Limit T-subalgebras in free associative algebras

Limit T-subalgebras in free associative algebras

Journal of Algebra 412 (2014) 264–280 Contents lists available at ScienceDirect Journal of Algebra www.elsevier.com/locate/jalgebra Limit T -subalg...

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Journal of Algebra 412 (2014) 264–280

Contents lists available at ScienceDirect

Journal of Algebra www.elsevier.com/locate/jalgebra

Limit T -subalgebras in free associative algebras Dimas José Gonçalves a,∗,1 , Alexei Krasilnikov b,2 , Irina Sviridova b,3 a Departamento de Matemática, Universidade Federal de São Carlos, 13565-905, São Carlos, SP, Brazil b Departamento de Matemática, Universidade de Brasília, 70910-900 Brasília, DF, Brazil

a r t i c l e

i n f o

Article history: Received 31 October 2012 Available online 2 June 2014 Communicated by Louis Rowen MSC: 16R10 16R40 16R99 16S10 8B20 8A30 Keywords: Free associative algebra Polynomial identity

a b s t r a c t Let F X be the free unitary associative algebra over a field F on a free generating set X. An unitary subalgebra R of F X is called a T -subalgebra if R is closed under all endomorphisms of F X. A T -subalgebra R∗ in F X is limit if every larger T -subalgebra W  R∗ is finitely generated (as a T -subalgebra) but R∗ itself is not. It follows easily from Zorn’s lemma that if a T -subalgebra R is not finitely generated then it is contained in some limit T -subalgebra R∗ . In this sense limit T -subalgebras form a “border” between those T -subalgebras which are finitely generated and those which are not. In the present article we give the first example of a limit T -subalgebra in F X, where F is an infinite field of characteristic p > 2 and |X| ≥ 4. Note that, by Shchigolev’s result, over a field F of characteristic 0 every T -subalgebra in F X is finitely

* Corresponding author. E-mail addresses: [email protected] (D.J. Gonçalves), [email protected] (A. Krasilnikov), [email protected] (I. Sviridova). 1 Supported by CNPq (480139/2012-1), DPP/UnB, CNPq-FAPDF PRONEX grant 2009/00091-0 (193.000.580/2009), and by CAPES. 2 Supported by CNPq (307328/2012-0 and 480139/2012-1), DPP/UnB, CNPq-FAPDF PRONEX grant 2009/00091-0 (193.000.580/2009), and by CAPES. 3 Supported by CNPq (480139/2012-1), DPP/UnB, FEMAT, CNPq-FAPDF PRONEX grant 2009/00091-0 (193.000.580/2009), and by CAPES. http://dx.doi.org/10.1016/j.jalgebra.2014.03.032 0021-8693/© 2014 Elsevier Inc. All rights reserved.

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generated; hence, over such a field limit T -subalgebras in F X do not exist. © 2014 Elsevier Inc. All rights reserved.

1. Introduction Let F be a field, X a non-empty set and let F X be the free unitary associative algebra over F on the free generating set X. Recall that a T-ideal of F X is an ideal closed under all endomorphisms of F X. Similarly, a T -subalgebra and a T -subspace (the latter is also called a T -space) are, respectively, an unitary subalgebra and a vector subspace in F X closed under all endomorphisms of F X. Let A be an unitary associative algebra over F . Recall that a polynomial f (x1 , . . . , xn ) ∈ F X is called a polynomial identity in A if f (a1 , . . . , an ) = 0 for all a1 , . . . , an ∈ A. It can be easily checked that, for a given algebra A, its polynomial identities form a T -ideal T (A) in F X. The converse also holds: for every T -ideal I in F X there is an algebra A such that I = T (A), that is, I is the ideal of all polynomial identities satisfied in A. A polynomial f (x1 , . . . , xn ) ∈ F X is a central polynomial of A if, for all a1 , . . . , an ∈ A, f (a1 , . . . , an ) is central in A. Clearly, f = f (x1 , . . . , xn ) is a central polynomial of A if and only if [f, xn+1 ] is a polynomial identity of A. For a given algebra A its central polynomials form a T -subalgebra C(A) in F X. However, not every T -subalgebra coincides with the T -subalgebra C(A) of all central polynomials of some algebra A. Let I be a T -ideal in F X. A subset S ⊂ I generates I as a T -ideal if I is the minimal T -ideal in F X containing S. A T -subalgebra and a T -subspace of F X generated by S (as a T -subalgebra and a T -subspace, respectively) are defined in a similar way. It is clear that the T -ideal (T -subalgebra, T -subspace) generated by S is the ideal (the subalgebra, the vector subspace) in F X generated by all polynomials f (g1 , . . . , gm ), where f = f (x1 , . . . , xm ) ∈ S and gi ∈ F X for all i. We refer to [7,9,17,26] for further terminology and basic results concerning T -ideals and algebras with polynomial identities and to [2,6,14,15,17] for an account of results concerning T -subspaces. If F is a field of characteristic 0 then every T -ideal in F X is finitely generated (as a T -ideal); this is a celebrated result of Kemer [18,19] that solves the Specht problem. Moreover, over such a field F each T -subspace (and, therefore, each T -subalgebra) in F X is finitely generated; this has been proved more recently by Shchigolev [29]. Very recently Belov [5] has proved that, for each Noetherian commutative and associative unitary ring K and each n ∈ N, each T -ideal in Kx1 , . . . , xn  is finitely generated. On the other hand, over a field F of characteristic p > 0 there are T -ideals in F x1 , x2 , . . . that are not finitely generated. This has been proved by Belov [3], Grishin [11] and Shchigolev [27] (see also [4,12,17,29]). The construction of such T -ideals

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makes use of the non-finitely generated T -subspaces in F x1 , x2 , . . . constructed by Grishin [11] for p = 2 and by Shchigolev [28] for p > 2 (see also [12]). Shchigolev [28] also constructed non-finitely generated T -subspaces and T -subalgebras in F x1 , . . . , xn , where n > 1 and F is a field of characteristic p > 2. A T -subalgebra R∗ in F X is called limit if every larger T -subalgebra W  R∗ is finitely generated (as a T -subalgebra) but R∗ itself is not. A limit T -ideal and a limit T -subspace are defined in a similar way. It follows easily from Zorn’s lemma that if a T -subalgebra R is not finitely generated then it is contained in some limit T -subalgebra R∗ . Similarly, each non-finitely generated T -ideal (T -subspace) is contained in a limit T -ideal (T -subspace). In this sense limit T -subalgebras (T -ideals, T -subspaces) form a “border” between those T -subalgebras (T -ideals, T -subspaces) which are finitely generated and those which are not. By [3,11,27], over a field F of characteristic p > 0 the algebra F x1 , x2 , . . . contains non-finitely generated T -ideals; therefore, it contains at least one limit T -ideal. No example of a limit T -ideal is known so far. Even the cardinality of the set of limit T -ideals in F x1 , x2 , . . . is unknown; it is possible that, for a given field F of characteristic p > 0, there is only one limit T -ideal. The non-finitely generated T -ideals constructed in [1,16] come closer to being limit than any other known non-finitely generated T -ideal. However, it is unlikely that the T -ideals found in [1,16] are limit. Note that a similar problem concerning limit verbal subgroups in a free group remains open for 40 years; there exist infinitely many such subgroups [24] but no example is known (see [22] for more details). About limit T -subspaces in F X we know more than about limit T -ideals. Brandão Jr., Koshlukov, Krasilnikov and Silva [6] have proved that the vector space C(G) of all central polynomials of the infinite dimensional Grassmann algebra G over an infinite field F of characteristic p > 2 is a limit T -subspace in F x1 , x2 , . . .. Very recently, Gonçalves, Krasilnikov and Sviridova [10] have found infinitely many other limit T -subspaces in F x1 , x2 , . . .. However, it can be easily verified that the limit T -subspaces described in [6,10] are not limit T -subalgebras: C(G) is finitely generated as a T -subalgebra and the limit T -subspaces of [10] are not subalgebras in F x1 , x2 , . . .. The aim of the present article is to construct the first example of a limit T -subalgebra in F x1 , x2 , . . .. We write [a, b] = ab − ba, [a, b, c] = [[a, b], c]. For each l ≥ 0, let q (l) (x1 , x2 ) = x1p −1 [x1 , x2 ]x2p −1 . l

l

Our main result is as follows. Theorem 1. Let F be an infinite field of characteristic p > 2. Let Γ be the (unitary) T -subalgebra in F x1 , x2 , . . . generated by the polynomials q (0) , q (1) , . . . , q (l) , . . .

(1)

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together with the polynomials x1 [x2 , x3 , x4 ] and x1 [x2 , x3 ][x4 , x5 ]. Then Γ is a limit T -subalgebra in F x1 , x2 , . . .. Note that the T -subspace in F x1 , x2 , . . . generated by the polynomials x1 [x2 , x3 , x4 ] and x1 [x2 , x3 ][x4 , x5 ] coincides with the T -ideal T (3,2) generated (as a T -ideal) by [x2 , x3 , x4 ] and [x2 , x3 ][x4 , x5 ]. Thus, Γ is the T -subalgebra generated by the T -ideal T (3,2) together with the polynomials (1). We also prove the following. Theorem 2. Let F be an infinite field of characteristic p > 2. Let Γn = Γ ∩F x1 , . . . , xn . Then Γn is a limit T -subalgebra in F x1 , . . . , xn  for every n ≥ 4. We make the following conjecture. Conjecture. Let F be an infinite field of characteristic p > 2. Then Γ is the only limit T -subalgebra in F x1 , x2 , . . . and, for each n ≥ 4, Γn is the only limit T -subalgebra in F x1 , . . . , xn . Recall that it follows from [3,11,27] that over a field F of characteristic p > 0 there exists at least one limit T -ideal in F x1 , x2 , . . .. Problem. For a field of characteristic p > 0, find an example of a limit T -ideal in F x1 , x2 , . . .. How many such limit T -ideals exist for a given (infinite) field F of characteristic p > 0? Remarks. 1. As a T -subspace Γ is generated by 1 together with the polynomials (1) and the polynomials x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ]. Γ is not finitely generated as a T -subspace in F x1 , x2 , . . . (otherwise it would be finitely generated as a T -subalgebra as well). However, Γ is not a limit T -subspace; it is contained in the (larger) limit T -subspace generated by xp1 and Γ itself (see [10]). 2. The limit T -subalgebra Γ is not equal to the T -subalgebra C(A) of all central polynomials of any algebra A. Indeed, suppose that Γ = C(A) for some A. Let T (A) be the T -ideal of all polynomial identities of A. Then, for each f ∈ C(A) and each g ∈ F X, we have [f, g] ∈ T (A). Since [x1 , x2 ] ∈ Γ = C(A), we have [x1 , x2 , x3 ] ∈ T (A). Note that, over a field of characteristic p > 0, [x2 , xp1 ] = [x2 , x1 , . . . , x1 ], where the commutator on the right hand side of the equality is of length p + 1. Since [x1 , x2 , x3 ] ∈ T (A), we have [x2 , x1 , . . . , x1 ] ∈ T (A), hence [x2 , xp1 ] ∈ T (A). It follows that xp1 ∈ C(A); / Γ . This contradiction proves that Γ = C(A) for any algebra A, on the other hand, xp1 ∈ as claimed. 3. Recall that G is the infinite dimensional Grassmann algebra over a field F and C(G) is the vector space of all central polynomials of G. Let F be an infinite field of characteristic p > 2. Then, by [6], C(G) is a limit T -subspace in F x1 , x2 , . . ., that is,

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C(G) is not a finitely generated T -subspace but every larger T -subspace W  C(G) is finitely generated. We note that the former part of this assertion (stating that C(G) is not a finitely generated T -subspace) was also proved independently by Bekh-Ochir and Rankin [2] and by Grishin [13]. 4. The first example of a limit T -subspace in the free non-unitary associative algebra of countable infinite rank over a field F of characteristic p > 0 was found by Kireeva [20]. If p > 2 then Kireeva’s limit T -subspace coincides with the vector space C(H) of all central polynomials of the infinite dimensional non-unitary Grassmann algebra H (see [6]). 2. Preliminaries Note that if I is a T -ideal in F X then T -ideals, T -subalgebras and T -subspaces can be defined in the quotient algebra F X/I in a natural way. In the rest of the paper F will be an infinite field of characteristic p > 2. Let X = {x1 , x2 , . . .} be an infinite countable set, and F X be the free unitary associative algebra over F generated by X. Let T (3) denote the T -ideal of F X generated by the polynomial [x1 , x2 , x3 ]. Then T (3) is the ideal of F X generated by all polynomials [g1 , g2 , g3 ] (gi ∈ F X). The next basic properties of T (3) are well-known (see for example [2,6,8,14,25]). Lemma 3. Let F be a field of characteristic p > 2. Let g1 , g2 ∈ F X. Then 1. g1p + T (3) is central in F X/T (3) ; 2. (g1 + g2 )p + T (3) = (g1p + g2p ) + T (3) ; 3. (g1 g2 )p + T (3) = g1p g2p + T (3) . Recall that T (3,2) is the T -ideal in F X generated by the polynomials [x1 , x2 , x3 ] and [x1 , x2 ][x3 , x4 ]. Since T (3) ⊂ T (3,2) , the statement of Lemma 3 remains valid if we replace T (3) by T (3,2) there. It can be easily seen that if g1 , g2 , . . . , gn−1 , h1 , h2 ∈ F X and gn = [h1 , h2 ] then g1 g2 . . . gn + T (3,2) = gσ(1) gσ(2) . . . gσ(n) + T (3,2)

(2)

for any permutation σ ∈ Sn . For integers i1 , i2 such that 1 ≤ i1 < i2 ≤ n and integers a1 , . . . , an ≥ 0 such that ai1 , ai2 ≥ 1, xa1 1 xa2 2 . . . xann xi1 xi2 denotes the monomial xb11 xb22 . . . xbnn ∈ F X, where bj = aj −1 if j ∈ {i1 , i2 } and bj = aj otherwise. The next lemma is a corollary of [7, Proposition 4.3.3].

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Lemma 4. If f (x1 , . . . , xn ) ∈ F X is a multihomogeneous polynomial of multidegree (a1 , . . . , an ) then f + T (3,2) = αxa1 1 . . . xann +



α(i1 ,i2 )

1≤i1
xa1 1 . . . xann [xi1 , xi2 ] + T (3,2) , xi1 xi2

where α, α(i1 ,i2 ) ∈ F . Recall that Γ is the (unitary) T -subalgebra generated by polynomials q (0) , q (1) , . . . , q (l) , . . . , x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ].

(3)

Note that T (3,2) ⊂ Γ . Indeed, x1 [x2 , x3 , x4 ]x5 = x1 [x2 , x3 , x4 x5 ] − x1 x4 [x2 , x3 , x5 ] so T (3) ⊆ Γ . Further, x1 [x2 , x3 ][x4 , x5 ]x6 + T (3) = x1 [x2 , x3 ]x6 [x4 , x5 ] + T (3) = x1 x6 [x2 , x3 ][x4 , x5 ] + T (3) so the T -ideal generated by [x1 , x2 ][x3 , x4 ] is also contained in Γ . Thus, T (3,2) ⊂ Γ , as claimed. For any g ∈ F X, let gTS denote the T -subspace in F X generated by g. The following lemma is a corollary of a result of Grishin and Tsybulya [14, Theorem 1.3, item 1)]. Lemma 5. (See [14].) Let F be an infinite field of characteristic p > 2. Let a1 , a2 ≥ 1 and let l ≥ 0 be the largest integer such that pl divides both a1 , a2 . Then 

TS  TS x1a1 −1 x2a2 −1 [x1 , x2 ] + T (3) = q (l) + T (3) .

Lemma 5 immediately implies the following. Lemma 6. Let F be an infinite field of characteristic p > 2. For any integers a1 , a2 ≥ 1, the polynomial x1a1 −1 x2a2 −1 [x1 , x2 ] belongs to Γ . 3. Proof of Theorem 1 Let f ∈ F X be a polynomial. We denote by Γf the T -subalgebra of F X generated by f and Γ and we write Γ (s) (s ≥ 0) for the T -subalgebra generated by Γ and by the polynomial s

f = xp1 q (s) (x2 , x3 ). In particular, Γ (0) is the T -subalgebra generated by x1 [x2 , x3 ] and Γ .

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Lemma 7. Γ (s) is generated as a T -subalgebra by the polynomials s

xp1 q (s) (x2 , x3 ), x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ].

(4)

In particular, Γ (0) is generated as a T -subalgebra by the polinomial x1 [x2 , x3 ]. Proof. Let Γ (s)∗ be the T -subalgebra of F X generated by the polynomials (4). It is clear that Γ (s)∗ ⊂ Γ (s). Let s = 0. It can be easily seen that Γ (0)∗ is generated as a T -subalgebra by the polynomial x1 [x2 , x3 ]. Since T (3,2) ⊂ Γ (0)∗ and q (l) (x1 , x2 ) + T (3,2) = x1p −1 x2p −1 [x1 , x2 ] + T (3,2) , l

l

we obtain q (l) ∈ Γ (0)∗ for all l ≥ 0. Hence, Γ (0)∗ = Γ (0). Let s > 0. If l > s, then by (2) we have  ps l−s q (l) (x1 , x2 ) + T (3,2) = (x1 x2 )(p −1) q (s) (x1 , x2 ) + T (3,2) . Hence q (l) ∈ Γ (s)∗ . s Suppose that l ≤ s. Take f (x1 , x2 , x3 ) = xp1 q (s) (x2 , x3 ). Note that the multihomogeneous component of multidegree (pl , pl ) of the polynomial s

f (1, x1 + 1, x2 + 1) = (x1 + 1)p

−1

s

[x1 + 1, x2 + 1](x2 + 1)p

−1

 s −12 is equal to αq (l) (x1 , x2 ) where α = ppl −1 . Since f (1, x1 + 1, x2 + 1) ∈ Γ (s)∗ and α = 0 in F , we have q (l) ∈ Γ (s)∗ . Thus, q (l) ∈ Γ (s)∗ for all l ≥ 0. Since T (3,2) ⊂ Γ (s)∗ , we have Γ (s)∗ = Γ (s), as required. 2 Lemma 8. Let f (x1 , . . . , xn ) ∈ F X be a multihomogeneous polynomial of multidegree (a1 , . . . , an ). If ai = 1 for some i, then either Γf = F X

or

Γf = Γ

or

Γf = Γ (0).

Proof. Note that we can assume without loss of generality, permuting the generators x1 , . . . , xn if necessary, that a1 = 1. We also can suppose that f = αxa1 1 . . . xann +

 1≤i1
α(i1 ,i2 )

xa1 1 . . . xann [xi1 , xi2 ], xi1 xi2

(5)

where α, α(i1 ,i2 ) ∈ F . This follows easily from Lemma 4 and from the fact that Γf +h = Γf for any polynomial h ∈ T (3,2) ⊂ Γ . If α = 0 in (5) then α−1 f (x1 , 1, . . . , 1) = x1 ∈ Γf , and Γf = F X.

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Suppose that α = 0. We claim that we may assume without loss of generality that f is of the form f (x1 , . . . , xn ) = x1 g(x2 , . . . , xn ), where g=



α(i1 ,i2 )

2≤i1
xa2 2 . . . xann [xi1 , xi2 ]. xi1 xi2

(6)

a

Indeed, consider a term m =

n x1 1 ...xa n xi1 xi2

m = x1

[xi1 , xi2 ] in (5). If i1 > 1 then

xa2 2 . . . xann [xi1 , xi2 ]. xi1 xi2

Suppose that i1 = 1; then m = m [x1 , xi2 ], where m =

(7) a

n x2 2 ...xa n xi2

. We have

    m + T (3,2) = m [x1 , xi2 ] + T (3,2) = m x1 , xi2 − x1 m , xi2 + T (3,2) . Note that [m x1 , xi2 ] ∈ Γ . Using the equality [uv, xi2 ] = [u, xi2 ]v + u[v, xi2 ] and the property (2), we obtain that [m , xi2 ] = m + h , where m is of the form (6) and h ∈ T (3,2) . Hence, we can write m = x1 g  + h

(8)

where g  is of the form (6) and h ∈ Γ . It follows easily from (7) and (8) that there exists a multihomogeneous polynomial g = g(x2 , . . . , xn ) ∈ F X of the form (6) such that f = x1 g + h, where h ∈ Γ . Since Γf = Γx1 g we can assume without loss of generality (replacing f with x1 g) that f = x1 g(x2 , . . . , xn ), where g is of the form (6), as claimed. It is clear that if f = 0 then Γf = Γ . Suppose that f = 0. Permuting the generators x2 , . . . , xn if necessary we can assume that α(2,3) = 0; then f is of the form xa2 . . . xann f = x1 α(2,3) 2 [x2 , x3 ] + x2 x3



α(i1 ,i2 )

(i1 ,i2 )=(2,3)

xa2 2 . . . xann [xi1 , xi2 ] . xi1 xi2

Let f1 (x1 , x2 , x3 ) = f (x1 , x2 , x3 , 1, . . . , 1) ∈ Γf ; then f1 = α(2,3) x1 x2a2 −1 x3a3 −1 [x2 , x3 ]. It can be easily seen that the multihomogeneous component of multidegree (1, 1, 1) of the polynomial f1 (x1 , x2 +1, x3 +1) is equal to α(2,3) x1 [x2 , x3 ]. It follows that x1 [x2 , x3 ] ∈ Γf and therefore Γ (0) ⊆ Γf . Let T (2) be the T -ideal in F X generated by [x1 , x2 ]. Clearly, f ∈ T (2) . On the other hand, since x1 [x2 , x3 ]x4 = x1 [x2 , x3 x4 ] − x1 x3 [x2 , x4 ],

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we have T (2) ⊂ Γ (0). Hence, f ∈ Γ (0) and Γf ⊆ Γ (0). It follows that Γf = Γ (0). The proof of Lemma 8 is completed. 2 Lemma 9. Let f = f (x1 , . . . , xn ) ∈ F X be a multihomogeneous polynomial of the form f=



s1

1≤i1
sn

xp . . . xpn α(i1 ,i2 ) 1 [xi1 , xi2 ], xi1 xi2

(9)

where α(i1 ,i2 ) ∈ F , si ≥ 1 for all i. Then Γf = Γ or Γf = Γ (s) where s is the minimum of s1 , s2 , . . . , sn . Proof. We can assume without loss of generality (permuting the free generators x1 , . . . , xn if necessary) that s = s1 ≤ si for all i = 1, . . . , n. s1 s2 If n = 1 then f = 0, and if n = 2 then f = α(1,2) x1p −1 x2p −1 [x1 , x2 ]. Thus, by Lemma 6, in both cases we have Γf = Γ . Suppose that n > 2. We claim that we may assume without loss of generality that f is of the form s

f (x1 , . . . , xn ) = xp1 g(x2 , . . . , xn ),

(10)

where g=



s2

2≤i1
Indeed, consider a term m =

xp 1

sn

...xp n xi1 xi2

[xi1 , xi2 ] in (9). If i1 > 1 then

s2

s

m = xp1

sn

xp . . . xpn α(i1 ,i2 ) 2 [xi1 , xi2 ]. xi1 xi2

sn

xp2 . . . xpn [xi1 , xi2 ]. xi1 xi2

(11)

Suppose that i1 = 1. Let ai = psi for all i. Then s

m + T (3,2) = x1p

ps2 −1 x2

sn

. . . xpn [x1 , xi2 ] + T (3,2) xi2 aj

= x1a1 −1 xj1j1 . . . xjl l [x1 , xi2 ]xi2i2 a

a

−1

+ T (3,2) ,

where {j1 , . . . , jl } = {1, . . . , n} \ {1, i2 }, l = n − 2 > 0. Suppose that a1 = aj1 = . . . = ajz

and ajz+1 , ajz+2 , . . . , ajl > a1 ,

Let aj

aj

z+1 u = x1 xj1 · · · xjz xjz+1 · · · x jl l ,

0 ≤ z ≤ l.

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where ai = ai /ps for all i. Let h = h(x1 , x2 ) = x1a1 −1 [x1 , x2 ]x2 i2 have h ∈ Γ . It follows that a

s

h(u, xi2 ) = up

−1

a

[u, xi2 ]xi2i2

−1

−1

273

. By Lemma 6, we

∈ Γ.

(12)

Since, by Lemma 3, [v1p , v2 ] ∈ T (3) ⊂ T (3,2) for all v1 , v2 ∈ F X, we have h(u, xi2 ) + T (3,2) s

= (x1 xj1 · · · xjz )p If z = 0 then h(u, xi2 ) ≡

−1 ajz+1 xjz+1

s1

xp 1

sn

...xp n x1 xi2

aj

a

· · · xjl l [x1 xj1 . . . xjz , xi2 ]xi2i2

−1

+ T (3,2) .

[x1 , xi2 ] (mod T (3,2) ). Thus, by (12), s1

sn

xp . . . xpn m= 1 [x1 , xi2 ] ∈ Γ. x1 xi2

(13)

Suppose that z > 0. Then h(u, xi2 ) + T (3,2) = h1 + h2 + T (3,2) , where aj

a

−1

aj

a

−1

s

−1 ajz+1 xjz+1

· · · xjl l [x1 , xi2 ]xj1 . . . xjz xi2i2

s

−1 ajz+1 xjz+1

· · · xjl l x1 [xj1 . . . xjz , xi2 ]xi2i2

h1 = (x1 xj1 · · · xjz )p h2 = (x1 xj1 · · · xjz )p

, .

It can be easily checked using (2) that s1

h1 + T

(3,2)

h2 + T

(3,2)

sn

xp . . . xpn = 1 [x1 , xi2 ] + T (3,2) = m + T (3,2) , x1 xi2 

=

s2

s β(i1 ,i2 ) xp1

2≤i1
sn

xp2 . . . xpn [xi1 , xi2 ] + T (3,2) xi1 xi2

for some β(i1 ,i2 ) ∈ F . Since T (3,2) ⊂ Γ , by (12) we obtain 

m+

s2

s

β(i1 ,i2 ) xp1

2≤i1
sn

xp2 . . . xpn [xi1 , xi2 ] ∈ Γ. xi1 xi2

(14)

By (11), (13) and (14), we can write f = f1 + f2 , where f1 =

s xp1



 2≤i1
s2

sn xp2 . . . xpn γ(i1 ,i2 ) [xi1 , xi2 ] xi1 xi2

is of the form (10) and f2 ∈ Γ . It is clear that Γf = Γf1 . Thus, we can assume (replacing f with f1 ) that the polynomial f is of the form (10), as claimed.

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274

If f = 0 then Γf = Γ . Suppose that f = 0. Then we can assume without loss of generality that α(2,3) = 0. It follows that the T -subalgebra Γf contains the polynomial s

s2

−1 h(x1 , x2 , x3 ) = α(2,3) f (x1 , x2 , x3 , 1, 1, . . . , 1) = xp1 x2p

−1 ps3 −1 x3 [x2 , x3 ].

Then Γf also contains the multihomogeneous component of the polynomial h(x1 , x2 + 1, x3 + 1) of degree ps in each variable xi (i = 1, 2, 3), that is equal, modulo T (3,2) , to s

s

γxp1 x2p

−1 ps −1 x3 [x2 , x3 ],

 s2 −1ps3 −1 ps (s) where γ = pps −1 (x2 , x3 ) ∈ Γf and hence ps −1 ≡ 1 (mod p). It follows that x1 q Γ (s) ⊆ Γf . Now let us prove that Γf ⊆ Γ (s). We have s

s2

xp1 x2p

−1 ps3 −1 ps4 x3 x4

sn

. . . xpn [x2 , x3 ] + T (3,2)

= xp1 xp2 −p xp3 −p xp4 . . . xpn x2p −1 [x2 , x3 ]x3p −1 + T (3,2)  s2 −s sn −s ps −1 ps3 −s −1 ps4 −s = x1 x2p x3 x4 . . . xpn q (s) (x2 , x3 ) + T (3,2) s

s2

s

s3

s

s4

sn

s

s

so s

s2

xp1 x2p

−1 ps3 −1 ps4 x3 x4

sn

. . . xpn [x2 , x3 ] ∈ Γ (s).

Similarly we can prove that, for all i1 , i2 such that 2 ≤ i1 < i2 ≤ n, we have s2

s xp1

sn

xp2 . . . xpn [xi1 , xi2 ] ∈ Γ (s). xi1 xi2

Since f has the form (10), we get f ∈ Γ (s). It follows that Γf ⊆ Γ (s) and therefore Γf = Γ (s). The proof of Lemma 9 is completed. 2 Lemma 10. Let f = f (x1 , . . . , xn ) be a multihomogeneous polynomial of multidegree (ps1 , . . . , psn ), where si ≥ 0 for all i. Denote by s the minimum of s1 , s2 , . . . , sn . Then one of the following holds: 1. 2. 3. 4.

Γf Γf Γf Γf

= F X; = Γ; = Γ (s); s is the T -subalgebra generated by xp1 and Γ (s).

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275

Proof. Since a T -subalgebra is invariant under automorphisms of F X, we may assume without loss of generality that s = s1 . If s = 0 then, by Lemma 8, we have either Γf = F X or Γf = Γ or Γf = Γ (0). Suppose that s ≥ 1. By Lemma 4 we can assume that s1

sn

f = αxp1 . . . xpn



+

s1

α(i1 ,i2 )

1≤i1
sn

xp1 . . . xpn [xi1 , xi2 ], xi1 xi2

where α, α(i1 ,i2 ) ∈ F . Denote by h(x1 , . . . , xn ) the following polynomial h=

 1≤i1
s1

α(i1 ,i2 )

sn

xp1 . . . xpn [xi1 , xi2 ]. xi1 xi2

If α = 0 then f = h, and by Lemma 9 we have Γf = Γ or Γf = Γ (s). s Suppose that α = 0. Then Γf is the T -subalgebra generated by the polynomials xp1 , h and by Γ . s1 s2 s2 −s sn sn −s s . . . xpn )p + T (3,2) , the monoIndeed, since xp1 xp2 . . . xpn + T (3,2) = (x1 xp2 s1 s2 s sn mial xp1 xp2 . . . xpn belongs to the T -subalgebra generated by xp1 and Γ . Hence, s1 s2 s s n f = αxp1 xp2 . . . xpn + h belongs to the T -subalgebra generated by xp1 , h and by Γ . s s1 s2 On the other hand, it is clear that α−1 f (x1 , 1, 1, . . . , 1) = xp1 ∈ Γf so xp1 xp2 . . . s1 s2 s sn sn xpn ∈ Γf and therefore h = f − αxp1 xp2 . . . xpn ∈ Γf . Thus, xp1 ∈ Γf and h ∈ Γf . s Since Γ ⊂ Γf , Γf coincides with the T -subalgebra generated by the polynomials xp1 , h and by Γ , as claimed. s Thus, Γf is the T -subalgebra in F X generated by xp1 and by Γh . By Lemma 9, one of the following holds: s

(a) Γf coincides with the T -subalgebra Γ  generated by xp1 and by Γ ; s (b) Γf coincides with the T -subalgebra Γ  generated by xp1 and Γ (s). s

s

Since xp1 and q (s) (x2 , x3 ) belongs to Γ  , we have xp1 q (s) (x2 , x3 ) ∈ Γ  so Γ  = Γ  . The proof of the lemma is completed. 2 Proposition 11. Let Ω be a T -subalgebra of F X such that Γ  Ω. Then one of the following holds: 1. Ω = F X; 2. Ω = Γ (μ) for some μ ≥ 0; λ 3. Ω is the T -subalgebra generated by xp1 and by Γ (μ) for some λ ≥ μ ≥ 0. Proof. The Vandermonde argument shows that over an infinite field of characteristic p > 0 each T -ideal (see for example [7,9,17]), each T -subspace (see [10,14,15]) and each T -subalgebra in F X is generated by its multihomogeneous polynomials f (x1 , . . . , xn )

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of multidegrees (ps1 , . . . , psn ) for some si ≥ 0 (i = 1, . . . , n). Let M be a subset of Ω formed by such multihomogeneous polynomials that generates Ω as a T -subalgebra. Using Lemma 10 we can describe the T -subalgebra Γf ⊆ Ω for any f ∈ M . If Γf = F X for some f ∈ M then Ω = F X. Suppose that Γf = F X for all f ∈ M . Since Ω  Γ , there exists f ∈ M such that Γf = Γ . In this case it follows from Lemma 10 that Ω ⊇ Γ (s) for some s ≥ 0. Let μ be the smallest non-negative integer such that Γ (μ) ⊆ Ω. Since xp1 x2p −1 [x2 , x3 ]x3p −1 + T (3,2)  s−μ s−μ pμ μ s−μ μ = xp1 x2p −1 xp3 −1 x2p −1 [x2 , x3 ]x3p −1 + T (3,2) s

s

s

then it is clear that for all s ≥ μ we have Γ (s) ⊆ Γ (μ).

(15)

λ

Define the polynomial ω(x1 ) by ω(x1 ) = xp1 , where λ is the smallest integer r ≥ 0 r r such that xp1 ∈ Ω. If such λ does not exist (that is, if xp1 ∈ / Ω for any r ≥ 0), then we define ω(x1 ) = 0. If ω(x1 ) = 0 then, by Lemma 10, we have Γf = Γ or Γf = Γ (s) (s ≥ 0) for any f ∈ M . By the definition of μ we have Γf ⊆ Γ (μ) for any f ∈ M . Hence, Ω = Γ (μ). λ

λ

Suppose that ω(x1 ) = xp1 . Let Ω  be the T -subalgebra generated by xp1 and by Γ (μ). It is clear that Ω  ⊆ Ω. We will prove that in this case Ω = Ω  . Let f ∈ M . If Γf = Γ or Γf = Γ (s) (s ≥ μ) then Γf ⊆ Ω  . Suppose that Γf is s generated by xp1 and Γ (s) for some s ≥ 0. Then it is clear that s ≥ μ, and Γ (s) ⊆ Ω  s s−λ λ by (15). Similarly, s ≥ λ and xp1 = (xp1 )p belongs to Ω  . Thus, Γf ⊆ Ω  for all f ∈ M , and hence Ω = Ω  as desired. λ Suppose now in order to get a contradiction that λ < μ. Since xp1 ∈ Ω and λ q (λ) (x2 , x3 ) ∈ Γ ⊆ Ω, we have xp1 q (λ) (x2 , x3 ) ∈ Ω. Therefore Γ (λ) ⊆ Ω, a contradiction with the definition of μ. Thus, λ ≥ μ. The proof of the proposition is completed. 2 Now we are in a position to complete the proof of Theorem 1. First we prove that Γ is not finitely generated as a T -subalgebra. Suppose the contrary: Γ is generated as a T -subalgebra by some finite subset Q. Then Q lies in the T -subalgebra generated by a set S = 1, q (0) , q (1) , q (2) , . . . , q (z) , x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ] for some integer z ≥ 0. Hence, Γ is generated by S as a T -subalgebra. It follows that each element of Γ is a linear combination of 1 and of products of the form f = f1 (g(1,1) , . . . , g(1,5) )f2 (g(2,1) , . . . , g(2,5) ) . . . fn (g(n,1) , . . . , g(n,5) ),

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277

where fi ∈ S \ {1}, and g(i,j) ∈ F X for all i, j. Note that if n ≥ 2 then f ∈ T (3,2) and if n = 1 then f belongs to the T -subspace STS generated by S. Since T (3,2) ⊆ STS , we have f ∈ STS in either case. Since 1 ∈ S, Γ coincides with the T -subspace STS generated by S. This contradicts to the fact that q (z+1) ∈ Γ does not belong to the T -subspace of F X generated by the polynomials xp1 , q (0) , . . . , q (z) , x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ] (see, for example, [14, Theorem 3.1]). Hence, Γ is not finitely generated as a T -subalgebra. Now, let Ω be a T -subalgebra in F X such that Γ  Ω. Then, by Proposition 11 and Lemma 7, Ω is finitely generated. Thus, Γ is a limit T -subalgebra in F X. The proof of Theorem 1 is completed. 4. Proof of Theorem 2 Recall that X = {x1 , x2 , . . .}. Define Xn = {x1 , . . . , xn } ⊂ X. Let F Xn  be the free unitary associative F -algebra on the free generating set Xn , F Xn  ⊂ F X. If W ⊂ F X then Wn denotes the intersection of W and F Xn , Wn = W ∩ F Xn . The following lemma can be easily verified. Lemma 12. If W is a T -subalgebra (T -ideal) in F X then Wn is a T -subalgebra (T -ideal) in F Xn . If W is generated as a T -subalgebra by a set H, then Wn is generated as an unitary subalgebra in F Xn  by the elements h(f1 , f2 , . . . , fm ), where h(x1 , . . . , xm ) ∈ H and fi ∈ F Xn  for all i. (3,2)

In particular, Tn

= T (3,2) ∩ F Xn  is a T -ideal of the algebra F Xn . (3,2)

Lemma 13. If U is a T -subalgebra in F Xn /Tn (3,2) F X such that T (3,2) ⊂ Ω and U = Ωn /Tn .

then there is a T -subalgebra Ω in

Proof. Consider the preimage V = π −1 (U ) of U with respect to the canonical homomor(3,2) phism π : F Xn  → F Xn /Tn . It is clear that V is a T -subalgebra in F Xn  such (3,2) that U = V /Tn . Let Ω be the T -subalgebra in F X generated V and T (3,2) . Since (3,2) Tn ⊂ V , by Lemma 12 we have Ωn = V . 2 (3,2)

Theorem 14. If n ≥ 3 then Γn /Tn (3,2)

(3,2)

is a limit T -subalgebra in F Xn /Tn (3,2)

.

is a T -subalgebra of F Xn /Tn . Let us prove that Proof. It is clear that Γn /Tn (3,2) Γn /Tn is not finitely generated as a T -subalgebra. Denote by H the set of polynomials (3). By Lemma 12, Γn is generated as an unitary subalgebra in F Xn  by the (3,2) elements h(f1 , f2 , . . . , fm ), where h ∈ H and fi ∈ F Xn  for all i. Hence, Γn /Tn is gen(3,2) erated as an unitary algebra by the elements q (l) (f1 , f2 ) + Tn , where f1 , f2 ∈ F Xn ,

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(3,2)

and l ≥ 0. Thus, Γn /Tn ments

(3,2)

is generated as a T -subalgebra in F Xn /Tn

by the ele-

q (0) + Tn(3,2) , q (1) + Tn(3,2) , . . . , q (l) + Tn(3,2) , . . . Using the argument similar to one used in the proof of Theorem 1 we can prove that if (3,2) Γn /Tn is finitely generated then it can be generated by q (0) + Tn(3,2) , q (1) + Tn(3,2) , . . . , q (z) + Tn(3,2) for some z ≥ 0. Note that Γ is generated by T (3,2) and by the polynomials q (l) (x1 , x2 ) ∈ F X2  (l ≥ 0). It follows that Γ is generated by q (0) , q (1) , . . . , q (z) , x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ]. This is a contradiction because, by Theorem 1, Γ is not a finitely generated T -subalgebra (3,2) in F X. Hence, Γn /Tn is not finitely generated as a T -subalgebra. (3,2) (3,2) Let U be a T -subalgebra in F Xn /Tn such that Γn /Tn  U . We will prove that U is finitely generated as a T -subalgebra. By Lemma 13, there is a T -subalgebra Ω (3,2) in F X such that T (3,2) ⊂ Ω and U = Ωn /Tn . Since Γn  Ωn and T (3,2) ⊂ Ω, for n ≥ 3 we obtain Γ  Ω. Then, by Proposition 11, one of the following holds: 1. Ω = F X; 2. Ω = Γ (μ) for some μ ≥ 0; λ 3. Ω is the T -subalgebra generated by xp1 and by Γ (μ) for some λ ≥ μ ≥ 0. (3,2)

(3,2)

If Ω = F X, then U = F Xn /Tn , and hence U is generated by x1 + Tn . Suppose Ω = Γ (μ) for some μ ≥ 0. By Lemma 7, Ω is generated by the polynomials μ (3,2) xp1 q (μ) (x2 , x3 ), x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ]. Hence U = Ωn /Tn is generated as an μ (3,2) p (μ) unitary algebra by the elements (f1 ) q (f2 , f3 ) + Tn , where fi ∈ F Xn  for all i. If μ (3,2) n ≥ 3 then U is generated as a T -subalgebra by xp1 q (μ) (x2 , x3 ) + Tn . λ Let us consider the case when Ω is the T -subalgebra generated by xp1 and by Γ (μ) for some λ ≥ μ ≥ 0. By the argument similar to one used above, we have that for λ (3,2) n ≥ 3 the T -subalgebra U is generated (as a T -subalgebra) by the elements xp1 + Tn , μ (3,2) xp1 q (μ) (x2 , x3 ) + Tn . In all cases we have that U is finitely generated as a T -subalgebra. The proof is completed. 2 Corollary 15. If n ≥ 3 then the T -subalgebra Γn is not finitely generated. Let Ωn be a T -subalgebra in F Xn  such that Γn  Ωn . To prove Theorem 2 it suffices to check that Ωn is a finitely generated T -subalgebra in F Xn .

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279

Let Ω be the T -subalgebra of F X generated by Ωn and T (3,2) . It’s easy to see that Ωn = Ω ∩ F Xn  and Γ  Ω for n ≥ 2. By Proposition 11, one of the following holds: 1. Ω = F X; 2. Ω = Γ (μ) for some μ ≥ 0; λ 3. Ω is the T -subalgebra generated by xp1 and by Γ (μ) for some λ ≥ μ ≥ 0. If Ω = F X, then Ωn = F Xn , and hence Ωn is generated by x1 . Suppose Ω = Γ (μ) for some μ ≥ 0. Then, by Lemma 7, Ω is generated by μ

xp1 q (μ) (x2 , x3 ), x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ].

(16)

If n ≥ 5 then Ωn is generated as a T -subalgebra of F Xn  by the elements (16) as well. Suppose n = 4. Let us consider a polynomial of the form f = f1 [f2 , f3 ][f4 , f5 ], where f1 , . . . , f5 ∈ F X4 . We have that f + T (3) is a linear combination of the elements m + T (3) , where m = x1a1 −1 [x1 , x2 ]x2a2 −1 x3a3 −1 [x3 , x4 ]x4a4 −1 , T (3) is the T -ideal generated by [x1 , x2 , x3 ], and ai ≥ 1 for all i (see for example [21,23] or [6,10,14,15,25]). Let l1 and l2 the largest integers such that pl1 divides both a1 , a2 and pl2 divides both a3 , a4 . By Lemma 5, m belongs to the T -subalgebra generated by q (l1 ) , q (l2 ) and T (3) . Modifying the proof of Lemma 7, one can prove that for any l the μ polynomial q (l) belongs to the T -subalgebra generated by xp1 q (μ) (x2 , x3 ) and T (3) . We leave the details to the reader to check that f belongs to the T -subalgebra genμ erated by xp1 q (μ) (x2 , x3 ) and x1 [x2 , x3 , x4 ]. Thus, Ω4 is generated as a T -subalgebra by polynomials μ

xp1 q (μ) (x2 , x3 ), x1 [x2 , x3 , x4 ]. λ

To finish, suppose that Ω is the T -subalgebra generated by xp1 and by Γ (μ) for some λ ≥ μ ≥ 0. By the argument similar to one used above, we have that if n ≥ 5 then Ωn is generated as a T -subalgebra in F Xn  by λ

μ

xp1 , xp1 q (μ) (x2 , x3 ), x1 [x2 , x3 , x4 ], x1 [x2 , x3 ][x4 , x5 ], and if n = 4 then Ω4 is generated as a T -subalgebra in F X4  by λ

μ

xp1 , xp1 q (μ) (x2 , x3 ), x1 [x2 , x3 , x4 ]. In all cases, we have that Ωn is finitely generated as a T -subalgebra. The proof of Theorem 2 is completed.

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