Prof. D. P.Patil, Department of Mathematics, Indian Institute of Science, Bangalore May-July 2003 Basic Algebra 7. Homomorphisms of modules Richard Dagobert Brauer† Albert Thoralf Skolem†† (1901-1977) (1887-1963) The exercises 7.10, 7.11, 7.15 are marked as ∗ and may be ignored in the first reading. 7.1. Let ϕ : A → B be a ring homomorphism. If V is a B–module and W is an A– module, there exists a natural group homomorphism (see T7.7 and T7.15 c)) HomA(V, W) → HomB (V, HomA(B, W)) defined by f → (v → (b → f (bv))) with inverse g → (v → g(v)(1B )).IfR is another ring and if V is a B–left–R–right–bimodule, then the above isomorphism is a R–isomorphism. (This holds for example in the case if R = B is commutative.) 7.2. Let A and B be rings. Let U be an A–left-module, V be a B–right-module and W be a 1 (A, B)–bimodule ) of Type AWB . Then there exists a natural isomorphism HomA(U, HomB (V, W)) → HomB (V, HomA(U, W)) defined by f → (v → (u → f (u)(v))), with the inverse g → (u → (v → g(v)(u))). 7.3. Let I be a finite set and let A be an indecomposable ring 2) I a). The canonical projections βi : A → A, i ∈ I are the only A–algebra–homomorphisms from AI → A. I b). The map S(I) → AutA–Alg A defined by σ → ((ai ) → (aσ −1(i))) is an isomorphism of groups. 7.4. Let f : V → W be a homomorphism of modules over a ring A.Ifkerf and im f are finite A–modules, then V is also a finite A–module. For the minimal number of generators we have: µA(V ) ≤ µA(ker f)+ µA(im f). 7.5. Let K be a division ring, V be a finite dimensional K–vector space, f : V → W be a linear mapping into an arbitrary K–vector space W and let U ⊆ V be a subspace. Then DimK V − DimK U ≥ DimK f(V)− DimK f(U). 7.6. Let f : V → W, g : W → X and let h : X → Y be linear maps of finite dimensional vector spaces over a division ring K. a). (Inequality of Sylvester) rank f +rank g−DimK W ≤ rank(gf ) ≤ Min{rank f, rank g} . (Hint : The first inequality easily follows from the exercise 7.5.) 1) See T7.14 2) A non-zero ring A is called indecomposable or connected ifA is not isomorphic to a product of two non-zero rings. A non-zero ring A is indecomposable if and only if the only idempotents in the center Z(A) are 0 and 1 (proof!). For example an integral domain is indecomposable. D. P. Patil July 16, 2003 ,3:38 p.m. 2 Basic Algebra ; May-July 2003 ; 7. Homomorphisms of modules b). (Inequality of Frobenius) rank(hg) + rank(gf ) ≤ rank g + rank(hgf ) . (Hint : We may assume that g is surjective and then apply part a).) 7.7. Let V and W be vector spaces over a division ring K. A linear map f : V → W has finite dimensional kernel if and only if there exists a linear map g : W → V such that gf = idV + h, where h ∈ EndK V is of finite rank. 7.8. Let V and W be vector spaces over a division ring K. Then a). The linear maps from V into W of finite rank form a Z(K)–submodule E of HomK (V, W). For f1,...,fn ∈ E and a1,...,an ∈ Z(K) we have: rank(a1f1 +···+anfn) ≤ rank f1 +···+rank fn . b). The endomorphisms of V of finite rank form a two-sided ideal in the ring EndK V . 7.9. (Characters) LetM be a (multiplicative) semigroup and let K be a division ring. A non- zero semi-group homomorphism from M in the multiplicative monoid of K is called a character of M with values in K. The constant map x → 1K is a character of M, whcih is called the t r ivial character. If M is a monoid, then every character of M is a monoid homomorphism.3) If −1 a ∈ K, a = 0, then the conjugation κa = (b → aba ) in K is a character of the multiplicative K K monoid of with values in . → | × = a). If χ : M K is a character, where M is finite andχ M is not trivial, then x∈M χ(x) 0. ∈ × = = = (Hint : Let x0 M with χ(x0) 1. Then x∈M χ(x) x∈M χ(x0x) χ(x0) x∈M χ(x).) b). (Lemma on characters)Letϕ1,...,ϕn be characters of M with values in K; Suppose M that ϕ1,...,ϕn are (as an elements of K ) linearly independent over K. If a linear combination = n ∈ = κ ϕ i=1 ai ϕi with coefficients ai K is a character of M, then ϕ ai ϕi for every i with = ∈ = ai 0. (Hint : For x,y M on one side ϕ(x)ϕ(y) i ϕ(x)ai ϕi (y), and on the other side = = ϕ(x)ϕ(y) ϕ(xy) i ai ϕi (x)ϕi (y).) c). (Lemma of Dedekind–Artin) LetK be a field, M be a non-empty semigroup and let ϕi ,i∈ I, be a family of distinct characters of M with values in K. Then ϕi ,i∈ I, are linearly independent over K in KM . (Hint : Use the lemma on characters.) d). Some applications of the lemma of Dedekind–Artin: 1). Let A, K be algebras over a field k, where A is finite dimensional and K is a field. Then there exist at A most Dimk A distinct k–algebra–homomorphisms of A in K.(Hint : Homk(A, K) is a K–subspace of K of the dimension Dimk A. More generally see T7.16.) 2). Let K be a field. The maps t → t n,n∈ N, are the only polynomial maps of K into itself corresponding characters of the multiplicative monoid of K with values in K. More generally: The functions t → t n,n∈ Z, are the only group homomorphisms of K× → K×, corresponding to the rational functions on K×. Deduce that if K is finite, then the multiplicative group K× is cyclic. (see also exercise T4.12.) 3). The functions t → exp at, a ∈ C,ofR in C are linearly independent over C. ν 4). Let K be a field. the sequences (a )ν∈N,a∈ K, are linearly independent over K. (see also exercsie 4.17.) e). (Inner automorphisms of a division rings) Let K be a division ring with the center k. We consider K as a k–algebra. ∈ κ ∈ 1). Let xi ,i I, be a family of non-zero elements K. Then the inner automorphisms xi ,i I, in KK are linearly independent K if and only if x−1,i∈ I, are linearly independent over k. i ∈ × −1 = n −1 ∈ −1 = n −1 = ( Hint : Let x0,x1,...,xn K and x0 i=1 αi xi , αi k. Then x0yx0 i=1 αi x0yxi n −1 −1 κ = n κ = −1 κ = n κ i=1 αi x0xi (xi yxi ) , i.e. x0 i=1 ai xi ,ai : αi x0xi . Conversely, from x0 i=1 ai xi , all 3) Every homomorphism of a monoid into a monoid in which cancellation holds is a monoid homomorphism, i.e. maps the neutral element onto the neutral element. Let ϕ : M → N be a homomorphism. Suppose that cancellation holds in N. Let e ∈ M and e ∈ N be neutral elements. Then ϕ(e) = ϕ(e2) = ϕ(e)ϕ(e) and on the other hand ϕ(e) = e ϕ(e). Therefore ϕ(e) = e . D. P. Patil July 16, 2003 ,3:38 p.m. Basic Algebra ; May-July 2003 ; 7. Homomorphisms of modules 3 a = 0 and lemma on characters, we get κ = κ κ = κ and therefore a x = α x with α ∈ k and i x0 ai xi ai xi i i i 0 i −1 = κ −1 = n −1 −1 = n −1 x0 x0 (x0 ) i=1 ai xi x0 xi i=1 αi xi .) 2). (Theorem of Noether–Skolem–Brauer) IfK is finite dimensional over k, then every −1 −1 k–algebra–endomorphism of K is an inner automorphism of K. ( Hint : Choose a k-basis y1 ,...,yn κ κ ∈ − of K. Then by 1) y1 ,..., yn is a K-basis of EndkK for dimensional reasons. Now, if ϕ Endk AlgK, = n κ = κ κ = κ then ϕ i=1 ai yi and hence using lemma on characters we get ϕ ai yi ai yi for every index i with ai = 0.) 3). Let n := Dimk K be finite. Further, let y1,...,yn be an arbitrary k– basis of K and λi resp. i be the left resp. right multiplications by yi in K. Then λi j = j λi , 1 ≤ i, j ≤ n, form a k–basis of Endk K. ( Hint : Let f ∈ EndkK = Kκ −1 +···+Kκ −1 by part 1). Then for appropriate elements y1 yn n n n n ∈ = κ − = κ − = − ∈ αij k we get f i=1 ai 1 i,j=1 αij yj 1 i,j=1 αij λyj λ 1 ρyi i,j=1 kλyj ρyi . yi yi yi Remark: The part 3) could be reformulated without any use of coordinates in the following way: The canonical k-algebra-homomorphism op K ⊗k K −→ EndkK, x⊗ y −→ λx ρy , is an isomorphism. This shows directly that Kop is the inverse of K in the Brauer group of k.