Presentation 1

Transcription

1 Presentation 1 Jesse Selover March 5, 2015 We re going to try to cover a pretty strange result. It might seem unmotivated if I do a bad job, so I m going to try to do my best. The overarching theme of my presentation is modules are almost as good as vector spaces. 1 Direct Sums First we ll define the direct sum of modules. This is almost the same as the direct product we saw on Tuesday: 1.1 Definition If M i, i I are all R-modules, we define the direct sum to be the set of sequences i I M i {(a i ) i I a i M i, finitely many nonzero a i } with finitely many nonzero terms. If the index set I is finite, the direct sum is exactly the same as the direct product. Now I will wave my hands and say that if all the M i are R-modules, i I M i is an R-module. The proof is the same as for the direct product, plus you note that two sequences with finitely many nonzero terms have a sum with finitely many terms. By abuse of notation, we ll write a + b instead of (a, b), when it s clear where a and b live. We re identifying the M i with submodules of the direct sum. Last time we covered the sum of submodules. In keeping with the theme of today (modules are like vector spaces), it s also true Proposition: If M 1, M 2 M are submodules, and M = M 1 + M 2, 1

2 then M 1 M 2 = 0 iff M = M 1 M 2. You know this from vector spaces, and the proof is essentially the same. An important tool for understanding a vector space is the basis. Modules don t always have bases, which is terrible. But we have similar concepts. 2 Generators Let m i M, for i I. If Rm i = M, i we say that the m i generate M. If finitely many are enough, then the module M is finitely generated. Now, there is always a minimal generating set for a module. Over a field, this would be a basis. A basis needs to also be linearly independent; this might not happen for a module! 2.1 Example Here s an example of a minimal generating set which is not linearly independent. We have the representation of Z/2 given by ( ) The ring of invariants is F [x 2, xy, y 2 ], as we ve seen before. Now, the ring of polynomials F [x, y] is a module over this ring of invariants: there s no problem multiplying polynomials by invariant polynomials, and that multiplication is associative so it will satisfy the module condition. One minimal generating set for F [x, y] as an F [x 2, xy, y 2 ]-module is {1, x, y}. If we remove 1, we can check that we ll no longer be able to generate 1; same for x and y, so this generating set is indeed minimal. But these elements aren t linearly independent over F [x 2, xy, y 2 ]: (xy)x (x 2 )y = 0, which is a nontrivial linear dependence relation. This is really bad. But some modules are understandable: 2

3 3 Free Cover A free module is a module that is a direct sum of copies of R. First off: all vector spaces are free. R n is a free R-module, and is also finitely generated. The canonical basis is given by the e i with 1 in the ith component and 0 elsewhere. It s linearly independent, as well as being a minimal generating set, so a free module is very easy to understand. Proposition: M is finitely generated iff it is a quotient of a free module. Another way of saying this is that finitely generated modules are exactly the same thing as free modules plus some relations. Proof: Suppose M is a finitely generated module, generated by a 1,..., a n. It suffices to show that there is some surjective map φ : R n M, because then by the first isomorphism theorem M is isomorphic to R n / ker(φ). We ll define a map from R n M by φ(e i ) = a i. Since the e i are independent generators, we can send them wherever we want to define a module homomorphism, so this is a module homomorphism. Let s look at the image of φ. Well, a i im (φ), so Ra i im (φ). i But the left-hand-side is already M, which is the maximum possible image, so the image is everything, and so R n surjects onto M! By the first isomorphism theorem, M is then isomorphic to some quotient of R n. Ok, so modules (at least the finitely generated ones) can t get too bad. Let s show graded modules are even better. 4 Graded Nakayama 4.1 Setup Let R be a non-negatively-graded F -algebra, with R 0 = F. Let m be its augmentation ideal: m = R i, i 1 the positive degree components. R/m = R 0, so we require that R 0 = F so that this quotient is a field and m is maximal. Now suppose M is a finitely generated graded R-module. It s an exercise in the book to show that M/(mM) is a module over R/m. Since R/m is a field, M/(mM is a vector space, and it makes sense to talk about a basis for it. 3

4 4.2 Nakayama Lemma The main result is as follows: Lemma (Nakayama): Let m 1,..., m n M, and suppose that the images of these elements under the map M M/(mM) form a basis for that vector space. Then m 1,..., m n generate M. Proof: We have q : M M/(mM), with kernel mm. Let a M. We have a basis for M/(mM, so we know we can write q(a) = a 1 m a n m n + mm. So a = a 1 m a n m n + r, with r mm. M is almost generated by the m i, but there is this remainder term; we could write M = i Rm i + mm, but we have to show that mm isn t necessary. Let s take a break from the proof to prove a strangely useful lemma: 4.3 Strangely Useful Lemma Suppose that mm = M. Then M = 0. Proof: The proof is by contradiction. M is finitely generated, and R is nonnegatively graded, so there must be some least graded component. Let k be the smallest number such that M k 0. We supposed already that mm = M, so all of their graded components must be equal. Let s look at (mm) k = M k. m is generated by elements of at least degree 1, so it shifts degrees up. Consequently, m(m i ) has no degree i part. Since k is the lowest degree, the left-hand-side (mm) k is zero, and M k = 0. But we just assumed M k was the smallest non-zero component, so we get a contradiction, and all the components of M are zero. There s a very useful corollary to this lemma, which is that 4

5 4.4 Strangely Useful Corollary If M = N + mm, then M = N. Proof: We will prove m(m/n) = M/N, and then by the lemma above M/N will be zero. This means M = N. So let s compare M/N to m(m/n). We ve supposed M = N + mm, so M/N = (N + mm)/n. Applying the second isomorphism theorem for modules, (N + mm)/n = (mm)/(mm N). Now, we ll try to show m(m/n) is equal to this. Maybe you can just see this, but let s do it rigorously. When we want to show a particular quotient is isomorphic to another module, one way is to use the first isomorphism theorem: there is a map given by sending φ : mm m(m/n), φ(a 1 m a k m k ) = a 1 (m 1 + N) + + a k (m k + N), and the kernel of this map is exactly those elements of mm which are in N. So m(m/n) = mm/(ker(φ)) = (mm)/(mm N). As we showed above, this is M/N, so m(m/n) = M/N. Back in the realm of the Graded Nakayama Lemma, we had shown M = i Rm i + mm. By this corollary, that means that M = i Rm i, and the m i are truly a generating set. Hooray! 5 Group Rings and Maschke s Theorem Stepping back from modules for a moment, let s go back to groups. Definition: The group ring of a finite group G over a field F is the set of formal sums { } a g g a g F. 5

6 We denote it F G. These are just linear combinations of formal symbols representing the group elements. The addition is just component-wise: ( ) ( ) a g g + b g g = g + b g )g. (a The multiplication is extended from the group structure: ( ) ( ) a g g b g g = (a g b h )gh. g,h G This ring is commutative if and only if G is an abelian group. Now, representations of G are going to be the same thing as modules over the ring F G: If we start with a representation ρ : G GL(n, F ), the module structure on F n is given by ( ) f g g v = f g (ρ(g)v). If we have an F G-module M, then M is a vector space over F (because F F G, so M is also an F -module). The G-action on M is given by ρ(g)v = g v, where the right-hand side uses the module multiplication. Now we can state a great theorem: Theorem (Maschke): Let G be a finite group, and V a module over CG. If U V is a CG-submodule, there is a complementary CG-submodule W V such that V = U W. The proof will be given by Jordan. 6