Fundamental theorem of modules over a PID and applications

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1 Fundamental theorem of modules over a PID and applications Travis Schedler, WOMP 2007 September 11, The fundamental theorem of modules over PIDs A PID (Principal Ideal Domain) is an integral domain (=ring without zero-divisors) such that every ideal is principal (=generated by a single element) Examples: any field k; k[x], Z, Z[i] (the latter three use the Euclidean algorithm and are more generally examples of Euclidean domains); Z[ ] (a PID that is not a Euclidean domain) Any PID is a Noetherian UFD (Unique Factorization Domain) Here: Noetherian means that every ideal is finitely generated; Equivalently, every finitely-generated module is a Noetherian module (all submodules are finitely-generated) The main goal of this talk is to prove and apply the theorem Theorem 011 If R is a PID, and M any finitely-generated R-module, then M = m R/s i, (012) i=1 where s 1 s 2 s m Corollary 013 We may alternatively write M as for some primes p i and integers r i 1, r 0 M = R r R/p r 1 1 R/prn n, (014) Proof of Corollary 013 Simply apply unique factorization, and observe that R/(p r 1 1 pr j j ) = R/p r 1 1 R/pr j j (015) 1

2 02 Applications of the fundamental theorem The following is immediate: Theorem 021 Any finitely-generated abelian group A may be represented as a direct sum A = m Z/s m, (022) i=1 with s 1 s 2 s m Alternatively, we may write A = Z r Z/p r 1 1 Z/prn n (023) The following consequences of Theorem 011 (and Corollary 013) are somewhat less obvious Theorem 024 (Rational canonical form) Let M be any square matrix over a field k Then, after changing basis, we may rewrite M as a block-diagonal matrix, with blocks M i of the form a i, a i, a i,2, (025) a i,di 1 for polynomials such that χ(m) = the characteristic polynomial of M s i = x d i + a i,di 1x d i a i,1 x + a i,0 (026) s 1 s 2 s m, χ(m) = s 1 s 2 s m, (027) Proof The main idea is as follows: View a matrix M as a linear transformation T : V V Now, for any polynomial P (x), we have the linear operator P (T ) : V V, (028) viewing T m = T T T, m times (ie, multiplication is composition of operators) Thus, we may view V as an k[x]-module, and thus may apply Corollary 013 to describe the structure of V We obtain V = k[x]/s 1 k[x]/s m, (029) for some polynomials s 1 s m Let us write for some vector subspaces such that V = V 1 V m, (0210) V i = k[x]/si (0211) Picking bases for each V i yields a basis of V in which M is block-diagonal: M M 2 0 0, (0212) 0 M n 2

3 where each M i is the matrix for T Vi It remains to describe each M i Let us suppose that d i = deg s i Then, V i = k[x]/si is a vector space of dimension d i, with basis 1, x, x 2,, x d i 1 (using the isomorphism) Using this basis, we obtain the matrix (025), whose characteristic polynomial is s i It remains to verify the second part of (027) This follows because the characteristic polynomial of a block-diagonal matrix is the product of the characteristic polynomials of the blocks Theorem 0213 (Jordan form) Let M be any square matrix over a field k Then, after changing basis, we may rewrite M as a block-diagonal matrix, with blocks of the form D Y D Y D 0 0, (0214) 0 0 Y D where D is a matrix whose characteristic polynomial is irreducible, and Y is the elementary matrix , (0215) 0 0 of the same size as D (We may choose D to be in rational canonical form, (025)) 1 Corollary 0216 If k is algebraically closed, then the Jordan blocks have the form λ λ λ 0 0, (0217) λ for some λ k Proof of Corollary 0216 If k is algebraically closed, any irreducible characteristic polynomial must be linear, and thus the matrix D must be a 1 1-matrix Remark 0218 We don t actually need k to be algebraically closed for the corollary, as long as the characteristic polynomial splits into linear factors (ie, all eigenvalues are in k) Proof of Theorem 0213 Using Corollary 013 (instead of Theorem 011), we obtain V = k[x] r k[x]/p r 1 1 k[x]/prn n, (0219) 1 We can think of D as a single number: a root α of its characteristic polynomial This makes sense if we view the vector space on which D acts as isomorphic to k[x]/χ(d) = k[α], with the matrix acting as multiplication by α If k is perfect, then we can also replace Y by I and view the whole matrix as acting on a k[x]/χ(d)-vector space (with a Jordan form like (0217)); but not in general if χ(d) is not separable over k 3

4 where p i are irreducible polynomials, and r i 1 We must have r = 0 since V is finite-dimensional Let us again choose subspaces V i V so that V i = k[x]/p r i i as a k[x]-module, and V = V 1 V n It remains only to show that (0214) is a matrix for T Vi with the appropriate choice of basis We use the basis {x j p l i} 0 j<di ;0 l<r i (0220) It is obvious that this is a basis because it consists of one polynomial for every degree between 0 and d i r i 1 = deg(p r i i ) 1 Now, acting by T on the element x j p l i just is multiplication by x It remains only to note that x (x d i 1 p l i) = p l+1 i (a 0 + a 1 x + + a di 1 x d i 1 ) p l i (0221) 03 Relationship to semisimple and nilpotent decomposition When k is algebraically closed, Jordan form is a decomposition of our square matrix M into M = S + N, (031) where S is semisimple (given by the diagonal in the basis (0217), with all other entries zero), and N is nilpotent (given by the subdiagonal in the basis (0217), with all other entries zero) Here, a matrix N is nilpotent iff N j = 0 for some j 1, and S is semisimple iff it is diagonalizable over the algebraic closure of k More generally, we only need k to be perfect: Theorem 032 Let k be a perfect field The matrix M may be uniquely written as M = S + N, (033) where S is semisimple and N is nilpotent, in such a way that S and N are k-polynomials of M The matrices M, S, and N all commute with each other We may interpret this theorem when M is written as (0214): the part with D s is the matrix S, and the part with Y s is the matrix N It is not difficult to check that this S is semisimple It is clear that S and N commute and hence that they commute with M To show that S is a polynomial in M is not really simplified using (0214) Thus, the following actually proves the theorem without needing Jordan form (but using (0214) perhaps gives a visual picture of what is going on): Proof First, suppose χ(m) = p s 1 1 psn n (034) for distinct irreducibles p i For each i, let q i be a polynomial such that p s 1 1 ps i 1 i 1 q ip s i+1 i+1 psn n 1 (mod p s i) (035) Let P i (x) be the above polynomial Then, P i (M) must be the diagonal matrix (as in (0212)) which has 0 s all along the diagonal except at the M i -block, where it is the identity So, we may reduce to the case when n = 1, and χ(m) = p s 1 1 Now, if k is algebraically closed, then it is clear that the semisimple part S is a scalar, which is a polynomial of M 4 i

5 If k is only assumed to be perfect, we use Galois theory (covered in the next lecture): Let K be the splitting field over k of the polynomial p q Then, p 1 = q 1 q 2 q l for some linear polynomials q i We may assume that q i = x λ i, (036) with {λ i } an orbit of the Galois group of K over k (this orbit has size [K : k]) Then, using the above procedure, we have S = P (M) for P (x) = σ(λ 1 Q 1 )(x) k[x] (037) σ Gal(K/k) For uniqueness, we note that if S + N = S + N for S, N, S, N all polynomials of M, then S S = N N Since N, N and S, S commute, the LHS is semisimple and the RHS is nilpotent, which shows they are both zero Remark 038 We don t actually need k to be perfect, as long as all the irreducible factors of χ(m) are separable (ie, have distinct roots over k) Otherwise, we can canonically decompose M = S +N over the separating field of χ(m) over k (the minimal extension such that all irreducible factors of χ(m) are separable) Without passing to such an extension, we can ( get a ) Jordan/rational 0 y canonical form, but it won t necessarily yield an S + N one: for example, has irreducible 1 0 ( ) y 0 characteristic polynomial x 2 + y over F 2 (y), but it is equivalent to over F 1 y 2 ( y) Remark 039 There is a multiplicative analogue: any invertible matrix M may be uniquely written as M = SU, where S is semisimple and U is unipotent, and S, U are k-polynomials of M (assuming χ(m) is separable over k) Indeed, simply set U = S 1 (S + N) in the above theorem 04 Proof of the fundamental theorem The fundamental theorem may be proved by using a generalization of Gaussian elimination of matrices First, let us reduce the problem to one that involves matrices Given any finitely-generated module M over a PID R, being finitely generated means that R d M, (041) for some d 0 Now, let the kernel of this map be K Because any PID is Noetherian, K must be finitely generated as well, so we have so that M is the cokernel of the composition R f K R d M, (042) R f R d (043) Claim 044 We can choose new R-bases of R f, R d such that the map R f R d has the form (if f = d): s s , (045) 0 s n for s 1 s 2 s n (if f d, then add zeros to the right or bottom of the above matrix) 5

6 From the claim we deduce the theorem, since the cokernel of the above matrix is n i=1 R/s i Proof of Claim 044 Given any d f-matrix M, changing bases on the domain and range is the same as replacing M by a new matrix M = AMB, (046) where A, B are invertible matrices of the appropriate size Now, given a 00 a 0f, (047) a d0 a df we would like to apply the Gaussian elimination (along rows and columns) together with permutations to obtain a matrix of the desired form To show this, let s 1 be any generator of the ideal (a ij ) 0 i d,0 j f Claim 048 Using row and column operations and permutations, we can ensure that a 00 s 1 Using this claim, we may kill off a i0, a 0i for i 0 Now, we are left with a matrix a a 11 a 1f, (049) 0 a f1 a fd where a 00 a ij for all i, j Then, by applying induction, we obtain a matrix of the desired form Proof of Claim 048 It is enough to show that, for any entries a ij, a kl, that we can change bases so that gcd(a ij, a kl ) 2 appears somewhere in the matrix (using permutations) It is enough to show that ( ) a (0410) b can be multiplied by invertible matrices so that gcd(a, b) appears Suppose that gcd(a, b) = x a+y b Clearly, (x, y) = (1) (otherwise we could divide this equation and get a gcd with fewer prime factors) Then, if x u + y v = 1, the matrix ( ) x y (0411) v u is invertible, and ( ) ( ) x y a v u b as desired = ( ) gcd(a, b), (0412) v a + u b 2 We use the abusive notation gcd(x, y) for any element such that (x, y) = (gcd(x, y)) This is unique up to multiplication by a unit 6

x y B =. v u Note that the determinant of B is xu + yv = 1. Thus B is invertible, with inverse u y v x On the other hand, d BA = va + ub 2

x y B =. v u Note that the determinant of B is xu + yv = 1. Thus B is invertible, with inverse u y v x On the other hand, d BA = va + ub 2 5. Finitely Generated Modules over a PID We want to give a complete classification of finitely generated modules over a PID. ecall that a finitely generated module is a quotient of n, a free module. Let