Lecture 19 Notes: Modules over a Ring

Lecture 19: Modules over a Ring 19 Modules over a Ring The motivation for modules is that we are trying to tell a story where rings are the protagonist, and for a story to be interesting, the protagonist must act. When we find a way for a ring to act, we get the definition of a module. Definition 19.1 Let R be a ring. A module M over R is an abelian group, together with an action map R × M → M (written as (r, m) 7→ r(m) or rm), subject to the following axioms: • 1R (m) = m for all m im M ; • r1 (r2 (m)) = (r1 r2 )(m) for all r1 , r2 ∈ R and m ∈ M ; • Distributivity in both variables: (r1 + r2 )m = r1 m + r2 m for all r1 , r2 ∈ R and m ∈ M , and r(m1 + m2 ) = rm1 + rm2 for all r ∈ R and m1 , m2 ∈ M . The first two axioms are very similar to the definition of a group action on a set. So a ring to a module is like a group G to a G-set (a set with an action by G). It’s not exactly the same because in a ring we have two operations instead of one, but they’re a similar flavor. 19.1 Examples Example 19.2 If R = F is a field, then a module is the same as a vector space. The axioms here are exactly the same. The textbook emphasizes this heavily, and this analogy can get you some mileage; but for general rings, things become more complicated. Note 19.3 The definition also applies to a noncommutative ring R, in the same way — our definition does not reference commutativity. Then we have some familiar examples of modules over noncommutative rings: for example, given any field F we can take R = Matn×n (F ) and M = F n , since matrices act on column vectors by multiplication. As another example, if R = C[G] is the group ring, then a R-module is the same as a complex representation of G. For any ring R, there is a uniquely defined homomorphism Z → R, where 1 7→ 1R . On a similar note, every abelian group has a unique structure of a Z-module: we know that 1 (in Z) must map m 7→ m, so then by distributivity, n = 1 + 1 + · · · + 1 must map v 7→ v + v + · · · + v . | {z } n Similarly, −n must map v to −(v + v + · · · + v). So a Z-module is the same as an abelian group. Example 19.4 What is a module over R = C[x]? Proof. First, a C[x]-module is a C-vector space V just scalars). But then we also need to see what where xv = Av. There are no constraints on this so a R-module is a vector space V , together with polynomial P (x) = an xn + · · · + a0 is given by by looking at the action of constant polynomials (which are x does. We know x must act by a linear map A : V → V , map, and this defines the action of every other polynomial: a linear map A : V → V . Explicitly, the action of a general P v = a0 v + a1 Av + · · · + an An v. Note that the vector space may or may not be finite-dimensional; if it is, then we end up in a situation studied in linear algebra, where we have a vector space and a linear operator. 87 Lecture 19: Modules over a Ring Example 19.5 What is a module over R = Z/nZ? Proof. The main point is that if R/I is a quotient of R, then every R/I-module is also a R-module, where we define r(m) to be r(m) (here r denotes r mod I). Meanwhile, in order to go backwards, I must act by 0. So a R/I module is the same as a R-module where every element of I acts in a trivial way (meaning that rv = 0 for all r ∈ I and v ∈ M ). So in this case, a Z/nZ-module is the same as an abelian group where the order of every element divides n — meaning na = 0 for all a in the group. Then more concretely, for every m (where we use m to denote m mod n), we can write mv = v + v + · · · + v . {z } | m In order for this to be well-defined, the sum should not depend on the choice of representative for the residue; but this is guaranteed by the condition na = 0. (This is the same reasoning as in the first paragraph, for this specific example.) For any ring R, there is a simple example of a module: Definition 19.6 The free module over R is M = R itself, where the action is multiplication (meaning that r(x) = rx). This is parallel to the observation that a group G acts on itself by left multiplication. 19.2 Submodules Definition 19.7 Given a module M , a submodule N ⊂ M is an abelian subgroup which is invariant under the R-action — meaning rx ∈ N for all x ∈ N and r ∈ R. If N ⊂ M is a submodule, we can define their quotient M/N , where we take the quotient in the sense of abelian groups. This quotient of abelian groups carries a module structure as well, given by the obvious rule rm = rm (where m denotes m mod N ). This is well-defined because N is a submodule — if m1 − m2 is in N , then rm1 − rm2 = r(m1 − r2 ) is in N as well. Then the homomorphism theorem and correspondence theorem work in the exact same way as in abelian groups. (For rings and ideals, we saw they work in a similar way; but here the parallel is closer.) Example 19.8 What are the submodules of the free module M = R? Proof. The answer is exactly the ideals of R — we’re looking for abelian subgroups of R which are invariant under multiplication by all terms in R, and by definition these are ideals. We’ll later see how to understand any module by looking at generators and relations — this turns out to be easier than the corresponding problem for a group. But first we’ll look at another example of a module, which will be useful for developing that theory. 88 Lecture 19: Modules over a Ring Definition 19.9 Given two modules M and N , their direct sum is M ⊕ N = {(m, n) | m ∈ M, n ∈ N } with the action r(m, n) = (rm, rn). Note 19.10 The direct sum is the same as the product M × N . This is true for any finite sum — we have M1 ⊕ · · · ⊕ Mn = M1 × · · · × Mn . But this isn’t true for infinite sums and products. Definition 19.11 The free module of rank n is Rn = R ⊕ R ⊕ · · · ⊕ R . {z } | n In the case where R = F is a field, the free module of rank n is exactly F n , the standard n-dimensional vector space. 19.3 Homomorphisms In linear algebra, we work with matrices in order to understand linear maps. Matrices are also relevant here — the terms are different, but the concept is very similar. Definition 19.12 A homomorphism from a module M to a module N is a homomorphism of abelian groups φ : M → N , which is compatible with the R-action — meaning φ(rm) = rφ(m) for all r ∈ R and m ∈ M . In vector spaces, this is the same as a linear map. We’ll use HomR (M, N ) to denote the set of all homomorphisms M → N . Note that homomorphisms can be added and rescaled, in the same way as linear maps: (φ1 + φ2 )(m) = φ1 (m) + φ2 (m), and (rφ)(m) = rφ(m). So then HomR (M, N ) is itself a R-module. Understanding homomorphisms in general may be hard, but it’s easy to understand homomorphisms from a free module. Given a homomorphism φ ∈ HomR (R, M ) for any module M , we can let m = φ(1R ). Then this determines the entire homomorphism — for any r ∈ R, we have φ(r) = φ(r · 1R ) = r · φ(1R ) = rm. So a homomorphism is determined by m = φ(1R ), and there are no restrictions on m — this is why R is called a free module. This means HomR (R, M ) is isomorphic to M : more explicitly, the bijection is given by mapping φ ∈ HomR (R, M ) to mφ = φ(1), and m ∈ M to the homomorphism φm : r 7→ rm. Similarly, HomR (Rn , M ) is equally easy to understand. Now Rn is generated by the elements 1i which have a 1 in their ith place, and 0’s everywhere else (so 1i = (0, . . . , 0, 1, 0, . . . , 0), where the 1 is in the n ith place). So HomR (Rn , M ) is isomorphic to M n , where the bijection sends φ ∈ HomR (R P , M ) to the element (φ(11 ), φ(12 ), . . . , φ(1n )), and (m1 , . . . , mn ) ∈ M to the homomorphism φ(x1 , . . . , xn ) = xi mi . In particular, we have HomR (Rn , Rm ) = (Rm )n = Matm×n (R) — we can write homomorphisms in the way we’re used to in linear algebra, where A ∈ Matm×n (R) sends (x1 , . . . , xn )t to A(x1 , . . . , xn )t . So as long as we work with free modules and homomorphisms, to a large extent we can operate as if we’re doing linear algebra. But in linear algebra, there’s various characterizations of nondegenerate matrices that no longer hold here — 89 Lecture 19: Modules over a Ring for instance, a linear operator that is injective (meaning it has zero kernel) is also surjective, but that’s not true for general modules. 19.4 Generators and Relations Definition 19.13 A collection of elements m1 , . . . , mn ∈ M forms a system of generators if every x ∈ M can be expressed P as ri mi for ri ∈ R. So in other words, φm1 ,...,mn : Rn → M is onto. If such a finite set exists, we say that M is finitely generated. Many modules we’re interested in are in fact finitely generated. If this map is also one-to-one, then it’s an isomorphism, and M is free. But usually this won’t happen, and we still want to describe M . To do this, we can look at K = ker(φ), which is a submodule in Rn . If K is itself finitely generated, then we can choose a set of generators for K, and get a somewhat explicit description of M — we can fix a system of k generators for K, and obtain another homomorphism ψ : Rk → K. Since K sits inside Rn , we can think of ψ instead as a homomorphism ψ : Rk → Rn , with image K. But such a homomorphism corresponds to a matrix A ∈ Matk×n , and we have M = Rn /ARn . Definition 19.14 If we can find a finite set of generators for M such that the kernel is also finitely generated, then M is called finitely presented. We’ll see that for many rings, the finitely presented modules are a very large class of modules — in fact, for many rings, any finitely generated module is finitely presented. We’ll also see how to make this description of a module very explicit when the ring is a Euclidean domain. In particular, taking the Euclidean domain to be Z will give us a classification of all finitely generated abelian groups, and taking the Euclidean domain to be F [x] will give us Jordan normal form! 90