. If ir = 1 and i1 < n then αc = (ai1 ai2 · · · air−1 )αa1 = 0.
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If ir > 1 and i1 = n then cα = an α(ai2 · · · air ) = 0. If ir = 1, i1 = n, and r < n then it may be verified that αc = (an an−1 · · · a2 a1 )(an−1 · · · a2 )(an ai2 · · · air−1 )a1 − (an−1 · · · a2 a1 )(an an−1 · · · a2 )(an ai2 · · · air−1 )a1 = (an an−1 · · · a2 a1 )(an−1 · · · a2 a1 )(an ai2 · · · air−1 ) − (an−1 · · · a2 a1 )(an an−1 · · · a2 a1 )(an ai2 · · · air−1 ) = 0. By the defining relations, it is clear that ai β = 0, for all 1 < i n, and γ aj = 0 for all 1 < j < n. By looking at the tableau form of the elements of Mn , it is now easy to see that αtβ = 0 for all t ∈ Mn . Therefore K[Mn ] is not prime. Next assume that n 4. Note that αγ = 0 and γ α = 0. Suppose that c = (ai1 · · · air ) is a column such that cα = αc
and cα = 0.
As we have seen above, this implies that r < n and ir = 1. Suppose also that γ c = 0. Since n 4, for all t ∈ supp(γ c) we have by Lemma 3 t∈ / (an an−1 · · · a2 a1 )Mn . Hence the first column in the tableau form of every element in supp(γ c) is of the form (aj1 aj2 · · · ajs a1 ), with s < n − 1. As observed above, we get α(aj1 aj2 · · · ajs a1 ) = 0 in this case. Therefore αγ c = 0. Hence, by looking at the tableau form of the elements of Mn , it is easy to see that αγ wα = 0 for all w ∈ Mn . Therefore αγ K[Mn ]αγ = 0, and thus K[Mn ] is not semiprime. 2
3. Rank 2 plactic algebras In this section we study the plactic algebra of rank 2 over K. Theorem 10. K[M2 ] is a prime K-algebra. Proof. Since M2 is the bicyclic monoid, by [3], see also Theorem 3.3 of [11], for any field L, L[M2 ] is right and left primitive. By Lemma 6, K[M2 ] is prime. 2 Theorem 11. K[M2 ] is a semiprimitive K-algebra.
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Proof. Suppose that J (K[M2 ]) = 0. Let α ∈ J (K[M2 ]) \ {0}. Let z = a2 a1 . We can write j
j
α = λ1 zi1 a11 a2k1 + · · · + λr zir a1r a2kr , where λi ∈ K \ {0} and is , js , ks 0 are such that all the triples (is , js , ks ) are different. Consider the natural projection ψ : K[M2 ] → K M2 . Since M2 is the bicyclic monoid, by [11], K[M2 ] is primitive. It follows that ψ(α) = 0. So we can write p
q
p
q
α = f1 a1 1 a2 1 + · · · + fm a1 m a2 m , for some m 1, pn , qn 0 with all (pn , qn ) different, where each fn is of the form fn = β1 zl1 + · · · + βjn zljn , for some β1 , . . . , βjn ∈ K \ {0}, 0 l1 < l2 < · · · < ljn , such that β1 + · · · + βjn = 0. Note that fn is not a monomial in z. It is known that the Jacobson radical of a Z-graded ring is homogeneous, see Theorem 30.28 of [5]. So J (K[M2 ]) is homogeneous for the gradation defined by the total degree on M2 . Since deg(z) = 2 and deg(a1 ) = deg(a2 ) = 1, we may assume that 2is + js + ks is constant, for s = 1, . . . , r. But then every fn is a monomial in z, a contradiction. Therefore J (K[M2 ]) = 0. 2 Our next aim is to describe the prime spectrum of the plactic algebra K[M2 ]. Let B be the bicyclic monoid B = p, q | pq = 1. It is well known and easy to check that ei = q i pi , i 0, form a chain of idempotents e0 > e1 > e2 > · · · , and fij = q i (1 − e1 )pj , i, j = 0, 1, 2, . . . , form an infinite set of matrix units in K[B], see Example 21.26 in [7]. Let I be the K-linear span of the set {fij }. It is clear that this is the span of all q a pb − q a+1 pb+1 , a, b 0, and therefore I is an ideal of K[B]. It follows that I is a simple ring. Since 1 − ei ∈ I for all i 1, we have that K[B]/I ∼ = K p, p−1 , the group ring of a cyclic infinite group.
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Suppose that P is a nonzero prime ideal of K[B]. If I ⊆ P , then P /I is a prime ideal of K[B]/I . If I ⊆ P , then P ∩ I = {0}, but K[B] is prime, a contradiction. Thus we have described all prime ideals in K[B] and the classical Krull dimension of K[B] is clKdim K[B] = 2. Next we study the prime ideals of K[M2 ]. Let P1 be the ideal of K[M2 ] generated by a1 a2 − a2 a1 . Then K[M2 ]/P1 ∼ = K[x, y], the polynomial ring in two variables. Since, by Theorem 10, K[M2 ] is prime, clKdim K[M2 ] 3. Let P be a prime ideal of K[M2 ]. We consider three cases. Case 1. P ∩ M2 = ∅. Let z = a2 a1 . By 3.8 of [10], P ∩ z = ∅. Since z is central and P is prime, z ∈ P . So the natural map K[M2 ] → K[M2 ]/P factors as K[M2 ] → K[M2 ]/zK[M2 ] → K[M2 ]/P . But the ring in the middle is R = Ka1 , a2 | a2 a1 = 0, so its only minimal prime ideals are a1 R and Ra2 . Thus either a1 ∈ P or a2 ∈ P . Hence, if P is minimal, then K[M2 ]/P ∼ = K[x]. In particular, a maximal chain of primes 0 ⊂ P ⊂ P2 ⊂ · · · can have only two nonzero elements. Case 2. P ∩ M2 = ∅ and P ∩ K[z] = {0}. In this case, there exists an irreducible polynomial f ∈ K[z] such that P ∩ K[z] = f K[z] and, since P ∩ M = ∅, z ∈ / f K[z]. So the natural map K[M2 ] → K[M2 ]/P factors as K[M2 ] → K[z]/f K[z] a1 , a2 | a2 a1 = z → K[M2 ]/P . Now L = K[z]/f K[z] is a field and, since z = 0 in K[z]/f K[z], we have
K[z]/f K[z] a1 , a2 | a2 a1 = z ∼ = L[B],
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where B is the bicyclic monoid. Since L[B] is prime, if P is a minimal nonzero prime then we get P = (P ∩ K[z])K[M2]. Since we know the prime ideals in L[B], this completes the description of prime ideals of K[M2 ] in this case. Case 3. P ∩ K[z] = {0}. Such primes are in 1–1 correspondence with primes in the localization −1 K[M2 ] K[z] \ {0} . By Lemma 6, −1 ∼ K[M2 ] K[z] \ {0} = K(x)[B]. So the description of such primes follows from the description of the primes of K(x)[B], that can be determined as before. As a consequence of the above study we get a complete list of prime ideals of K[M2 ] and we obtain the following result. Theorem 12. clKdim K[M2 ] = 3. 4. Semiprimitivity of K[M3 ] Let M = a, b, c be the plactic monoid of rank 3, with the generators ordered as follows: a < b < c. Our main aim is to show that J (K[M]) = 0. Lemma 13. Let R be a Z-graded ring. Then the homogeneous elements of positive degree in J (R) are nilpotent. Proof. Let α ∈ J (R) be a homogeneous nonzero element of degree r > 0. There exists β ∈ R such that α + β + αβ = 0. Let β = β1 + · · · + βn , where β1 , . . . , βn are the (nonzero) homogeneous components of β of degrees z1 < · · · < zn respectively. Then it is easy to see that z1 = r,
z2 = 2r,
...,
and α + β1 = 0, αβ1 + β2 = 0,
zn = nr,
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...... αβn−1 + βn = 0, αβn = 0. Hence βi = (−α)i and α n+1 = 0.
2
Lemma 14. Let α be a nonzero element of the plactic algebra K[M] of rank 3. Then (i) If α is homogeneous with respect to degc and supp(α) ⊆ a, b, cba then α ∈ / J (K[M]). / (ii) If α is homogeneous with respect to degb and supp(α) ⊆ a, c, cba then α ∈ J (K[M]). / (iii) If α is homogeneous with respect to dega and supp(α) ⊆ b, c, cba then α ∈ J (K[M]). Proof. Suppose that supp(α) ⊆ a, b, cba and α is homogeneous with respect to degc . Let degc (α) = q. Then clearly α = (cba)q α , for some α ∈ K[a, b]. Suppose that α ∈ J (K[M]). Let β ∈ K[a, b]. Then αβ ∈ J (K[M]) and it is homogeneous with respect to degc . If degc (αβ) = 0 then α ∈ K[a, b], and hence α ∈ J (K[a, b]). If degc (αβ) > 0 then by Lemma 13, αβ is nilpotent. Since cba is central and regular, α K[a, b] is a nil right ideal of K[a, b]. But K[a, b] ∼ = K[M2 ] and, by Theorem 11, it is semiprimitive. So in both cases we get a contradiction. Therefore α ∈ / J (K[M]). This proves (i). The proofs of (ii) and (iii) are similar. 2 Lemma 15. Let w, w ∈ M be different elements. Then cbw = cbw if and only if one of the following cases holds: (i) (ii) (iii) (iv) (v) (vi) (vii)
{w, w } = {(cba)q (ba)p (ca)r (a)(c)l+1, (cba)q (ba)p (ca)r+1(c)l }, {w, w } = {(cba)q (ba)p (a)(b)k+1(c)l , (cba)q (ba)p+1(b)k (c)l }, {w, w } = {(cba)q (ba)p (ca)r+1(cb)s (b)(c)l+1, (cba)q (ba)p (ca)r+1 (cb)s+1(c)l }, {w, w } = {(cba)q (ba)p (ca)(cb)s (b)k+2 (c)l , (cba)q (ba)p+1 (cb)s+1(b)k (c)l }, {w, w } = {(cba)q (ca)(cb)s (b)k+1 (c)l , (cba)q+1(cb)s (b)k (c)l }, {w, w } = {(cba)q (ca)(cb)s+1(c)l , (cba)q+1(cb)s (c)l+1 }, {w, w } = {(cba)q (ba)(cb)s+1(b)k (c)l , (cba)q+1(cb)s (b)k+1 (c)l }.
In all the above possible cases, q, p, r, s, k, l are nonnegative integers. Proof. It is easy to check that if one of the cases (i)–(vii) holds then cbw = cbw . Suppose that cbw = cbw . The elements of M have two possible tableau forms: (cba)q (ba)p (ca)r (a)j +1 (b)k (c)l or (cba)q (ba)p (ca)r (cb)s (b)k (c)l ,
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where q, p, r, s, j, k, l are nonnegative integers. The convention here is that the words in parentheses form the subsequent columns of the tableau. Suppose that w = (cba)q (ba)p (ca)r (a)j +1 (b)k (c)l . Then cbw = (cba)q+1(ba)p (ca)r (a)j (b)k (c)l . Suppose that w = (cba)q (ba)p (ca)r (cb)s (b)k (c)l . Then (cba)q+1(ba)p (ca)r−1(cb)s+1(b)k−1 (c)l (cba)q+1(ba)p (ca)r−1(cb)s (c)l+1 cbw = q+1 p−1 s k+1 l (cba) (ba) (cb) (b) (c) q s+1 k l (cba) (cb) (b) (c)
if r > 0 and k > 0, if r > 0 and k = 0, if r = 0 and p > 0, if r = 0 and p = 0.
We say that w is of type (1)–(5) if cbw is of one of these five forms, respectively. Since w = w and cbw = cbw , it is easy to see that w is of type (1)
⇒
w is of type (3) or (4),
w is of type (2)
⇒
w is of type (3), (4) or (5),
w is of type (3)
⇒
w is of type (1), (2), or (5),
w is of type (4)
⇒
w is of type (1), (2) or (5),
w is of type (5)
⇒
w is of type (2), (3) or (4).
So we have the following possible cases. Types (1) and (3): w, w = (cba)q (ba)p (ca)r (a)(c)l+1 , (cba)q (ba)p (ca)r+1(c)l . Types (1) and (4):
w, w = (cba)q (ba)p (a)(b)k+1(c)l , (cba)q (ba)p+1(b)k (c)l .
Types (2) and (3):
w, w = (cba)q (ba)p (ca)r+1(cb)s (b)(c)l+1 , (cba)q (ba)p (ca)r+1(cb)s+1(c)l .
Types (2) and (4): w, w = (cba)q (ba)p (ca)(cb)s (b)k+2 (c)l , (cba)q (ba)p+1(cb)s+1(b)k (c)l . Types (2) and (5): w, w = (cba)q (ca)(cb)s (b)k+1 (c)l , (cba)q+1(cb)s (b)k (c)l .
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Types (3) and (5): w, w = (cba)q (ca)(cb)s+1(c)l , (cba)q+1(cb)s (c)l+1 . Types (4) and (5): w, w = (cba)q (ba)(cb)s+1(b)k (c)l , (cba)q+1(cb)s (b)k+1 (c)l . In all the above possible cases, q, p, r, s, k, l are nonnegative integers. 2 The following is an easy consequence of Lemma 15. Lemma 16. Let w, w ∈ M be different elements. Then cbw = cbw and wba = w ba if and only if one of the following cases holds: (i) {w, w } = {(cba)q (ba)p (ca)r (a)(c), (cba)q (ba)p (ca)r+1}, (ii) {w, w } = {(cba)q (ba)p (ca)r+1(cb)s (b)(c), (cba)q (ba)p (ca)r+1(cb)s+1}, (iii) {w, w } = {(cba)q (ba)(cb)(b)k , (cba)q+1(b)k+1 }, where q, p, r, s, k are nonnegative integers. Before stating the main theorem of this section, we first prove a quite strong and surprising result on homogeneous ideals of K[M], that seems to be of independent interest. Theorem 17. Let K[M] be the plactic algebra of rank 3. Let I be a nonzero ideal that is homogeneous with respect to the degrees in a, b and c. Then there exists a homogeneous nonzero α ∈ I (with respect to these degrees) such that |supp(α)| 2. Proof. Let α ∈ I be a nonzero homogeneous element with respect to the degrees in a, b and c. Write α = ni=1 αi wi , where wi ∈ M and 0 = αi ∈ K. We choose α with a minimal n. If wi u = wj u (or uwi = uwj ), for some u ∈ M and i = j , then we must have αu = 0 (uα = 0 respectively). This will be often used without further comment. From Lemma 3 it follows that if i = degc (α) then α(ba)i ∈ K[a, b, cba]. Suppose that α(ba)i = 0. The elements in supp(α(ba)i ) are of the form (cba)i (ba)p (a)j (b)k , where p + j = dega (α) and p + k = degb (α). As before, the words in parentheses form the subsequent columns of the tableau. Let i = degb (α). Since α is homogeneous, ∀w ∈ supp α(ba)i ,
wa i = (cba)i (ba)i (a)j ,
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where j = dega (α), so that |supp(α(ba)i a i )| 1. Hence, we may assume that α(ba)i a i = 0. Since baa = aba, replacing α by some αa t , we may assume that α(ba)i a = 0. In this case, it is easy to see that α(ba)i = λ (cba)i (ba)i −1 (a)j −i +1 (b) − (cba)i (ba)i (a)j −i , for some nonzero λ ∈ K. Therefore, we may assume that α(ba)i = 0 and replacing α by some α(ba)t , we may assume that αba = 0. A symmetric argument shows that we may also assume that cbα = 0. Since cb and c commute, replacing α by some cj α, we may assume that the tableau of each w ∈ supp(α) contains at most one column (ba). Since αba = 0, the tableau of each w ∈ supp(α) contains at most one column (c). Since ba and a commute, replacing α by some αa t , we may assume that the tableau of each w ∈ supp(α) contains at most one column (cb). Since cbα = 0, we get α=
γj wj − wj
j
where γj ∈ K and every wj , wj is a pair of one of the forms listed in Lemma 15. There are the following possibilities for these pairs: (i) (ii) (iii) (iv) (v) (vi) (vii)
{w1 , w1 } = {(cba)q1 (ba)p1 (ca)r1 (a)(c), (cba)q1 (ba)p1 (ca)r1 +1 }, {w2 , w2 } = {(cba)q2 (a)(b)k2+1 (c)l2 , (cba)q2 (ba)(b)k2 (c)l2 }, {w3 , w3 } = {(cba)q3 (ba)p3 (ca)r3 +1 (b)(c), (cba)q3 (ba)p3 (ca)r3+1 (cb)}, {w4 , w4 } = {(cba)q4 (ca)(b)k4+2 (c)l4 , (cba)q4 (ba)(cb)(b)k4 (c)l4 }, {w5 , w5 } = {(cba)q5 (ca)(cb)s5 (b)k5 +1 (c)l5 , (cba)q5+1 (cb)s5 (b)k5 (c)l5 }, {w6 , w6 } = {(cba)q6 (ca)(cb), (cba)q6+1 (c)}, {w7 , w7 } = {(cba)q7 (ba)(cb)(b)k7 (c)l7 , (cba)q7+1 (b)k7 +1 (c)l7 },
where qi , ri , ki are nonnegative integers and pi , si , li ∈ {0, 1}. Moreover supp(α) can contain at most one pair of each type, except for type (v), where there may be two pairs, corresponding to l5 = 0 and l5 = 1. The following obvious facts will be often used without further comment. (i) dega (w1 ) = q1 +p1 +r1 +1, dega (w1 )−degb (w1 ) = r1 +1, dega (w1 )−degc (w1 ) = p1 , (ii) dega (w2 ) = q2 + 1, dega (w2 ) − degb (w2 ) = −k2 , dega (w2 ) − degc (w2 ) = 1 − l2 , (iii) dega (w3 ) = q3 + p3 + r3 + 1, dega (w3 ) − degb (w3 ) = r3 , dega (w3 ) − degc (w3 ) = p3 − 1, (iv) dega (w4 ) = q4 + 1, dega (w4 ) − degb (w4 ) = −k4 − 1, dega (w4 ) − degc (w4 ) = −l4 , (v) dega (w5 ) = q5 + 1, dega (w5 ) − degb (w5 ) = −s5 − k5 , dega (w5 ) − degc (w5 ) = −s5 − l5 , (vi) dega (w6 ) = q6 + 1, dega (w6 ) − degb (w6 ) = 0, dega (w6 ) − degc (w6 ) = −1,
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(vii) dega (w7 ) = q7 + 1, dega (w7 ) − degb (w7 ) = −k7 − 1, dega (w7 ) − degc (w7 ) = −l7 . Thus there exist γi ∈ K such that α = γ1 (cba)q1 (ba)p1 (ca)r1 (a)(c) − (ca) + γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c)l2 + γ3 (cba)q3 (ba)p3 (ca)r3+1 (b)(c) − (cb) + γ4 (cba)q4 (ca)(b)2 − (ba)(cb) (b)k4 (c)l4 + γ5 (cba)q5 (ca)(b) − (cba) (cb)s5 (b)k5 + γ5 (cba)q5 (ca)(b) − (cba) (cb)s5 (b)k5 (c) + γ6 (cba)q6 (ca)(cb) − (cba)(c) + γ7 (cba)q7 (ba)(cb) − (cba)(b) (b)k7 (c)l7 . We shall proceed in several steps. Step 1. Suppose that γ1 = 0. Since α is homogeneous and dega (w1 ) − degb (w1 ) > 0, we may assume that α = γ1 (cba)q1 (ba)p1 (ca)r1 (a)(c) − (ca) + γ3 (cba)q3 (ba)p3 (ca)r3 +1 (b)(c) − (cb) . We may assume that γ3 = 0. Then, since α is homogeneous, q1 + p1 + r1 = q3 + p3 + r3 ,
r1 + 1 = r3
and p1 = p3 − 1.
Thus p1 = 0, p3 = 1 and α = γ1 (cba)q3+2 (ca)r1 (a)(c) − (ca) + γ3 (cba)q3 (ba)(ca)r1+2 (b)(c) − (cb) . Now cα = γ3 (cba)q3+1 (ca)r1+2 ((b)(c) − (cb)) = 0. Hence we may further assume that γ1 = 0. Step 2. Suppose that γ6 = 0. Since α is homogeneous, dega (w6 ) − degb (w6 ) = 0 and dega (w6 ) − degc (w6 ) = −1, we may assume that α = γ3 (cba)q6 (ca) (b)(c) − (cb) + γ5 (cba)q6 (ca)(b) − (cba) (c) + γ6 (cba)q6 (ca)(cb) − (cba)(c) . Since αba = 0 and αba = γ5 + γ6 (cba)q6+1 (ca)(b) − (cba) , we get γ5 = −γ6 = 0. It follows that α = (γ3 − γ6 )(cba)q6 (ca) (b)(c) − (cb) .
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Therefore we may further assume that γ6 = 0. Step 3. Suppose that γ3 = 0. Since α is homogeneous and dega (w3 ) − degb (w3 ) 0, we may assume that α = γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c)l2 + γ3 (cba)q3 (ba)p3 (ca)r3+1 (b)(c) − (cb) + γ5 (cba)q5 (ca)(b) − (cba) (cb)s5 (b)k5 + γ5 (cba)q5 (ca)(b) − (cba) (cb)s5 (b)k5 (c). We may assume that (γ2 , γ5 , γ5 ) = (0, 0, 0). Since α is homogeneous, r3 = 0 and it is easy to see that we may write α = γ2 (cba)q3+1 (a)(b) − (ba) (c) + γ3 (cba)q3 (ba)p3 (ca) (b)(c) − (cb) + γ (cba)q3+p3 (ca)(b) − (cba) (c)1−p3 , with γ ∈ {γ5 , γ5 } and (γ2 , γ ) = (0, 0). Since αba = 0 and αba = γ2 (cba)q3+2 (a)(b) − (ba) + γ (cba)q3+p3 (ca)(b) − (cba) (c)1−p3 ba, γ2 γ = 0 (because the second summand is nonzero if γ = 0). Thus, since α is homogeneous, p3 = 1, γ = −γ2 and α = γ2 (cba)q3+1 (a)(b) − (ba) (c) + γ3 (cba)q3 (ba)(ca) (b)(c) − (cb) − γ2 (cba)q3+1 (ca)(b) − (cba) . Now we have bα = −γ2 (cba)q3+1 (ba)(cb) − (cba)(b) = 0. Therefore we may further assume that γ3 = 0. Step 4. Suppose that γ2 = 0. Since α is homogeneous, we may assume that α = γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c)l2 + γ4 (cba)q2 (ca)(b)2 − (ba)(cb) (b)k2 −1 + γ5 (cba)q2 (ca)(b) − (cba) (b)k2 + γ7 (cba)q2 (ba)(cb) − (cba)(b) (b)k2 −1 if k2 > 0 and α = γ2 (cba)q2 (a)(b) − (ba) (c)l2 + γ5 (cba)q2 (ca)(b) − (cba)
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if k2 = 0. We may also assume that (γ4 , γ5 , γ7 ) = (0, 0, 0) and γ5 = 0 in these two cases, respectively. Then l2 = 1. If k2 = 0 then bα = γ5 (cba)q2 (ba)(cb) − (cba)(b) = 0. Therefore we may assume that k2 > 0. Since αba = 0 and αba = (γ2 + γ4 + γ5 )(cba)q2+1 (a)(b) − (ba) (b)k2 , we have that γ2 + γ4 + γ5 = 0. Since bα = γ5 (cba)q2 (ba)(cb) − (cba)(b) (b)k2 + γ7 (cba)q2 (ba)(cb) − (cba)(b) (b)k2 , we may assume that γ7 = −γ5 . Now we have α = γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c) + γ4 (cba)q2 (ca)(b)2 − (ba)(cb) (b)k2 −1 + γ5 (cba)q2 (ca)(b)2 − (ba)(cb) (b)k2 −1 = γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c) − γ2 (cba)q2 (ca)(b)2 − (ba)(cb) (b)k2 −1 , αa = γ2 (cba)q2 (ba) (a)(b) − (ba) (b)k2 −1 (c) − γ2 (cba)q2+1 (a)(b) − (ba) (b)k2 −1 , αa k2 = γ2 (cba)q2 (ba)k2 (a)(b) − (ba) (c) − γ2 (cba)q2+1 (ba)k2−1 (a)(b) − (ba) , ck2 αa k2 = γ2 (cba)q2+k2 (a)(b) − (ba) (c) − γ2 (cba)q2+k2 (ca)(b) − (cba) = γ2 (cba)q2+k2 (a)(b)(c) − (ba)(c) − (ca)(b) + (cba) and bck2 αa k2 = γ2 (cba)q2+k2 −(ba)(cb) + (cba)(b) = 0. Therefore we may further assume that γ2 = 0. Step 5. Suppose that γ4 = 0. Since α is homogeneous, α = γ4 (cba)q4 (ca)(b)2 − (ba)(cb) (b)k4 (c)l4 + γ5 (cba)q4 (ca)(b) − (cba) (cb)s5 (b)k5 + γ5 (cba)q4 (ca)(b) − (cba) (cb)s5 (b)k5 (c) + γ7 (cba)q4 (ba)(cb) − (cba)(b) (b)k4 (c)l4 .
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Suppose that l4 = 1. In this case, α = γ4 (cba)q4 (ca)(b)2 − (ba)(cb) (b)k4 (c) + γ5 (cba)q4 (ca)(b) − (cba) (cb)(b)k4 + γ5 (cba)q4 (ca)(b) − (cba) (b)k4 +1 (c) + γ7 (cba)q4 (ba)(cb) − (cba)(b) (b)k4 (c). Since αba = 0 and αba = γ4 (cba)q4+1 (ca)(b)2 − (ba)(cb) (b)k4 + γ5 + γ5 (cba)q4+1 (ca)(b) − (cba) (b)k4 +1 + γ7 (cba)q4+1 (ba)(cb) − (cba)(b) (b)k4 = γ4 + γ5 + γ5 (cba)q4+1 (ca)(b)k4+2 − (γ4 − γ7 )(cba)q4+1 (ba)(cb)(b)k4 − γ5 + γ5 + γ7 (cba)q4+2 (b)k4 +1 , we have γ4 = γ7 = −γ5 − γ5 . Thus α = γ4 + γ5 (cba)q4 (ca)(b) − (cba) (b)k4 +1 (c) + γ5 (cba)q4 (ca)(b) − (cba) (cb)(b)k4 . Hence we may assume that (γ4 + γ5 )γ5 = 0. Now αa k4 +2 = γ4 + γ5 (cba)q4+1 (ba)k4 +1 (a)(c) − (ca) = 0. Therefore, we may assume that l4 = 0. Then α = γ4 (cba)q4 (ca)(b)2 − (ba)(cb) (b)k4 + γ5 (cba)q4 (ca)(b) − (cba) (b)k4 +1 + γ7 (cba)q4 (ba)(cb) − (cba)(b) (b)k4 . Since αba = 0 and αba = (γ4 + γ5 )(cba)q4+1 (a)(b) − (ba) (b)k4 +1 , γ5 = −γ4 . Now we have α = γ4 (cba)q4 −(ba)(cb) + (cba)(b) (b)k4 + γ7 (cba)q4 (ba)(cb) − (cba)(b) (b)k4 = (γ7 − γ4 )(cba)q4 (ba)(cb) − (cba)(b) (b)k4 . Therefore we may further assume that γ4 = 0.
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Step 6. Suppose that γ7 = 0. Since α is homogeneous, we may write α = γ5 (cba)q7 (ca)(b) − (cba) (cb)s5 (b)k7 +1−s5 + γ5 (cba)q7 (ca)(b) − (cba) (b)k7 +1 (c) + γ7 (cba)q7 (ba)(cb) − (cba)(b) (b)k7 (c)l7 . Suppose that l7 = 1. Then we may assume s5 = 1 and we may write α = γ5 (cba)q7 (ca)(b) − (cba) (cb)(b)k7 + γ5 (cba)q7 (ca)(b) − (cba) (b)k7 +1 (c) + γ7 (cba)q7 (ba)(cb) − (cba)(b) (b)k7 (c). Since αba = 0 and αba = γ5 + γ5 (cba)q7+1 (ca)(b) − (cba) (b)k7 +1 + γ7 (cba)q7+1 (ba)(cb) − (cba)(b) (b)k7 , we have a contradiction. Hence l7 = 0 and we may assume α = γ5 (cba)q7 (ca)(b) − (cba) (b)k7 +1 + γ7 (cba)q7 (ba)(cb) − (cba)(b) (b)k7 . Since αba = 0 and αba = γ5 (cba)q7+1 (a)(b) − (ba) (b)k7 +1 , γ5 = 0. Therefore we may further assume that γ7 = 0. Step 7. It remains to consider the case where only γ5 , γ5 can be nonzero. Hence we may assume that γ5 γ5 = 0 and α = γ5 (cba)q5 (ca)(b) − (cba) (cb)(b)k5 + γ5 (cba)q5 (ca)(b) − (cba) (b)k5 +1 (c). Since αba = 0 and αba = γ5 + γ5 (cba)q5+1 (ca)(b) − (cba) (b)k5 +1 , γ5 = −γ5 . Now we have α = γ5 (cba)q5 (ca)(b) − (cba) (cb)(b)k5 − γ5 (cba)q5 (ca)(b) − (cba) (b)k5 +1 (c), αa k5 +1 = γ5 (cba)q5+1 (ca)(b) − (cba) (ba)k5 − γ5 (cba)q5+1 (a)(b) − (ba) (ba)k5 (c), ck5 αa k5 +1 = γ5 (cba)q5+k5 +1 (ca)(b) − (cba) − γ5 (cba)q5+k5 +1 (a)(b) − (ba) (c)
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and bck5 αa k5 +1 = γ5 (cba)q5+k5 +1 (ba)(cb) − (cba)(b) = 0. This completes the proof of the theorem. 2 Theorem 18. Let K[M] be the plactic algebra of rank 3. Then K[M] is semiprimitive. Proof. As mentioned before, the Jacobson radical of a Z-graded ring is homogeneous, see [5, Theorem 30.28]. Therefore if J (K[M]) = 0 then we may choose a nonzero element 0 = α ∈ J (K[M]) which is homogeneous with respect to the gradation on K[M] defined by the length of words in M, and also by the degree in each of the generators. By Theorem 17, we may assume that α = α1 w1 + α2 w2 , where wi ∈ M and 0 = αi ∈ K. By Lemma 13, w1 = w2 . Furthermore, if w1 u = w2 u (or uw1 = uw2 ), for some u ∈ M, then we must have αu = 0 (uα = 0, respectively). This will be used without further comment. From Lemma 3 it follows that if i = degc (α) then α(ba)i ∈ K[a, b, cba]. By Lemma 14, α(ba)i = 0. Therefore, replacing α by some α(ba)j , we may assume that αba = 0. Similarly, replacing α by some (cb)j α, we may assume that cbα = 0. By Lemma 16, we may assume that α is one of the following elements: (cba)q (ba)p (ca)r (a)(c) − (cba)q (ba)p (ca)r+1, (cba)q (ba)p (ca)r+1(cb)s (b)(c) − (cba)q (ba)p (ca)r+1(cb)s+1, (cba)q (ba)(cb)(b)k − (cba)q+1(b)k+1 , where q, p, r, s, k are nonnegative integers. If α = (cba)q (ba)p (ca)r (a)(c) − (cba)q (ba)p (ca)r+1 then cp α = (cba)q+p (ca)r (a)(c) − (ca)r+1 = 0, which contradicts Lemma 14. If α = (cba)q (ba)p (ca)r+1(cb)s (b)(c) − (cba)q (ba)p (ca)r+1(cb)s+1 then cp αa s+1 = (cba)q+p+s+1(ca)r (a)(c) − (ca) = 0, which contradicts Lemma 14. If α = (cba)q (ba)(cb)(b)k − (cba)q+1(b)k+1 then, by induction on n, it is easy to see that α n = (−1)n−1 (cba)nq+n−1 (ba)(cb) − (cba)(b) (b)nk+n−1 = 0, for all n, in contradiction with Lemma 13. Hence K[M] is semiprimitive. 2
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Acknowledgments This work was partially supported by DGESID PB98-0873, Spain and the Generalitat de Catalunya, and by KBN research grant 2 P03A 030 18, Poland. The second author is also grateful for the warm hospitality of the Centre de Recerca Matematica (Barcelona), where a part of the work was done.
References [1] A. Arnold, M. Kanta, D. Krob, Recognizable subsets of the two letter plactic monoid, Inform. Process. Lett. 64 (1997) 53–59. [2] A.H. Clifford, G.B. Preston, The Algebraic Theory of Semigroups, vol. 1, Amer. Math. Soc., Providence, RI, 1964. [3] O.I. Domanov, Semisimplicity and identities of inverse semigroup algebras, Rings and Modules, Mat. Issled. 38 (1976) 123–137. [4] W. Fulton, Young Tableaux, Cambridge University Press, New York, 1997. [5] G. Karpilovsky, The Jacobson Radical of Classical Rings, Longman, New York, 1991. [6] D.E. Knuth, Permutations, matrices and generalized Young tableaux, Pacific J. Math. 34 (1970) 709–727. [7] T.Y. Lam, A First Course in Noncommutative Rings, in: Grad. Texts in Math., vol. 131, Springer-Verlag, New York, 1991. [8] A. Lascoux, B. Leclerc, J.Y. Thibon, Crystal Graphs and q-analogues of weight multiplicities for the root system An , Lett. Math. Phys. 35 (1995) 359–374. [9] A. Lascoux, B. Leclerc, J.Y. Thibon, The plactic monoid, in: Algebraic Combinatorics on Words, Cambridge Univ. Press, Cambridge, 2002. [10] A. Lascoux, M.P. Schützenberger, Le monoïde plaxique, in: Noncommutative Structures in Algebra and Geometric Combinatorics (Naples, 1978), in: Quad. “Ricerca Sci.”, vol. 109, CNR, Rome, 1981, pp. 129– 156. [11] W.D. Munn, On contracted semigroup rings, Proc. Roy. Soc. Edinburgh Ser. A 115 (1990) 109–117. [12] J. Okni´nski, Semigroup Algebras, Dekker, New York, 1991. [13] D.S. Passman, The Algebraic Structure of Group Rings, Wiley–Interscience, New York, 1977. [14] C. Schensted, Longest increasing and decreasing subsequences, Canad. J. Math. 13 (1961) 179–191. [15] M.P. Schützenberger, Pour le monoïde plaxique, Math. Inform. Sci. Humaines 140 (1997) 5–10.
Plactic algebras Ferran Cedó a and Jan Okni´nski b,∗ a Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra (Barcelona), Spain b Institute of Mathematics, Warsaw University, 02-097 Warsaw, Poland
Received 21 March 2002 Communicated by Michel van den Bergh
Abstract The structure of the plactic algebra K[M] of rank 2 or 3 is studied. It is shown that these algebras are semiprimitive. Moreover, the prime spectrum is completely described in the rank 2 case. Also it is proved that the plactic algebra of rank n 4 is not semiprime and the plactic algebra of rank 3 is not prime. 2004 Elsevier Inc. All rights reserved.
The construction of the plactic monoid originated from the papers of Schensted [14] and Knuth [6] on the ground of certain combinatorial problems and operations on Young tableaux. It was systematically studied by Lascoux and Schützenberger in [10] and became a very important tool in the theory of symmetric functions, Young tableaux and various aspects of representation theory [4,9]. The importance of these monoids was also indicated by Schützenberger in [15]. Plactic monoids appeared recently in the theory of modular representations of the symmetric group and in quantum groups, via the representation theory of quantum enveloping algebras [8]. The plactic monoid of rank 2 has also been studied from the point of view of language theory [1]. Our aim in this paper is to study basic ring theoretic properties of the plactic algebra K[M] over a field K. For the general background on semigroup algebras we refer to [12].
* Corresponding author.
E-mail address: [email protected] (J. Okni´nski). 0021-8693/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2003.12.004
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1. Preliminaries In this section, we recall the definition and some of the basic properties of the plactic monoids. These results will be often used in the subsequent sections. Definition 1. The plactic monoid of rank n is the monoid presented with generators: a1 , a2 , . . . , an , subject to the relations: ak ai aj = ai ak aj
if i j < k,
as at ar = as ar at
if r < s t.
We will denote by Mn the plactic monoid of rank n. Definition 2. Since the defining relations of Mn are homogeneous with respect to the degree in every ai , this degree is inherited by Mn . So we denote by di (t) the degree of t ∈ Mn with respect to ai . By deg(t) we mean the total degree of t ∈ Mn . By a row in Mn we mean an element which is a word ai1 · · · air such that r 1 and i1 i2 · · · ir . A column is defined as a word with strictly decreasing indices. We say that a row u = ai1 · · · air dominates a row v = ak1 · · · aks if r s and ij > kj for every j = 1, . . . , r. We write u v in this case. A tableau is a word w = u1 · · · ut , where ui are rows such that u1 u2 · · · ut . One of the main features of Mn is that every element w ∈ Mn , w = 1, is equal to a unique tableau. For example w = a5 a3 a4 a4 a2 a3 a3 a3 a1 a1 a2 a2 a2 a3 is a tableau with the subsequent rows v1 = a5 ,
v2 = a3 a4 a4 ,
v3 = a2 a3 a3 a3 ,
v4 = a1 a1 a2 a2 a2 a3 .
Such tableaux can be interpreted as planar objects. a5 a a a w= 3 4 4 a2 a3 a3 a3 a1 a1 a2 a2 a2 a3 and the subsequent columns of w are w1 = a5 a3 a2 a1 ,
w2 = a4 a3 a1 ,
w3 = a4 a3 a2 ,
w4 = a3 a2 ,
w5 = a2 ,
w6 = a3 .
Here the element is read row by row from left to right. Moreover, if w1 , . . . , wk denote the columns of a tableau, read from the top to the bottom, then the element w1 · · · wk
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called the ‘column reading’ of w satisfies w = w1 · · · wk in Mn . For example, w = a5 a3 a2 a1 a4 a3 a1 a4 a3 a2 a3 a2 a2 a3 for the tableau w displayed above. For a presentation of these and many other features of the plactic monoid we refer to [9]. In the following lemma, we recall some other basic properties of these monoids proved in [10]. Lemma 3. The following conditions are satisfied in Mn (1) the center Z(Mn ) of Mn is equal to the cyclic monoid z, where z = an · · · a1 , (2) if zw = zv for some w, v ∈ Mn then w = v, (3) if w ∈ Mn then w ∈ zM if and only if w = c1 an c2 an−1 · · · a2 cn a1 cn+1 for some ci ∈ Mn .
2. Plactic algebras Throughout K will denote a field. The semigroup algebra K[Mn ] is called the plactic algebra of rank n over K. In this section we begin with a description of the center of K[Mn ]. Then we obtain certain negative results concerning the primeness and semiprimeness of these algebras. As a consequence of the tableau presentation of elements of Mn we get that the growth function of Mn is polynomially bounded, so the Gelfand–Kirillov dimension of the algebra K[Mn ] is finite. We refer to [9, Corollary 6.3.11], for the exact formula for the cardinality of the set of elements of total degree k in Mn , for any given k 1. In particular GKdim K[M2 ] = 3 and GKdim K[M3 ] = 6. Let α be a nonzero element of K[Mn ]. We denote by supp(α) the support of α. We denote by degan (α) the degree of α with respect to ai , that is degai (α) = max di (w) | w ∈ supp(α) . We denote by deg(α) the (total) degree of α deg(α) = max deg(w) | w ∈ supp(α) . Proposition 4. Let z = an · · · a1 ∈ Mn . Then the center of K[Mn ] is Z K[Mn ] = K[z] and the nonzero elements of K[z] are regular elements of the algebra K[Mn ]. Proof. By Lemma 3, we know that Z(Mn ) = z. Hence K[z] ⊆ Z(K[Mn ]). To prove the equality, weproceed by induction on n. The case n = 1 is clear. So assume that n > 1. Let α = m i=1 αi wi ∈ Z(K[Mn ]), with m 1, αi ∈ K \ {0} and distinct elements wi ∈ Mn ; in particular, α = 0. It is clear that the homogeneous components of α also
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belong to Z(K[Mn ]). Hence we may assume that α is homogeneous with respect to every generator ai . Now j an α(an−1 · · · a1 )j = zj α ∈ Z K[Mn ] , for every j 1. Let j = degan (α). Write b = an−1 · · · a1 and β = αbj . Since z is regular in K[Mn ] by Lemma 3, it follows that β = 0. It is easy to see that supp(β) ⊆ a1 , . . . , an−1 , z. Since β is homogeneous with respect to an , we get β = zj γ for some γ ∈ K[a1 , . . . , an−1 ]. Since zj γ = αbj and b ∈ Z(a1 , . . . , an−1 ), it follows that zj γ centralizes K[a1 , . . . , an−1 ]. This implies that γ ∈ Z(K[a1 , . . . , an−1 ]) and γ = 0 because β = 0. Clearly we have supp(α) = supp zj α = supp αbj = supp zj γ = supp(γ ). Then by the induction hypothesis γ ∈ K[b]. Therefore, the homogeneity of γ implies that γ = λbk , for some k 0 and λ ∈ K. Since |supp(α)| = |supp(γ )|, it follows that m = 1 and w1 ∈ Z(Mn ) = z. Therefore α ∈ K[z], as desired. For the second assertion, we need to prove that αβ = 0, with α ∈ K[z], β ∈ K[M] and α = 0, implies that β = 0. Suppose otherwise. Write α = α0 + α1 z + · · · + αt zt , for some t 0 and αj ∈ K. Choose minimal p such that αp = 0. Let γ be the homogeneous component of β of the smallest (total) degree. Clearly, αp zp γ = 0, whence zp γ = 0. From Lemma 3 it follows that γ = 0, a contradiction. This yields the assertion. 2 Now we can construct the central localization of K[Mn ] with respect to K[z] \ {0}. Thus we have −1 K[Mn ] ⊆ K[Mn ] K[z] \ {0} . We denote by Mn the quotient of Mn modulo its center z, and by w the class of w ∈ Mn modulo z. In other words, Mn is obtained from Mn by adding the relation z = 1. By 3.8 of [10], Mn is a simple monoid. For n = 2, it is the bicyclic monoid, see [2]. Lemma 5. Let x be an indeterminate over K. Let ϕ : K[Mn ] → K[x] Mn be defined by ϕ
m
i=1
αi wi
=
m
αi x dn (wi ) wi .
i=1
Then ϕ is an injective homomorphism of K-algebras.
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Proof. It is clear that ϕ is a homomorphism of K-algebras. Furthermore, if
x dn (w) w = x dn (w ) w , for w, w ∈ Mn , then dn (w) = dn w and there exists nonnegative integers m, m such that
zm w = zm w .
Since m + dn (w) = dn (zm w) = dn (zm w ) = m + dn (w ), we have m = m . By Proposition 4, w = w . Thus ϕ is injective. 2 Lemma 6. −1 ∼ K[Mn ] K[z] \ {0} = K(x) Mn . Proof. Let ϕ be the map defined in Lemma 5. It is clear that ϕ(z) = x. Now it is easy to see that the natural extension of ϕ: −1 K[Mn ] K[z] \ {0} → K(x) Mn is an isomorphism. 2 Note that this result allows us to study some properties of K[Mn ] via the properties of K(x)[Mn ]. A first question that we can ask is whether the plactic algebra is a right (or left) primitive ring. The following result gives a partial answer. Theorem 7. Let F be an algebraically closed field. Let A be an F -algebra such that dimF A < |F | and Z(A) = F . Then A is not left primitive. Proof. Suppose that A is left primitive. Let V be a faithful simple left A-module. Let D = EndA (V ). By Density Theorem, D is a division F -algebra which is a homomorphic image of an F -subalgebra of A. So dimF D dimF A < |F |. Since F is algebraically closed, D = F by Lemma 7.1.2(iii) of [13]. Let α ∈ Z(A) \ F . Consider the homomorphism λ : A → EndF (V ),
β → λβ
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such that λβ (v) = βv for all v ∈ V . Since V is a faithful left A-module, Ker(λ) = {0}. We shall see that λα ∈ Z(EndF (V )). Let f ∈ EndF (V ). Suppose that f ◦ λα = λα ◦ f . Then there exists v ∈ V such that f λα (v) = λα f (v) . By Density Theorem, there exists β ∈ A such that f (v) = βv and f λα (v) = βλα (v). Hence λα f (v) = αβv = βαv = βλα (v) = f λα (v) , a contradiction, therefore λα ∈ Z(EndF (V )). Thus there exists k ∈ F such that λα = k IdV . But then α − k ∈ Ker(λ) = {0} and α ∈ / F , a contradiction, therefore A is not left primitive. 2 Since Z(Mn ) is not trivial (for n > 0), we have the following consequence of Theorem 7. Corollary 8. If F is an algebraically closed uncountable field then F [Mn ] is neither left primitive nor right primitive. In fact we have the following result. Theorem 9. Let K[Mn ] be the plactic algebra of rank n. Then (1) if n 3 then K[Mn ] is not prime, (2) if n 4 then K[Mn ] is not semiprime. Proof. Suppose that n 3. Consider the nonzero elements α = (an an−1 · · · a2 a1 )(an−1 · · · a2 ) − (an−1 · · · a2 a1 )(an an−1 · · · a2 ), β = (an a1 ) − (a1 )(an ),
γ = (an a2 ) − (a2 )(an )
in K[Mn ]. Note that the words between parentheses are columns. Let c = (ai1 ai2 · · · air ) ∈ Mn be a column, i.e., i1 > i2 > · · · > ir . By Lemma 3, if i1 < n and 1 < ir or if i1 = n, ir = 1 and r = n then αc = cα. If ir = 1 and i1 < n then αc = (ai1 ai2 · · · air−1 )αa1 = 0.
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If ir > 1 and i1 = n then cα = an α(ai2 · · · air ) = 0. If ir = 1, i1 = n, and r < n then it may be verified that αc = (an an−1 · · · a2 a1 )(an−1 · · · a2 )(an ai2 · · · air−1 )a1 − (an−1 · · · a2 a1 )(an an−1 · · · a2 )(an ai2 · · · air−1 )a1 = (an an−1 · · · a2 a1 )(an−1 · · · a2 a1 )(an ai2 · · · air−1 ) − (an−1 · · · a2 a1 )(an an−1 · · · a2 a1 )(an ai2 · · · air−1 ) = 0. By the defining relations, it is clear that ai β = 0, for all 1 < i n, and γ aj = 0 for all 1 < j < n. By looking at the tableau form of the elements of Mn , it is now easy to see that αtβ = 0 for all t ∈ Mn . Therefore K[Mn ] is not prime. Next assume that n 4. Note that αγ = 0 and γ α = 0. Suppose that c = (ai1 · · · air ) is a column such that cα = αc
and cα = 0.
As we have seen above, this implies that r < n and ir = 1. Suppose also that γ c = 0. Since n 4, for all t ∈ supp(γ c) we have by Lemma 3 t∈ / (an an−1 · · · a2 a1 )Mn . Hence the first column in the tableau form of every element in supp(γ c) is of the form (aj1 aj2 · · · ajs a1 ), with s < n − 1. As observed above, we get α(aj1 aj2 · · · ajs a1 ) = 0 in this case. Therefore αγ c = 0. Hence, by looking at the tableau form of the elements of Mn , it is easy to see that αγ wα = 0 for all w ∈ Mn . Therefore αγ K[Mn ]αγ = 0, and thus K[Mn ] is not semiprime. 2
3. Rank 2 plactic algebras In this section we study the plactic algebra of rank 2 over K. Theorem 10. K[M2 ] is a prime K-algebra. Proof. Since M2 is the bicyclic monoid, by [3], see also Theorem 3.3 of [11], for any field L, L[M2 ] is right and left primitive. By Lemma 6, K[M2 ] is prime. 2 Theorem 11. K[M2 ] is a semiprimitive K-algebra.
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Proof. Suppose that J (K[M2 ]) = 0. Let α ∈ J (K[M2 ]) \ {0}. Let z = a2 a1 . We can write j
j
α = λ1 zi1 a11 a2k1 + · · · + λr zir a1r a2kr , where λi ∈ K \ {0} and is , js , ks 0 are such that all the triples (is , js , ks ) are different. Consider the natural projection ψ : K[M2 ] → K M2 . Since M2 is the bicyclic monoid, by [11], K[M2 ] is primitive. It follows that ψ(α) = 0. So we can write p
q
p
q
α = f1 a1 1 a2 1 + · · · + fm a1 m a2 m , for some m 1, pn , qn 0 with all (pn , qn ) different, where each fn is of the form fn = β1 zl1 + · · · + βjn zljn , for some β1 , . . . , βjn ∈ K \ {0}, 0 l1 < l2 < · · · < ljn , such that β1 + · · · + βjn = 0. Note that fn is not a monomial in z. It is known that the Jacobson radical of a Z-graded ring is homogeneous, see Theorem 30.28 of [5]. So J (K[M2 ]) is homogeneous for the gradation defined by the total degree on M2 . Since deg(z) = 2 and deg(a1 ) = deg(a2 ) = 1, we may assume that 2is + js + ks is constant, for s = 1, . . . , r. But then every fn is a monomial in z, a contradiction. Therefore J (K[M2 ]) = 0. 2 Our next aim is to describe the prime spectrum of the plactic algebra K[M2 ]. Let B be the bicyclic monoid B = p, q | pq = 1. It is well known and easy to check that ei = q i pi , i 0, form a chain of idempotents e0 > e1 > e2 > · · · , and fij = q i (1 − e1 )pj , i, j = 0, 1, 2, . . . , form an infinite set of matrix units in K[B], see Example 21.26 in [7]. Let I be the K-linear span of the set {fij }. It is clear that this is the span of all q a pb − q a+1 pb+1 , a, b 0, and therefore I is an ideal of K[B]. It follows that I is a simple ring. Since 1 − ei ∈ I for all i 1, we have that K[B]/I ∼ = K p, p−1 , the group ring of a cyclic infinite group.
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Suppose that P is a nonzero prime ideal of K[B]. If I ⊆ P , then P /I is a prime ideal of K[B]/I . If I ⊆ P , then P ∩ I = {0}, but K[B] is prime, a contradiction. Thus we have described all prime ideals in K[B] and the classical Krull dimension of K[B] is clKdim K[B] = 2. Next we study the prime ideals of K[M2 ]. Let P1 be the ideal of K[M2 ] generated by a1 a2 − a2 a1 . Then K[M2 ]/P1 ∼ = K[x, y], the polynomial ring in two variables. Since, by Theorem 10, K[M2 ] is prime, clKdim K[M2 ] 3. Let P be a prime ideal of K[M2 ]. We consider three cases. Case 1. P ∩ M2 = ∅. Let z = a2 a1 . By 3.8 of [10], P ∩ z = ∅. Since z is central and P is prime, z ∈ P . So the natural map K[M2 ] → K[M2 ]/P factors as K[M2 ] → K[M2 ]/zK[M2 ] → K[M2 ]/P . But the ring in the middle is R = Ka1 , a2 | a2 a1 = 0, so its only minimal prime ideals are a1 R and Ra2 . Thus either a1 ∈ P or a2 ∈ P . Hence, if P is minimal, then K[M2 ]/P ∼ = K[x]. In particular, a maximal chain of primes 0 ⊂ P ⊂ P2 ⊂ · · · can have only two nonzero elements. Case 2. P ∩ M2 = ∅ and P ∩ K[z] = {0}. In this case, there exists an irreducible polynomial f ∈ K[z] such that P ∩ K[z] = f K[z] and, since P ∩ M = ∅, z ∈ / f K[z]. So the natural map K[M2 ] → K[M2 ]/P factors as K[M2 ] → K[z]/f K[z] a1 , a2 | a2 a1 = z → K[M2 ]/P . Now L = K[z]/f K[z] is a field and, since z = 0 in K[z]/f K[z], we have
K[z]/f K[z] a1 , a2 | a2 a1 = z ∼ = L[B],
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where B is the bicyclic monoid. Since L[B] is prime, if P is a minimal nonzero prime then we get P = (P ∩ K[z])K[M2]. Since we know the prime ideals in L[B], this completes the description of prime ideals of K[M2 ] in this case. Case 3. P ∩ K[z] = {0}. Such primes are in 1–1 correspondence with primes in the localization −1 K[M2 ] K[z] \ {0} . By Lemma 6, −1 ∼ K[M2 ] K[z] \ {0} = K(x)[B]. So the description of such primes follows from the description of the primes of K(x)[B], that can be determined as before. As a consequence of the above study we get a complete list of prime ideals of K[M2 ] and we obtain the following result. Theorem 12. clKdim K[M2 ] = 3. 4. Semiprimitivity of K[M3 ] Let M = a, b, c be the plactic monoid of rank 3, with the generators ordered as follows: a < b < c. Our main aim is to show that J (K[M]) = 0. Lemma 13. Let R be a Z-graded ring. Then the homogeneous elements of positive degree in J (R) are nilpotent. Proof. Let α ∈ J (R) be a homogeneous nonzero element of degree r > 0. There exists β ∈ R such that α + β + αβ = 0. Let β = β1 + · · · + βn , where β1 , . . . , βn are the (nonzero) homogeneous components of β of degrees z1 < · · · < zn respectively. Then it is easy to see that z1 = r,
z2 = 2r,
...,
and α + β1 = 0, αβ1 + β2 = 0,
zn = nr,
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...... αβn−1 + βn = 0, αβn = 0. Hence βi = (−α)i and α n+1 = 0.
2
Lemma 14. Let α be a nonzero element of the plactic algebra K[M] of rank 3. Then (i) If α is homogeneous with respect to degc and supp(α) ⊆ a, b, cba then α ∈ / J (K[M]). / (ii) If α is homogeneous with respect to degb and supp(α) ⊆ a, c, cba then α ∈ J (K[M]). / (iii) If α is homogeneous with respect to dega and supp(α) ⊆ b, c, cba then α ∈ J (K[M]). Proof. Suppose that supp(α) ⊆ a, b, cba and α is homogeneous with respect to degc . Let degc (α) = q. Then clearly α = (cba)q α , for some α ∈ K[a, b]. Suppose that α ∈ J (K[M]). Let β ∈ K[a, b]. Then αβ ∈ J (K[M]) and it is homogeneous with respect to degc . If degc (αβ) = 0 then α ∈ K[a, b], and hence α ∈ J (K[a, b]). If degc (αβ) > 0 then by Lemma 13, αβ is nilpotent. Since cba is central and regular, α K[a, b] is a nil right ideal of K[a, b]. But K[a, b] ∼ = K[M2 ] and, by Theorem 11, it is semiprimitive. So in both cases we get a contradiction. Therefore α ∈ / J (K[M]). This proves (i). The proofs of (ii) and (iii) are similar. 2 Lemma 15. Let w, w ∈ M be different elements. Then cbw = cbw if and only if one of the following cases holds: (i) (ii) (iii) (iv) (v) (vi) (vii)
{w, w } = {(cba)q (ba)p (ca)r (a)(c)l+1, (cba)q (ba)p (ca)r+1(c)l }, {w, w } = {(cba)q (ba)p (a)(b)k+1(c)l , (cba)q (ba)p+1(b)k (c)l }, {w, w } = {(cba)q (ba)p (ca)r+1(cb)s (b)(c)l+1, (cba)q (ba)p (ca)r+1 (cb)s+1(c)l }, {w, w } = {(cba)q (ba)p (ca)(cb)s (b)k+2 (c)l , (cba)q (ba)p+1 (cb)s+1(b)k (c)l }, {w, w } = {(cba)q (ca)(cb)s (b)k+1 (c)l , (cba)q+1(cb)s (b)k (c)l }, {w, w } = {(cba)q (ca)(cb)s+1(c)l , (cba)q+1(cb)s (c)l+1 }, {w, w } = {(cba)q (ba)(cb)s+1(b)k (c)l , (cba)q+1(cb)s (b)k+1 (c)l }.
In all the above possible cases, q, p, r, s, k, l are nonnegative integers. Proof. It is easy to check that if one of the cases (i)–(vii) holds then cbw = cbw . Suppose that cbw = cbw . The elements of M have two possible tableau forms: (cba)q (ba)p (ca)r (a)j +1 (b)k (c)l or (cba)q (ba)p (ca)r (cb)s (b)k (c)l ,
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where q, p, r, s, j, k, l are nonnegative integers. The convention here is that the words in parentheses form the subsequent columns of the tableau. Suppose that w = (cba)q (ba)p (ca)r (a)j +1 (b)k (c)l . Then cbw = (cba)q+1(ba)p (ca)r (a)j (b)k (c)l . Suppose that w = (cba)q (ba)p (ca)r (cb)s (b)k (c)l . Then (cba)q+1(ba)p (ca)r−1(cb)s+1(b)k−1 (c)l (cba)q+1(ba)p (ca)r−1(cb)s (c)l+1 cbw = q+1 p−1 s k+1 l (cba) (ba) (cb) (b) (c) q s+1 k l (cba) (cb) (b) (c)
if r > 0 and k > 0, if r > 0 and k = 0, if r = 0 and p > 0, if r = 0 and p = 0.
We say that w is of type (1)–(5) if cbw is of one of these five forms, respectively. Since w = w and cbw = cbw , it is easy to see that w is of type (1)
⇒
w is of type (3) or (4),
w is of type (2)
⇒
w is of type (3), (4) or (5),
w is of type (3)
⇒
w is of type (1), (2), or (5),
w is of type (4)
⇒
w is of type (1), (2) or (5),
w is of type (5)
⇒
w is of type (2), (3) or (4).
So we have the following possible cases. Types (1) and (3): w, w = (cba)q (ba)p (ca)r (a)(c)l+1 , (cba)q (ba)p (ca)r+1(c)l . Types (1) and (4):
w, w = (cba)q (ba)p (a)(b)k+1(c)l , (cba)q (ba)p+1(b)k (c)l .
Types (2) and (3):
w, w = (cba)q (ba)p (ca)r+1(cb)s (b)(c)l+1 , (cba)q (ba)p (ca)r+1(cb)s+1(c)l .
Types (2) and (4): w, w = (cba)q (ba)p (ca)(cb)s (b)k+2 (c)l , (cba)q (ba)p+1(cb)s+1(b)k (c)l . Types (2) and (5): w, w = (cba)q (ca)(cb)s (b)k+1 (c)l , (cba)q+1(cb)s (b)k (c)l .
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Types (3) and (5): w, w = (cba)q (ca)(cb)s+1(c)l , (cba)q+1(cb)s (c)l+1 . Types (4) and (5): w, w = (cba)q (ba)(cb)s+1(b)k (c)l , (cba)q+1(cb)s (b)k+1 (c)l . In all the above possible cases, q, p, r, s, k, l are nonnegative integers. 2 The following is an easy consequence of Lemma 15. Lemma 16. Let w, w ∈ M be different elements. Then cbw = cbw and wba = w ba if and only if one of the following cases holds: (i) {w, w } = {(cba)q (ba)p (ca)r (a)(c), (cba)q (ba)p (ca)r+1}, (ii) {w, w } = {(cba)q (ba)p (ca)r+1(cb)s (b)(c), (cba)q (ba)p (ca)r+1(cb)s+1}, (iii) {w, w } = {(cba)q (ba)(cb)(b)k , (cba)q+1(b)k+1 }, where q, p, r, s, k are nonnegative integers. Before stating the main theorem of this section, we first prove a quite strong and surprising result on homogeneous ideals of K[M], that seems to be of independent interest. Theorem 17. Let K[M] be the plactic algebra of rank 3. Let I be a nonzero ideal that is homogeneous with respect to the degrees in a, b and c. Then there exists a homogeneous nonzero α ∈ I (with respect to these degrees) such that |supp(α)| 2. Proof. Let α ∈ I be a nonzero homogeneous element with respect to the degrees in a, b and c. Write α = ni=1 αi wi , where wi ∈ M and 0 = αi ∈ K. We choose α with a minimal n. If wi u = wj u (or uwi = uwj ), for some u ∈ M and i = j , then we must have αu = 0 (uα = 0 respectively). This will be often used without further comment. From Lemma 3 it follows that if i = degc (α) then α(ba)i ∈ K[a, b, cba]. Suppose that α(ba)i = 0. The elements in supp(α(ba)i ) are of the form (cba)i (ba)p (a)j (b)k , where p + j = dega (α) and p + k = degb (α). As before, the words in parentheses form the subsequent columns of the tableau. Let i = degb (α). Since α is homogeneous, ∀w ∈ supp α(ba)i ,
wa i = (cba)i (ba)i (a)j ,
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where j = dega (α), so that |supp(α(ba)i a i )| 1. Hence, we may assume that α(ba)i a i = 0. Since baa = aba, replacing α by some αa t , we may assume that α(ba)i a = 0. In this case, it is easy to see that α(ba)i = λ (cba)i (ba)i −1 (a)j −i +1 (b) − (cba)i (ba)i (a)j −i , for some nonzero λ ∈ K. Therefore, we may assume that α(ba)i = 0 and replacing α by some α(ba)t , we may assume that αba = 0. A symmetric argument shows that we may also assume that cbα = 0. Since cb and c commute, replacing α by some cj α, we may assume that the tableau of each w ∈ supp(α) contains at most one column (ba). Since αba = 0, the tableau of each w ∈ supp(α) contains at most one column (c). Since ba and a commute, replacing α by some αa t , we may assume that the tableau of each w ∈ supp(α) contains at most one column (cb). Since cbα = 0, we get α=
γj wj − wj
j
where γj ∈ K and every wj , wj is a pair of one of the forms listed in Lemma 15. There are the following possibilities for these pairs: (i) (ii) (iii) (iv) (v) (vi) (vii)
{w1 , w1 } = {(cba)q1 (ba)p1 (ca)r1 (a)(c), (cba)q1 (ba)p1 (ca)r1 +1 }, {w2 , w2 } = {(cba)q2 (a)(b)k2+1 (c)l2 , (cba)q2 (ba)(b)k2 (c)l2 }, {w3 , w3 } = {(cba)q3 (ba)p3 (ca)r3 +1 (b)(c), (cba)q3 (ba)p3 (ca)r3+1 (cb)}, {w4 , w4 } = {(cba)q4 (ca)(b)k4+2 (c)l4 , (cba)q4 (ba)(cb)(b)k4 (c)l4 }, {w5 , w5 } = {(cba)q5 (ca)(cb)s5 (b)k5 +1 (c)l5 , (cba)q5+1 (cb)s5 (b)k5 (c)l5 }, {w6 , w6 } = {(cba)q6 (ca)(cb), (cba)q6+1 (c)}, {w7 , w7 } = {(cba)q7 (ba)(cb)(b)k7 (c)l7 , (cba)q7+1 (b)k7 +1 (c)l7 },
where qi , ri , ki are nonnegative integers and pi , si , li ∈ {0, 1}. Moreover supp(α) can contain at most one pair of each type, except for type (v), where there may be two pairs, corresponding to l5 = 0 and l5 = 1. The following obvious facts will be often used without further comment. (i) dega (w1 ) = q1 +p1 +r1 +1, dega (w1 )−degb (w1 ) = r1 +1, dega (w1 )−degc (w1 ) = p1 , (ii) dega (w2 ) = q2 + 1, dega (w2 ) − degb (w2 ) = −k2 , dega (w2 ) − degc (w2 ) = 1 − l2 , (iii) dega (w3 ) = q3 + p3 + r3 + 1, dega (w3 ) − degb (w3 ) = r3 , dega (w3 ) − degc (w3 ) = p3 − 1, (iv) dega (w4 ) = q4 + 1, dega (w4 ) − degb (w4 ) = −k4 − 1, dega (w4 ) − degc (w4 ) = −l4 , (v) dega (w5 ) = q5 + 1, dega (w5 ) − degb (w5 ) = −s5 − k5 , dega (w5 ) − degc (w5 ) = −s5 − l5 , (vi) dega (w6 ) = q6 + 1, dega (w6 ) − degb (w6 ) = 0, dega (w6 ) − degc (w6 ) = −1,
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(vii) dega (w7 ) = q7 + 1, dega (w7 ) − degb (w7 ) = −k7 − 1, dega (w7 ) − degc (w7 ) = −l7 . Thus there exist γi ∈ K such that α = γ1 (cba)q1 (ba)p1 (ca)r1 (a)(c) − (ca) + γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c)l2 + γ3 (cba)q3 (ba)p3 (ca)r3+1 (b)(c) − (cb) + γ4 (cba)q4 (ca)(b)2 − (ba)(cb) (b)k4 (c)l4 + γ5 (cba)q5 (ca)(b) − (cba) (cb)s5 (b)k5 + γ5 (cba)q5 (ca)(b) − (cba) (cb)s5 (b)k5 (c) + γ6 (cba)q6 (ca)(cb) − (cba)(c) + γ7 (cba)q7 (ba)(cb) − (cba)(b) (b)k7 (c)l7 . We shall proceed in several steps. Step 1. Suppose that γ1 = 0. Since α is homogeneous and dega (w1 ) − degb (w1 ) > 0, we may assume that α = γ1 (cba)q1 (ba)p1 (ca)r1 (a)(c) − (ca) + γ3 (cba)q3 (ba)p3 (ca)r3 +1 (b)(c) − (cb) . We may assume that γ3 = 0. Then, since α is homogeneous, q1 + p1 + r1 = q3 + p3 + r3 ,
r1 + 1 = r3
and p1 = p3 − 1.
Thus p1 = 0, p3 = 1 and α = γ1 (cba)q3+2 (ca)r1 (a)(c) − (ca) + γ3 (cba)q3 (ba)(ca)r1+2 (b)(c) − (cb) . Now cα = γ3 (cba)q3+1 (ca)r1+2 ((b)(c) − (cb)) = 0. Hence we may further assume that γ1 = 0. Step 2. Suppose that γ6 = 0. Since α is homogeneous, dega (w6 ) − degb (w6 ) = 0 and dega (w6 ) − degc (w6 ) = −1, we may assume that α = γ3 (cba)q6 (ca) (b)(c) − (cb) + γ5 (cba)q6 (ca)(b) − (cba) (c) + γ6 (cba)q6 (ca)(cb) − (cba)(c) . Since αba = 0 and αba = γ5 + γ6 (cba)q6+1 (ca)(b) − (cba) , we get γ5 = −γ6 = 0. It follows that α = (γ3 − γ6 )(cba)q6 (ca) (b)(c) − (cb) .
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Therefore we may further assume that γ6 = 0. Step 3. Suppose that γ3 = 0. Since α is homogeneous and dega (w3 ) − degb (w3 ) 0, we may assume that α = γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c)l2 + γ3 (cba)q3 (ba)p3 (ca)r3+1 (b)(c) − (cb) + γ5 (cba)q5 (ca)(b) − (cba) (cb)s5 (b)k5 + γ5 (cba)q5 (ca)(b) − (cba) (cb)s5 (b)k5 (c). We may assume that (γ2 , γ5 , γ5 ) = (0, 0, 0). Since α is homogeneous, r3 = 0 and it is easy to see that we may write α = γ2 (cba)q3+1 (a)(b) − (ba) (c) + γ3 (cba)q3 (ba)p3 (ca) (b)(c) − (cb) + γ (cba)q3+p3 (ca)(b) − (cba) (c)1−p3 , with γ ∈ {γ5 , γ5 } and (γ2 , γ ) = (0, 0). Since αba = 0 and αba = γ2 (cba)q3+2 (a)(b) − (ba) + γ (cba)q3+p3 (ca)(b) − (cba) (c)1−p3 ba, γ2 γ = 0 (because the second summand is nonzero if γ = 0). Thus, since α is homogeneous, p3 = 1, γ = −γ2 and α = γ2 (cba)q3+1 (a)(b) − (ba) (c) + γ3 (cba)q3 (ba)(ca) (b)(c) − (cb) − γ2 (cba)q3+1 (ca)(b) − (cba) . Now we have bα = −γ2 (cba)q3+1 (ba)(cb) − (cba)(b) = 0. Therefore we may further assume that γ3 = 0. Step 4. Suppose that γ2 = 0. Since α is homogeneous, we may assume that α = γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c)l2 + γ4 (cba)q2 (ca)(b)2 − (ba)(cb) (b)k2 −1 + γ5 (cba)q2 (ca)(b) − (cba) (b)k2 + γ7 (cba)q2 (ba)(cb) − (cba)(b) (b)k2 −1 if k2 > 0 and α = γ2 (cba)q2 (a)(b) − (ba) (c)l2 + γ5 (cba)q2 (ca)(b) − (cba)
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if k2 = 0. We may also assume that (γ4 , γ5 , γ7 ) = (0, 0, 0) and γ5 = 0 in these two cases, respectively. Then l2 = 1. If k2 = 0 then bα = γ5 (cba)q2 (ba)(cb) − (cba)(b) = 0. Therefore we may assume that k2 > 0. Since αba = 0 and αba = (γ2 + γ4 + γ5 )(cba)q2+1 (a)(b) − (ba) (b)k2 , we have that γ2 + γ4 + γ5 = 0. Since bα = γ5 (cba)q2 (ba)(cb) − (cba)(b) (b)k2 + γ7 (cba)q2 (ba)(cb) − (cba)(b) (b)k2 , we may assume that γ7 = −γ5 . Now we have α = γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c) + γ4 (cba)q2 (ca)(b)2 − (ba)(cb) (b)k2 −1 + γ5 (cba)q2 (ca)(b)2 − (ba)(cb) (b)k2 −1 = γ2 (cba)q2 (a)(b) − (ba) (b)k2 (c) − γ2 (cba)q2 (ca)(b)2 − (ba)(cb) (b)k2 −1 , αa = γ2 (cba)q2 (ba) (a)(b) − (ba) (b)k2 −1 (c) − γ2 (cba)q2+1 (a)(b) − (ba) (b)k2 −1 , αa k2 = γ2 (cba)q2 (ba)k2 (a)(b) − (ba) (c) − γ2 (cba)q2+1 (ba)k2−1 (a)(b) − (ba) , ck2 αa k2 = γ2 (cba)q2+k2 (a)(b) − (ba) (c) − γ2 (cba)q2+k2 (ca)(b) − (cba) = γ2 (cba)q2+k2 (a)(b)(c) − (ba)(c) − (ca)(b) + (cba) and bck2 αa k2 = γ2 (cba)q2+k2 −(ba)(cb) + (cba)(b) = 0. Therefore we may further assume that γ2 = 0. Step 5. Suppose that γ4 = 0. Since α is homogeneous, α = γ4 (cba)q4 (ca)(b)2 − (ba)(cb) (b)k4 (c)l4 + γ5 (cba)q4 (ca)(b) − (cba) (cb)s5 (b)k5 + γ5 (cba)q4 (ca)(b) − (cba) (cb)s5 (b)k5 (c) + γ7 (cba)q4 (ba)(cb) − (cba)(b) (b)k4 (c)l4 .
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Suppose that l4 = 1. In this case, α = γ4 (cba)q4 (ca)(b)2 − (ba)(cb) (b)k4 (c) + γ5 (cba)q4 (ca)(b) − (cba) (cb)(b)k4 + γ5 (cba)q4 (ca)(b) − (cba) (b)k4 +1 (c) + γ7 (cba)q4 (ba)(cb) − (cba)(b) (b)k4 (c). Since αba = 0 and αba = γ4 (cba)q4+1 (ca)(b)2 − (ba)(cb) (b)k4 + γ5 + γ5 (cba)q4+1 (ca)(b) − (cba) (b)k4 +1 + γ7 (cba)q4+1 (ba)(cb) − (cba)(b) (b)k4 = γ4 + γ5 + γ5 (cba)q4+1 (ca)(b)k4+2 − (γ4 − γ7 )(cba)q4+1 (ba)(cb)(b)k4 − γ5 + γ5 + γ7 (cba)q4+2 (b)k4 +1 , we have γ4 = γ7 = −γ5 − γ5 . Thus α = γ4 + γ5 (cba)q4 (ca)(b) − (cba) (b)k4 +1 (c) + γ5 (cba)q4 (ca)(b) − (cba) (cb)(b)k4 . Hence we may assume that (γ4 + γ5 )γ5 = 0. Now αa k4 +2 = γ4 + γ5 (cba)q4+1 (ba)k4 +1 (a)(c) − (ca) = 0. Therefore, we may assume that l4 = 0. Then α = γ4 (cba)q4 (ca)(b)2 − (ba)(cb) (b)k4 + γ5 (cba)q4 (ca)(b) − (cba) (b)k4 +1 + γ7 (cba)q4 (ba)(cb) − (cba)(b) (b)k4 . Since αba = 0 and αba = (γ4 + γ5 )(cba)q4+1 (a)(b) − (ba) (b)k4 +1 , γ5 = −γ4 . Now we have α = γ4 (cba)q4 −(ba)(cb) + (cba)(b) (b)k4 + γ7 (cba)q4 (ba)(cb) − (cba)(b) (b)k4 = (γ7 − γ4 )(cba)q4 (ba)(cb) − (cba)(b) (b)k4 . Therefore we may further assume that γ4 = 0.
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Step 6. Suppose that γ7 = 0. Since α is homogeneous, we may write α = γ5 (cba)q7 (ca)(b) − (cba) (cb)s5 (b)k7 +1−s5 + γ5 (cba)q7 (ca)(b) − (cba) (b)k7 +1 (c) + γ7 (cba)q7 (ba)(cb) − (cba)(b) (b)k7 (c)l7 . Suppose that l7 = 1. Then we may assume s5 = 1 and we may write α = γ5 (cba)q7 (ca)(b) − (cba) (cb)(b)k7 + γ5 (cba)q7 (ca)(b) − (cba) (b)k7 +1 (c) + γ7 (cba)q7 (ba)(cb) − (cba)(b) (b)k7 (c). Since αba = 0 and αba = γ5 + γ5 (cba)q7+1 (ca)(b) − (cba) (b)k7 +1 + γ7 (cba)q7+1 (ba)(cb) − (cba)(b) (b)k7 , we have a contradiction. Hence l7 = 0 and we may assume α = γ5 (cba)q7 (ca)(b) − (cba) (b)k7 +1 + γ7 (cba)q7 (ba)(cb) − (cba)(b) (b)k7 . Since αba = 0 and αba = γ5 (cba)q7+1 (a)(b) − (ba) (b)k7 +1 , γ5 = 0. Therefore we may further assume that γ7 = 0. Step 7. It remains to consider the case where only γ5 , γ5 can be nonzero. Hence we may assume that γ5 γ5 = 0 and α = γ5 (cba)q5 (ca)(b) − (cba) (cb)(b)k5 + γ5 (cba)q5 (ca)(b) − (cba) (b)k5 +1 (c). Since αba = 0 and αba = γ5 + γ5 (cba)q5+1 (ca)(b) − (cba) (b)k5 +1 , γ5 = −γ5 . Now we have α = γ5 (cba)q5 (ca)(b) − (cba) (cb)(b)k5 − γ5 (cba)q5 (ca)(b) − (cba) (b)k5 +1 (c), αa k5 +1 = γ5 (cba)q5+1 (ca)(b) − (cba) (ba)k5 − γ5 (cba)q5+1 (a)(b) − (ba) (ba)k5 (c), ck5 αa k5 +1 = γ5 (cba)q5+k5 +1 (ca)(b) − (cba) − γ5 (cba)q5+k5 +1 (a)(b) − (ba) (c)
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and bck5 αa k5 +1 = γ5 (cba)q5+k5 +1 (ba)(cb) − (cba)(b) = 0. This completes the proof of the theorem. 2 Theorem 18. Let K[M] be the plactic algebra of rank 3. Then K[M] is semiprimitive. Proof. As mentioned before, the Jacobson radical of a Z-graded ring is homogeneous, see [5, Theorem 30.28]. Therefore if J (K[M]) = 0 then we may choose a nonzero element 0 = α ∈ J (K[M]) which is homogeneous with respect to the gradation on K[M] defined by the length of words in M, and also by the degree in each of the generators. By Theorem 17, we may assume that α = α1 w1 + α2 w2 , where wi ∈ M and 0 = αi ∈ K. By Lemma 13, w1 = w2 . Furthermore, if w1 u = w2 u (or uw1 = uw2 ), for some u ∈ M, then we must have αu = 0 (uα = 0, respectively). This will be used without further comment. From Lemma 3 it follows that if i = degc (α) then α(ba)i ∈ K[a, b, cba]. By Lemma 14, α(ba)i = 0. Therefore, replacing α by some α(ba)j , we may assume that αba = 0. Similarly, replacing α by some (cb)j α, we may assume that cbα = 0. By Lemma 16, we may assume that α is one of the following elements: (cba)q (ba)p (ca)r (a)(c) − (cba)q (ba)p (ca)r+1, (cba)q (ba)p (ca)r+1(cb)s (b)(c) − (cba)q (ba)p (ca)r+1(cb)s+1, (cba)q (ba)(cb)(b)k − (cba)q+1(b)k+1 , where q, p, r, s, k are nonnegative integers. If α = (cba)q (ba)p (ca)r (a)(c) − (cba)q (ba)p (ca)r+1 then cp α = (cba)q+p (ca)r (a)(c) − (ca)r+1 = 0, which contradicts Lemma 14. If α = (cba)q (ba)p (ca)r+1(cb)s (b)(c) − (cba)q (ba)p (ca)r+1(cb)s+1 then cp αa s+1 = (cba)q+p+s+1(ca)r (a)(c) − (ca) = 0, which contradicts Lemma 14. If α = (cba)q (ba)(cb)(b)k − (cba)q+1(b)k+1 then, by induction on n, it is easy to see that α n = (−1)n−1 (cba)nq+n−1 (ba)(cb) − (cba)(b) (b)nk+n−1 = 0, for all n, in contradiction with Lemma 13. Hence K[M] is semiprimitive. 2
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Acknowledgments This work was partially supported by DGESID PB98-0873, Spain and the Generalitat de Catalunya, and by KBN research grant 2 P03A 030 18, Poland. The second author is also grateful for the warm hospitality of the Centre de Recerca Matematica (Barcelona), where a part of the work was done.
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