Lifting to non-integral idempotents
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1 Journal of Pure and Applied Algebra 162 (2001) Lifting to non-integral idempotents Georey R. Robinson School of Mathematics and Statistics, University of Birmingham, Edgbaston, Birmingham B15 2TT, UK Received 15 February 2000; received in revised form5 May 2000 Communicated by Broue Abstract In the rst section of this paper, we illustrate (for a nite group G and the eld of fractions K of a suitable complete dvr R) how to produce all central idempotents of Z(KG) fromknowledge of the images of Z(RG) modulo certain powers of J (R): In the second section, we outline another approach to lifting elements closely related to idempotents in more general rings, which may be of wider interest. c 2001 Elsevier Science B.V. All rights reserved. MSC: Primary 20C20 0. Introduction Let G be a nite group, p be a prime divisor of the order of G: Let R be a complete discrete valuation ring (with valuation group Z) whose eld of fractions K is a splitting eld (of characteristic 0) for G and all its subgroups. Let () bethe unique maximal ideal of R: Set F = R=(): We assume that F has characteristic p: Set A = Z(RG); and let J = {a A: a n A for some n N}: Then J is an ideal of A containing A; and is the full pre-image of the Jacobson radical of A=A: When we discuss convergence in A; we mean with respect to the metric naturally induced on A by the metric on R obtained via the valuation : The procedure for lifting idempotents of Z(FG) to idempotents of Z(RG) is wellknown. In Section 1, we illustrate how to produce all central idempotents of KG; from knowledge of certain distinguished elements of A= 2t J; where t is a positive integer less than or equal to ( G ): It will be evident fromthe description below that the method of the rst section would work in much more general situations. However, our present motivation lies address: g.r.robinson@bham.ac.uk (G.R. Robinson) /01/$ - see front matter c 2001 Elsevier Science B.V. All rights reserved. PII: S (00)
2 360 G.R. Robinson / Journal of Pure and Applied Algebra 162 (2001) in the fact that knowledge of the primitive central idempotents of KG is equivalent to knowledge of the irreducible characters of KG; and these correspond naturally to the complex irreducible characters of G: Consequently, in the rst section, we content ourselves with explicitly describing only this particular case and remarking that these constructions work block-by-block with trivial modications. In Section 2, we outline another approach to lifting elements closely related to idempotents in more general contexts, which may be of wider interest. Among other things, we regard Section 1 of this paper as a contribution to the attempt to understand heights and defects of complex irreducible characters. The results there illustrate that the greater the defect of an irreducible character, the larger the factor ring of A that is necessary in order to determine the associated idempotent. On the other hand, we note that R may be chosen so that A= s J is nite for every non-negative integer s: Let us consider an idempotent e Z(KG): We call the smallest non-negative integer t such that t e A the deciency of e: By consideration of irreducible characters, G e A; so that t ( G ): We set X = t e: Then X 2 = t X: Conversely, if we have an element Y A\A such that Y 2 = t Y for some non-negative integer t; then e = t Y is an idempotent of deciency t of Z(KG): Remark. We point out that, more precisely, if e is primitive in Z(KG); and is the associated irreducible character, then the deciency of e is ( G ) ((1)) min{ ((g)): g G}: This may also be written as d (p) min{ ((g)) : g G}; where d is the defect (with respect to p) of in the now usual sense. Notice, however, that this deciency can be strictly less than d (p) for example, it always will be when G is a p-group and is non-linear. More generally, strict inequality holds precisely when (g) for all g G (in fact, it suces to consider p-regular g). This happens if and only if all decomposition numbers associated to are divisible by p: In particular, if is an irreducible character of maximal defect in its p-block (equivalently, has height 0), then the associated idempotent does have deciency d (p): This is the maximal possible deciency for any primitive idempotent (of Z(KG)) in that block, and only primitive idempotents associated to irreducible characters of height zero in that block can have that deciency. This raises the question of what happens if we work with the case A = Z(B) for some block B instead of A = Z(RG): Since blocks of very dierent groups can have isomorphic centres, we may wish to consider the commutative algebra A as an algebra in its own right, without reference to a particular group. In those circumstances, what information about the block can be read from invariants of A? Clearly, the number of irreducible characters in any block B with centre isomorphic to A is the R-rank of A; as was already known to Brauer. The discussion above shows that all blocks B which have centre (isomorphic to) A have defect groups of the same order, and that the number of irreducible characters
3 G.R. Robinson / Journal of Pure and Applied Algebra 162 (2001) of height zero in any such block B is determined by A; as has already been observed (with a somewhat dierent proof) by Cli et al. [1]. As Dade brought to our attention, it does not seem immediately obvious that the defects (or heights) of other irreducible characters in the block can be determined from the knowledge of the isomorphism type of A alone. 1. The main result We say that the elements X 0 + 2t J and Y 0 + 2t J of A= 2t J are orthogonal if X 0 Y 0 2t J: It will be useful in the proof below to dene (for each natural number n N) the ideal J n of A as the set of elements a A such that a n for each linear character of Z(KG): We note that J 1 = J; and that J n J n ; but equality need not hold in general. However, setting h = ( G ); we note that J h+r r A for each non-negative integer r: Hence a sequence (a n ) of elements of A is Cauchy with respect to the given metric if and only if for each positive integer n, there is an integer N(n) such that a r a s J n whenever r; s N(n): Our main result is: Theorem. Let t be a non-negative integer and let X 0 + 2t J and Y 0 + 2t J be elements of A= 2t J with neither X 0 nor Y 0 in A: Suppose that (X 0 + 2t J ) 2 = t X 0 + 2t J; and that (Y 0 + 2t J ) 2 = t Y 0 + 2t J: Then there is a unique element; X;of A \ A such that X X 0 (mod J t+1 ) and X 2 = t X: Similarly; there is a unique element Y of A\A such that Y Y 0 (mod J t+1 ) and Y 2 = t Y: Furthermore, t X and t Y are orthogonal idempotents if X 0 + 2t J and Y 0 + 2t J are orthogonal. Proof. Suppose that we rst have elements X and Y which lie in A\A such that X 2 = t X; Y 2 = t Y; X Y J t+1 : Set E 1 = t X; E 2 = t Y; both of which are idempotents of deciency t in Z(KG): Let be any linear character of Z(KG): Now E i {0; 1} for i =1; 2; while E 1 E 2 (mod ) by hypothesis, so that E 1 = E 2 : Since was arbitrary, E 1 = E 2 : Hence X = Y: Now we have an element X 0 A\A such that X 2 0 t X 0 (mod 2t J ). Suppose that for some non-negative integer n we have found an element X n A\A such that Notice that X 2 n t X n (mod J 2t+2 n); X 2 n t X n (mod 2t J ) and X n X 0 (mod J t+1 ): (2X n t 1 A ) 2 = 2t 1 A +4(X 2 n t X n ); an element of J 2t \J 2t+1 : This last element is also 2t [1 A +4 2t (X 2 n t X n )]: By hypothesis, 1 A +4 2t (Xn 2 t X n ) is a unit of A; which we denote by U n convenience. Furthermore, we note for later convenience that 2X n t 1 A J t : for
4 362 G.R. Robinson / Journal of Pure and Applied Algebra 162 (2001) Now (X 2 n t X n )(2X n t 1 A ) is certainly in 2t J t+1 : We set X n+1 = X n Un 1 2t (Xn 2 t X n )(2X n t 1 A ); an element of A: Furthermore, X n+1 X n (mod J t+2 n). Notice that when t =0; this inductively implies that X n+1 J: In particular, X n+1 A: When p = 2 and t 0; we note that X n+1 X n (mod A). Now consider Xn+1 2 t X n+1 : Notice that X n+1 = 2t Un 1 [ 2t X n U n (Xn 2 t X n )(2X n t 1 A )] = 2t Un 1 ([2X n t 1 A ][X n (2X n t 1 A ) (Xn 2 t X n )]); making use of the fact that (2X n t 1 A ) 2 = 2t U n : Hence X n+1 = 2t Un 1 [(2X n t 1 A )Xn 2 ]: Consequently, (Xn+1 2 t X n+1 )= 4t Un 2 [Xn 4 (2X n t 1 A ) 2 ] t Un 1 [(2X n t 1 A )Xn 2 ]: Again, we recall that (2X n t 1 A ) 2 = 2t U n ; so the above expression simplies to Un 1 2t (Xn 2 t X n ) 2 : By the hypotheses on X n ; this last expression lies in J 2t+2 n+1; and lies in 2t J: Hence the sequence (X n ) so constructed is Cauchy by the remarks above, so has a limit X A: Furthermore, X X 0 (mod J t+1 ); and X satises X 2 = t X: We claimthat X n+1 lies outside A: When p =2 or t =0; we have seen this already, so we suppose that p is odd, that t 0; and that X n+1 A; (we recall that X n A): We note that, in this case, 1 A + U n is a unit of A; and hence so is 1 A + Un 1 : Notice that since X n+1 = 2t Un 1 [(2X n t 1 A )Xn 2 ]; we may conclude that X n + Un 1 t X n (2X n t 1 A ) A: Since Xn 2 t X n 2t J; this last expression is congruent to (1 A + Un 1 )X n (mod A), a contradiction, as X n A and (1 A + Un 1 ) is a unit of A: Hence X A in any case. Returning to the proof, the argument at the beginning shows that if X (1) + 2t J and X (2) + 2t J satisfy (X (i) ) 2 t X (i) 2t J and X (i) A; but X (1) X (2) J t+1 ; then X (1) and X (2) must each lift to the same solution X of X 2 = t X; (X A): Conversely, elements which are incongruent (mod J t+1 ) must certainly lift to distinct elements. Finally, suppose that we have elements X 0 and Y 0 which lie in A\A such that X 2 0 t X 0 2t J; Y 2 0 t Y 0 2t J; X 0 Y 0 2t J: We inductively construct the convergent sequences (X n ) and (Y n ) as before, with respective limits X and Y: We prove by induction that X n Y n J 2t+2 n; this being the case by assumption when n =0: Suppose then that n 0 and that the assertion is known
5 G.R. Robinson / Journal of Pure and Applied Algebra 162 (2001) to hold for n: We have seen that X n+1 = 2t Un 1 [(2X n t 1 A )Xn 2 ] and similarly, Y n+1 = 2t Vn 1 [(2Y n t 1 A )Yn 2 ] for an appropriate unit, V n of A: Thus, X n+1 Y n+1 = 4t Un 1 Vn 1 (2X n t 1 A )(2Y n t 1 A )Xn 2 Yn 2 : We have seen that (2X n t 1 A )(2Y n t 1 A ) J 2t : Hence, by the inductive hypothesis, X n+1 Y n+1 lies in J 2t+2 n+1: Hence, the claimis established by induction, so that, taking limits, XY = 0 and the proof is complete. We remark that from a computational point of view, the formula X n+1 = X n Un 1 2t (Xn 2 t X n )(2X n t 1 A ) may appear unsatisfactory, since it involves inverting the mysterious element U n : It is not dicult to verify that the sequence (X n) still has the required convergence properties, where X 0 = X 0; X n+1 = X n V n 2t (X n 2 t X n)(2x n t 1 A ); and V n =1 A 4 2t (X n 2 t X n): It is a consequence of the results of the next section that convergence of the sequence (X n ) could be accelerated by using the formula X n+1 = Un 2 4t Xn 3 (2X n t 1 A )(3X n 2 t 1 A ) (with the same denition of U n as before). Even faster convergence could be obtained by setting X n+1 = Un 3 6t Xn 4 (2X n t 1 A )(10Xn 2 14 t X n +5 2t 1 A ) (with the same denition of U n ). Remarks. It is easy to recognise which lifted solutions X with X 2 = t X; (X A) are such that t X is a primitive idempotent of Z(KG); since this is the case precisely when XA has R-rank 1. We remark that we have shown that for each non-negative integer t; the number of primitive idempotents of Z(KG) which have deciency t (or less) is completely determined by the structure of the ring A= 2t J: Hence to nd the primitive idempotents associated to irreducible characters of a given defect s of KG; it is certainly sucient to consider the ring A=p 2s+1 A:
6 364 G.R. Robinson / Journal of Pure and Applied Algebra 162 (2001) Polynomials and idempotents The method of the rst section is a modication of a particular case of Newton s method. The results of this section evolved in attempting to accelerate convergence of an earlier version of the method of the rst section. Now let R denote the ring Z[x; y]; where x and y are commuting, (algebraically independent) indeterminates. We inductively dene polynomials p n (x; y) as follows: p 1 (x; y)=1; and for n 0; xp n+1 (x; y)=(2x y) 2 p n (x; y) p n (0; 1)(y x) n+1 : (1) It follows easily by induction that p n (xy; y)=y n 1 p n (x; 1) for all n: In particular, we see that p n (0;y)=y n 1 p n (0; 1) for each n (notice that this fact for p n is necessary for the denition of p n+1 ; since it is this which ensures that the right-hand side of the equation dening p n+1 vanishes when evaluated at (0;y)): We set q n (x; y)=( 1) n+1 p n (y x; y) for each n: Then replacing x by y x in the above shows that (x y)q n+1 (x; y)=(2x y) 2 q n (x; y) p n (0; 1)( x) n+1 ; (2) which may also be written as (x y)q n+1 (x; y)=(2x y) 2 q n (x; y) q n (1; 1)x n+1 : (3) We have x 2 p 1 (x; y)=(2x y)x (x 2 yx); while for n 0 we have Similarly, x n+2 p n+1 (x; y)=(2x y) 2 x n+1 p n (x; y) p n (0; 1)(xy x 2 ) n+1 : (4) (x y) n+2 q n+1 (x; y)=(2x y) 2 (x y) n+1 q n (x; y) p n (0; 1)(xy x 2 ) n+1 : (5) It follows by induction that we have x n+2 p n+1 (x; y)=y(2x y) 2n+1 +(x y) n+2 q n+1 (x; y) (6) for all n 0: Hence, we see that x n+2 p n+1 (x; y) (y x) n+2 p n+1 (y x; y)=y(2x y) 2n+1 (7) for all such n: Now suppose that we are given a ring S with 1. Suppose further that S has an ideal J such that n=1 J n = {0} and such that S is complete with respect to the J -adic topology. Let a be an element of S which is not a zero divisor in the subring C S (a). We say that a non-zero element s of C S (a) isa-potent if s 2 = as (notice that if S embeds in some larger ring in which a becomes invertible, then a 1 s becomes an idempotent, which, as will be apparent fromthe rst section, is our motivation for considering such elements). We are interested in locating non-trivial a-potent elements of C S (a) and we rst show that it suces consider images (mod a 4 J ) to do this.
7 G.R. Robinson / Journal of Pure and Applied Algebra 162 (2001) Suppose that we have an element b C S (a) with b 2 ab (mod a 4 J ). We will use b to produce an a-potent element of C S (a): It suces to work within the closure of the subring generated by a; b and a 2 (b 2 ab), which is commutative (note that the last element listed may be interpreted as a well-dened element of C S (a)). Hence we assume from now on that S is commutative, and that a is not a zero divisor in S: Notice that the hypotheses tell us that (2b a) 2 = a 2 (1 + j) for some j J; and that 1 + j is a unit of S: Hence 2b a is not a zero divisor in S: Further notice that (2b a) 2 (b 2 ab) is a well-dened element of S; and lies in the ideal a 2 J: Hence, for all positive integers n; (2b a) 2n (b 2 ab) n is an element of a 2n J n : Multiplying this element by (2b a); we see that (2b a) 1 2n (b 2 ab) n is a well-dened element of J for all such n: We have b 2 =(2b a)b (b 2 ab); while for n 0; b n+2 p n+1 (b; a)=(2b a) 2 b n+1 p n (b; a) p n (0; 1)(ab b 2 ) n+1 ; so it follows by induction that (2b a) 1 2n b n+2 p n+1 (b; a) is a well-dened element of S whenever n is a non-negative integer. We label this element b n+1 : We remark that n b n+1 = b (2b a) 1 (b 2 ab) p k (0; 1)(ab b 2 ) k+1 (2b a) 1 2k for all n 0: Since k=1 x n+2 p n+1 (x; y)=y(2x y) 2n+1 +(x y) n+2 q n+1 (x; y) for all n 0; we conclude that b n+1 = b n+2 p n+1 (b; a)(2b a) 1 2n = a +(b a) n+2 q n+1 (b; a)(2b a) 1 2n for all n 0: Notice then that b 2 n+1 ab n+1 =(b 2 ab) n+2 p n+1 (b; a)q n+1 (b; a)(2b a) 2 4n : Now (2b a) 2+4n has the form a 4n+2 u for some unit u of S: Consequently, b 2 n+1 ab n+1 lies in a 6 J n+2 : The sequence (b n ) is easily seen to be Cauchy with respect to the J -adic topology, so has a limit in S; say b 0 ; which satises b 2 0 = ab 0; and is thus a-potent. We have b 0 = b (2b a) 1 (b 2 ab) p k (0; 1)(ab b 2 ) k+1 (2b a) 1 2k ; k=1
8 366 G.R. Robinson / Journal of Pure and Applied Algebra 162 (2001) so it also follows that b 0 b (mod a 2 J ). Notice that we have given an explicit innite series expression for b 0 in terms of a and b (though the series would still converge under the weaker assumption that b 2 ab a 2 J; it is not clear to us that the limit would always be a-potent). We note however that if c is an element of C S ( a; b ) with c 2 ac (mod a 4 J ), and we have bc a 4 J; then we have b n+1 c n+1 a 6 J n+2 for all n: Hence we see that the respective limits b 0 and c 0 satisfy b 0 c 0 =0: Returning to the question of accelerated convergence, we point out that the above methods allow us to produce iterative procedures which converge at any desired rate. Recall that and that b n+1 =(2b a) 1 2n b n+2 p n+1 (b; a) = b (2b a) 1 (b 2 ab) n p k (0; 1)(ab b 2 ) k+1 (2b a) 1 2k b 2 n+1 ab n+1 =(b 2 ab) n+2 p n+1 (b; a)q n+1 (b; a)(2b a) 2 4n : k=1 Let us x a choice of n; and dene a Cauchy sequence (b (m) ) as follows: b (1) = b; and b (m+1) =(b (m) ) n+1 ; with notation as above. Then an easy induction argument shows that and that for all m: b (m+1) b (m) (mod a 2 J (n+2)m 1 ) (b (m) ) 2 ab (m) a 4 J (n+2)m 1 References [1] G. Cli, W. Plesken, A. Weiss, Order-theoretic properties of the center of a block, in: P. Fong (Ed.), The Arcata Conference on Representations of Finite Groups, Proc. Symp. Pure Math., AMS, Providence, 47(1) (1987)
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