Lecture 29: Free modules, finite generation, and bases for vector spaces

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1 Lecture 29: Free modules, finite generation, and bases for vector spaces Recall: 1. Universal property of free modules Definition Let R be a ring. Then the direct sum module is called the free R-module of rank n. R n := R... R Chit-chat Why is this called a free R-module? Behold: Proposition Let M be an R-module. Then any ordered n-tuple of elements x 1,...,x n M uniquely determines an R-module homomorphism given by where the 1 is in the ith coordinate. X : R n M (0,...,0, 1, 0,...,0) x i Remark This is the same property as for the free group on n generators: Any ordered n-tuple of elements of a group G determines a unique map from F n to G. Proof. Given (x 1,...,x n ), define X : R n M by X(a 1,...,a n ):=a 1 x a n x n M. This is a group homomorphism because X((a 1,...,a n )+(b 1,...,b n )) = (a 1 + b 1 )x (a n + b n )x n =(a 1 x a n x n )+(b 1 x b n x n ) = X(a 1,...,a n )+X(b 1,...,b n ). 37

2 LECTURE 38 29: FREE MODULES, FINITE GENERATION, AND BASES FOR VECTOR SPACES where the middle equality is using the property of M being an R-module. This is also an R-module homomorphism because X(r(a 1,...,a n )) = X((ra 1,...,ra n )) =(ra 1 )x (ra n )x n = r(a 1 x a n x n ) = rx((a 1,...,a n )) Again, the penultimate equality is using the fact that M is an R-module. 2. Spans and linear independence and bases Definition Fix x 1,...,x n M. (1) We say this collection spans M if the map X : R n M is a surjection. (2) We say that this collection is linearly independent in M if the map X : R n M is an injection. (3) We say this collection is a basis for M if X is both an injection and a surjection. Chit-chat You ll recognize these terms from linear algebra. And in terms of equations, these definitions mean exactly what you d imagine: Proposition Let M be a left R-module, and let x 1,...,x n M be an ordered collection. (1) The collection spans M if and only if for every y M, thereexistsa collection a 1,...,a n R so that y = a 1 x a n x n. (2) The collection is linearly independent if and only if the equation 0=a 1 x a n x n has one and only one solution: (a 1,...,a n )=(0,...,0). (3) The collection is a basis if and only if for any y M, theequation y = a 1 x a n x n has one and only one collection (a 1,...,a n )solvingit. Proof. The first is the definition of surjection. The latter claim follows because a homomorphism is injective if and only if the kernel is trivial, and (0,...,0) R n is the additive identity of R n. The last claim is the definition of a bijection. Definition We say that a module is finitely generated if there is some number n Z 0 and a surjective R-module homomoprhism R n M.

3 4. SPANNING SETS ARE BIGGER THAN INDEPENDENT SETS 39 Chit-chat This is also in analogy to groups. A group G is finitely generated if and only if there is some finite collection of elements g i such that all other elements can be expressed as products of g i and their inverses. Likewise, M is finitely generated if there is a finite collection x i such that every element of M can be obtained by taking linear combinations of x i. Non-example Not every module over R admits a basis. This is in contrast to vector spaces. For example, if R = Z and M = Z/nZ, thenfor any x M, theequation ax =0 has many solutions a could equal n, 2n,.... Take-away: Not every finitely generated R-module admits a basis. Recall: 3. Vector spaces and subspaces Definition Acommutativeringiscalledafield if R {0} is a group under multiplication. Definition over F. Let F be a field. A module over F is called a vector space Definition Let V be a vector space. Then a submodule of V is called a linear subspace of V. 4. Spanning sets are bigger than independent sets The following is the most importance consequences of being a field, as opposed to a ring: Theorem Let F be a field, and let M be a vector space over F. If v 1,...,v n span and w 1,...,w m are linearly independent, then n m. Proof of the Theorem. Let y 1,...,y m be linearly independent, and let v 1,...,v n be spanning. By re-ordering v i if necessary, we can assume that y 1 = a 1 v a n v n for a 1 =0. Theny 1,v 2,...,v n is also spanning, for we can obtain v 1 as a linear combination of the y 1 and the v i just divide the above equation by a 1 =0 and rearrange terms. Let M 1 M be the submodule generated by y 1 i.e., the image of R M defined by 1 y 1 and consider the quotient M/M 1.

4 LECTURE 40 29: FREE MODULES, FINITE GENERATION, AND BASES FOR VECTOR SPACES (You ll prove this is also an R-module i.e., a vector space in your homework.) Then y 2,...,y m are still linearly independent, for a linear combination of them equals zero if and only if a 1 y 1 = a 2 y a m y m for some a 1 F,andsuchanequationcanholdonlywhenallthea i =0, since the y i are assumed linearly independent. Note y 1 =0, v 2,...,v n are still spanning, so v 2,...,v n is spanning. So we have n 1vectorsspanningM/M 1, and we have m 1linearlyindependentvectorsinit. By repeating the trick above, if we have m linearly independent elements in avectorspacespannedbyn elements, we can obtain m k linearly independent elements in a quotient vector space spanned by n k elements. So which of these numbers will hit 0 first? If n k =0first,weareinaquotientvector space spanned by 0 elements i.e., the zero vector space so we must conclude m k =0aswell,fortherearenolinearlyindependentvectorsinthezero vector space. And in this case, m = n. If m k reaches zero before n k does, we have that m n. 5. Corollaries Corollary If M is a finitely generated vector space, then any two bases of M have the same number of elements in it. Proof. If the {v i } and {w i } above are both spanning and linearly independent, we have n m and m n. Hence m = n. Definition Let M be a finitely generated vector space over F.Then the number of elements in a basis for M is called the dimension of M over F. Remark This is the single most important fact in linear algebra: That we have a notion of dimension. It took us thousands of years to know what we mean by an n-dimensional space, so don t take this lightly! Corollary If M is a finitely generated vector space, any linearly independent collection w 1,...,w m can be completed to a basis that is, we can find w m+1,...,w n so that the resulting collection w 1,...,w n is both linearly independent and spanning. Proof. Since M is finitely generated, there is some N for which we have a surjection F N M. So any set of linearly independent vectors must have size N by the theorem. If X m : F m M is the map determined by w 1,...,w m, and if X m is not surjective, choose an element w m+1 not in im(x m ). Note the

5 5. COROLLARIES 41 resulting collection w 1,...,w m+1 is still linearly independent, for if then we have a 1 w a m+1 w m+1 =0 a 1 w a m w m = a m+1 w m+1. If a m+1 =0,bylinearindependenceofthew i,weknowalla i =0. Onthe other hand, if a m+1 =0wearriveatacontradictionbydividing: a 1 w a m w m = w m+1. a m+1 a m+1 The lefthand side is in the image of X m,butw m+1 was chosen not to be. So we have an injective homomorphism X m+1 : F m+1 M. If X m+1 is not surjective, we repeat the argument. It must become a surjective map at some m + k N by the theorem. So let k be the integer at which X m+k first becomes a surjection. By the above argument, it is still an injection, so we have a basis determined by the generators w 1,...,w m+k. Corollary to F n for some n. Any finitely generated module over a field F is isomorphic Proof. Begin with the linearly independent set 0 and complete to a basis. AbasisdefinesanisomorphismfromF n to your module. Remark This is definitely not true for R-modules if R is not a field after all, any finite abelian group is a Z-module, but any free Z-module is the zero module or an infinite module. Corollary Let V V be a subspace. Then dim V +dimv/v = dim V. Proof. Let v 1,...,v dim V be a basis for V.Letu 1,...,u dim V/V be a bsis for V/V.Thenchoosingrepresentativesu i for u i,theset v 1,...,v dim V,u 1,...,u dim V/V is a basis for V.Itobviouslyspanssinceforeacha V, a is a linear combination of u i,hencea is in the V -orbit of some linear combination of the u i. It is linearly independent because if we have that then 0=a 1 v a dim V v dim V + b 1 u b dim V/V u dim V/V 0=a 1 v a dim V v dim V + b 1 u b dim V/V u dim V/V. The a i terms go to zero since v i =0,hencewegetanequationsayingalinear combination of the u i is zero. This means each b i must be zero by linear

6 LECTURE 42 29: FREE MODULES, FINITE GENERATION, AND BASES FOR VECTOR SPACES independence of the u i. The original equation then says that 0 = a i v i,so by linear independence of the v i,thea i must be zero. Corollary (Rank-nullity theorem). Let f : V W be a map of F -modules and assume V is finitely generated. Then dim ker f +dimimf = dim V. Proof. By the first isomorphism theorem, we know there is a group isomorphism V/ker f = im f. ButthishomomorphismisalsoanF -module map, as you can check by hand. Thus im f = V/ker f.