MINIMAL POLYNOMIALS AND CHARACTERISTIC POLYNOMIALS OVER RINGS

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1 JP Journal of Algebra, Number Theory and Applications Volume 0, Number 1, 011, Pages Published Online: March, 011 This paper is available online at Pushpa Publishing House MINIMAL POLYNOMIALS AND CHARACTERISTIC POLYNOMIALS OVER RINGS Department of Mathematics Josai University Sakado, Saitama 350-0, Japan Let F be a field and V be a finite dimensional vector space over F. Let End F V be the endomorphism ring of V over F, and let be in End F V. Also let F [] t be the polynomial ring in t over F. Then, it is well known the relationship between 010 Mathematics Subject Classification: 15A04, 15A3, 15A33. Keywords and phrases: minimal polynomial, characteristic polynomial, endomorphisms ring of modules, classical groups. Received October, 010 Abstract Let R be a commutative ring with 1, and M be a free module of a finite rank over R. End R M is the endomorphism ring of M over R, is an element in End R M, and the matrix of diagonalizable. Our purpose is to investigate the relationship between the characteristic polynomial χ of and the minimal polynomial p of. If R is an integral domain, then we shall show that p is uniquely determined as a monic polynomial dividing χ. Also, the difference between the two sets of zeros of p and χ, respectively, is only the multiplicity of their roots. If R is not an integral domain, then we shall construct such that p is not necessarily monic nor divides χ. 1. Introduction

2 50 χ () t F[] t the characteristic polynomial of and p ( t) F[ t] the minimal polynomial of. For instance, χ ( a ) = 0 for a in F if and only if p ( a) = 0, that is, the difference between the two sets of roots of χ and p, respectively, is only the multiplicity of their roots. Further, if F is sufficiently large, say, an algebraically closed field, then χ () t is a product of linear equations. Moreover, if these roots of χ ( t) are different each other, then is diagonalizable. Above observation about linear endomorphism of a vector space V over a field F pose us a question what will occur if we replace F the field to R a commutative ring, V the vector space to M a free module over R, and in End F V to in End R M of which matrix is diagonalizable. The purpose of this note is to answer partially to the question. Estes and Guralnick [] investigated what the possible minimal polynomials are for integral symmetric matrices. Augot and Camion [1] presented algorithms connected with computation of the minimal polynomial of an n n matrix over a field K. Fiedler [3] showed that for a given polynomial, we can construct a symmetric matrix whose characteristic polynomial is the given polynomial. Schmeisser [5] proved that for a given polynomial f ( x) with only real zeros, we can construct a real symmetric tridiagonal matrix whose characteristic polynomial is n ( 1) f ( x) with n = deg f. We will refer Lang [4] as a standard text book in algebra, in which the reader will find necessary concepts and materials.. Preliminaries Throughout in the paper, R is a commutative ring with the identity 1, R [] t is the polynomial ring in t over R, M is a free module over R of rank n with X = { x 1, x,..., x n } a basis for M, and End R M is the endomorphism ring of M over R. For an element in End R M, we write ~ A X if A in M n ( R) is the matrix of relative to X, where M n ( R) denotes the ring of matrices of n n over R. We define the characteristic polynomial χ ( t ), χ or χ () t of A ( or ) to be the determinant t I A

3 MINIMAL POLYNOMIALS AND CHARACTERISTIC POLYNOMIALS 51 in R []. t By definition, it is independent to the choice of the basis X for M. Also, it is monic and unique for. An element a in R is called an eigenvalue or a characteristic root of in R if it is a root of χ, i.e., χ ( a ) = 0. For in End R M, we have a canonical ring homomorphism π : R[ t] End M defined by π( f () t ) = f ( ) for f ( t) in R [ t]. Therefore, M may be viewed as an R[] t -module, defining the operation of R [ t] on M by letting f ( t) x = f ( )x for f () t in R [] t and x in M. Lemma.1. ( ) = 0. χ Proof. See Theorem 3.1 (Caley-Hamilton) in Lang [4, p. 561]. We note that ker π {}, 0 for it contains at least χ 0 by Lemma.1. Let P be the set of monic polynomials in R [ t], for which we define K 0 and K 1, two subsets of ker π as follows: K 0 = The set of non-zero polynomials in ker π of which degree is the lowest in ker π. K 1 = The set of non-zero polynomials in P ker π of which degree is the lowest in P ker π. Clearly, K 0 φ and K 1 φ, for ker π contains a monic polynomial χ. We call any polynomial in K 0 a minimal polynomial of and denote it by p (), t also any one in K 1 a small polynomial of and write it as q ( t). As we know if R is a field, then there exists always a unique minimal polynomial which is monic. In particular, in such a case we may take p ( t) = q( t). As a matter of course, in general, p and q are not necessarily unique for. Indeed, if p is a minimal polynomial, so is cp for any c in R with cp 0. Also, if q is a small polynomial with deg p < deg q, so is p + q. On the other hand, it is clear that both deg p and deg q are unique for, and we have R deg p deg q.

4 5 Lemma.. (a) The following conditions ( a 1) and ( a ) are equivalent: ( ) 1 a p = deg q deg for any (or some) p K0 and any (or some) q K1 and ( a ) there is a monic minimal polynomial p. (b) In case of (a), p is a unique for. Proof. Since (a) is clear, we prove (b). Let u and v be both monic minimal polynomials. Then since deg u = deg v and they are monic, we have deg ( u v) < deg u. On the other hand, since u, v, are in ker π, so is u v. Hence u v = 0 by the minimality of u. Thus u = v and we have proved (b). 3. Statements of Theorems A, B and C Let be in End R M. For χ, p and q in R [ t], where p and q are arbitrary chosen in K 0 and K 1, respectively, we define three subsets of R as S χ = the set of roots of χ, and S p = the set of roots of p S = the set of roots of q. q In Theorem A, we shall show that if R is an integral domain and is diagonalizable, then S χ and S p coincide with each other, hence the difference between them is only the multiplicity of the roots. Theorem A. Let R be an integral domain and the matrix of diagonalizable. Then, we have the following: (a) there is a unique monic minimal polynomial p, (b) p divides χ, End R M be (c) S S, that is, the difference between roots of χ and p is only the χ = p multiplicity of each root, and

5 MINIMAL POLYNOMIALS AND CHARACTERISTIC POLYNOMIALS 53 (d) if χ has n distinct roots in R, we have χ p. = Theorems B and C show that if R is not an integral domain, then Theorem A is not necessarily valid. Theorem B. There is a finite commutative ring R, a module M over R, and an endomorphism in End R M of which matrix is diagonal and for which we have (a) χ = q with deg χ = is unique for and has no multiple roots, and is decomposed in two ways into a product of linear factors. (b) roots. (c) (d) p with deg p = 1 is unique for, but not monic, and has no multiple S χ S p with S = 4 χ and S p = 8. p does not divide χ. Theorem C. There is a finite commutative ring R, a module M over R, and in End RM of which matrix is diagonal and for which we have (a) p is unique for but not monic, whereas q is not unique for. (b) deg p < deg q = 3 < deg χ = 4 = rank M < R = 6. = (c) χ has two -ple roots and four simple roots, whereas p and q have all simple roots, and Sχ = S p = Sq = R, that is, for any a in R, t a divides each of χ, p and q Proof for Theorem A 4. Proof for Theorems A, B and C Since is diagonalizable, we have a matrix A in M n ( R) such that ~ A = diag( a1, a,..., an ), X for some basis X = { x1, x,..., x n } for M. Therefore, by the definition of

6 54 characteristic polynomial of, we have χ () t = ti A = ( t a 1 ) ( t a ) ( t a n ) (1) with a, a,..., an 1 in R. First, we prove (a) and (b) of the theorem. Let K be the quotient field of R, M = K R M be the coefficient extension of M, and be the prolongation of on M. Define the canonical ring homomorphism ϕ : K[ t] EndK ( M ) by ϕ( f () t ) = f ( ) for f () t in K [ t]. Then, ker ϕ is an ideal of K [ t]. Since K [] t is a PID, ker ϕ is generated by an element f ( t) in K ( t), that is, we have ker ϕ = ( f ( t) ) for some f ( t) in K [ t], () where we may assume that f ( t) is monic, since K is a field. Consequently, f () t is unique for. On the other hand, since χ ( ) = 0 by Lemma.1, χ ( t) belongs to R [] t ker ϕ. Hence, by (), we have Therefore, (1) yields that χ () t = f ( t) g( t) for some g( t) K[ t]. (3) f () t g() t = ( t a )( t a ) ( t a ), (4) 1 n for some a..., 1, a, am in R. Decomposing f and g as products of prime elements in K [ t], respectively, say, f = f f and g = g g g, (4) implies that 1 f r 1 s f f f g g g = ( t a ) ( t a ) ( t a ). 1 r 1 s 1 n Since K [] t is UFD, comparing the both sides of the above equation, we find a subset { a i, ai,..., a } of { a 1, a,..., a n } in R, and a subset { c 1, c,..., c r } in 1 i r

7 MINIMAL POLYNOMIALS AND CHARACTERISTIC POLYNOMIALS 55 K {} 0 such that f j = c ( t a ), c K, a R j i j j i j for j = 1,,..., r, which yields that f () t = c c c ( t a ) ( t a ) ( t a ), a R 1 r i1 i i r i j for j = 1,,..., r. However, since f ( t) is monic, we obtain c c 1 and thus 1 c r = f () t = ( t a )( t a ) ( t a ), a R (5) i1 i i r i j for j = 1,,..., r, which guarantees that f ( t) is contained in R [ t]. Thus, we may choose f () t as p () t by Lemma. and χ ( t) = p( t) g( t). Consequently, p is monic, divides χ and any zero of p ( t) is that of χ ( t), which proves (a) and (b) of the theorem. By (b) any root of p ( t) is that of χ ( t). To show (c) we have to prove the converse of this fact. Clearly, X the basis for M over R is also that of M over K. Further, since is the prolongation of to M we understand that = on X. Hence, for any i = 1,,..., n, 0 = = p () t x i f () t x i = ( a i )( a i ) ( a i ) x i 1 = ( a a )( a a ) ( a a ) ( a a ) x, i i1 i i i i i ir i which implies that for any i in { 1,,..., n }, we have a =, for some j in r i a i j { 1,,..., r }, since R is an integral domain and X is a basis for M over K. Thus, we have shown the converse, namely, a zero of χ is that of p. Consequently, two sets of zeros of χ and p, respectively, coincides with each other. This shows that the difference between the roots of χ and p is only the multiplicity, which is (c). (d) is clear by (c).

8 Proof for Theorem B Let R = Z 16 = { 0, 1,..., 15} with a = a + 16Z for = 0, 1,..., 15, Rx with a basis X = { x 1, x } over R, and a M = Rx1 ~ A = X To show (a), first, we will treat to factorize χ and get its roots. By the definition of the characteristic polynomial, we have the unique monic polynomial () t = ( t )( t 4). Substituting each element in Z 16 for t in χ ( t), we have χ S = {, 4, 10, 1}. Therefore, we have exactly two factorizations χ χ () t = ( t )( t 4) = ( t 10)( t 1), which also shows that χ () t has no multiple roots. The rest of (a), χ ( t) = q( t) will be treated later. Next, we deal with p ( t) and q ( t). It is obvious to see that 8 t is in ker π, since 8 = 0. Our claim is that this is the unique minimal polynomial. Suppose that f () t = at + b 0 belongs to ker π for a, b in Z 16. Then we have and 0 = f ( ) x 1 = ( a + b ) x 1 f ( ) x = ( 4a + b ), 0 = x which implies that a = 8 and b = 0, hence f ( t) = 8t. Thus we have shown that p () t = 8t is the unique minimal polynomial of. Further, this shows that there are no monic polynomial of degree one in ker π, and so we have χ = q. Moreover, p () t = 8t gives us S p = { 0,,..., 14}, i.e., Z 16. The rest of the proof is straightforward and we have completed the proof of the theorem.

9 MINIMAL POLYNOMIALS AND CHARACTERISTIC POLYNOMIALS Proof for Theorem C We claim that = Z 6 = { 0, 1,..., 5}, R M Rx1 Rx Rx3 Rx4 X = { x, x, x x } a basis for M over R, and 1 3, 4 ~ A = diag ( 1,, 3, 4) X = with satisfies all necessary conditions of the theorem. Recall that we have the canonical ring homomorphism π : R[ t] End M defined by π( f () t ) = f ( ) for f ( t) R[ t] and End M. Also, we have (1) () t = ( t 1)( t )( t 3) ( t 4). χ First, we show that for f () t = ( t 1) ( t ) ( t 3) and g ( t) = 3t( t 1), we have and () f, g are contained in ker π, i.e., R f ( A) = g( A) = 0. Indeed, for the identity matrix I = diag ( 1, 1, 1, 1), f ( A) = ( A 1 I )( A I ) ( A 3 I ) = diag ( 0, 1,, 3) diag ( 1, 0, 1, ) diag (, 1, 0, 1) = 0 I which verify (). Further, g( A) = 3diag ( 1,, 3, 4) diag ( 0, 1,, 3) = 0 I, (3) for 0 h () t ker π, we have deg h > 1. To show this, let 0 h () t = at + b ker π for a, b R. Then, 0 = a A + b I = diag ( a + b, a + b, 3a + b, 4a + b ), which implies that a = b = 0, a contradiction. Thus, deg h > 1 and we have (3). R

10 58 By () and (3), we find that g ( t) = 3t( t 1) is a polynomial of the lowest degree in ker π. Therefore, we may write p ( t) = 3t( t 1). Our next purpose is to show the uniqueness of p () t for. Namely, we prove that c (4) if 0 k () t = at + bt + c belongs to ker π, then we have a = b = 3 and = 0. Since k () t is in ker π, 0 = k( A) = aa + ba + c I. Substituting A = = diag ( 1,, 3, 4) and A diag ( 1, 4, 3, 4) in the above equation, we get 0 = a diag ( 1, 4, 3, 4) + b diag ( 1,, 3, 4) + c( 1, 1, 1, 1), which implies that a = b = 3 and c = 0 as was to be shown. Thus we have proved that p () t = 3 t( t + 1) is unique for. Also, (4) shows that ker π does not contain a monic polynomial of degree two. This together with f ( A) = 0 for f () t = ( t 1) ( t )( t 3) allows us to write q ( t) = ( t 1) ( t )( t 3). However, since p + q is in ker π, q ( t) = ( t 1) ( t ) ( t 3) is not unique for. Thus, we have proved (5) p is unique for, but not q. Since deg p =, deg q = 3, deg χ = rank M = 4 and R = 6, we have proved that (6) deg q deg p < deg χ = rank M < R. < By (5) and (6), we have proved (a) and (b) of the theorem. Now, we show (c) and (d). Since we have another factorization () t = t( t 1) ( t ) = ( t 3)( t 4 ) χ ( t 5), 1 and 4 are multiple roots. On the other hand, ( t 1) ( t ) does not have 0 as zero, and t ( t 1) not. Therefore, 0 and are simple roots of χ (). t Similarly, substituting 3 and 5 for t in ( t 4) ( t 5) and ( t 3) ( t 4), respectively, we have no zeros and also find that both 3 and 5 are simple roots of χ (). t Thus, we have proved

11 MINIMAL POLYNOMIALS AND CHARACTERISTIC POLYNOMIALS 59 (7) χ () t has two -ple roots and four simple roots. Finally, substituting any element α in R for t in each of respectively, we get zero. Further, we see that So, we have χ, and p, q p and q have no multiple roots. (8) χ, p, q have the same root set R, and p and q have only simple roots, and (9) for any α in R, t α devides each of χ, p and q, which gives us (c) and (d) of the theorem. Thus we have completed the proof for Theorem (C). Proposition. Let R be a ring and E be a left module over R. Then, the following (a) and (b) hold: (a) If two elements a, b in R satisfy (i) abe = bae = 0 and (ii) ar + br = R, then we have (1) E = E a + Eb for E a = { x E ax = 0} and = { y E by = 0}. E b (b) Further, if an additional condition (iii) a, b are central elements of R is satisfied, then we have () E = E a E b. References [1] D. Augot and P. Camion, On the computation of minimal polynomials, cyclic vectors, and Frobenius forms, Linear Algebra Appl. 60(15) (1997), [] D. R. Estes and R. M. Guralnick, Minimal polynomials of integral symmetric matrices, Linear Algebra Appl. 19 (1993),

12 60 [3] M. Fiedler, Expressing a polynomial as the characteristics polynomial of a symmetric matrix, Linear Algebra Appl. 141 (1990), [4] S. Lang, Algebra, 3rd ed., Addison-Wesley, Tokyo, [5] G. Schmeisser, A real symmetric tridiagonal matrix with a given characteristic polynomial, Linear Algebra Appl. 93(1) (1993),

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime