notes-archive/docs/math/abstract-algebra/DF-10-modules.md

Dummit & Foote Chapter 10 - Modules

Section 10.1 - Basic Definitions and Examples

Definition. Let R be a ring. A left $R$-module or a left module over $R$ is a nonempty set M together with

1. A binary operation + on M under which M is an abelian group
2. An action \times of R on M, that is, a map or function R \times M \rightarrow M, denoted rm, that for all r, s \in R, m, n \in M satisfies
   • (r + s)m = rm + sm
   • (rs)m = r(sm)
   • r(m + n) = rm + rn
   • If R has identity 1, then 1m = m

Theorem. If R is commutative, any left-module is also a right-module.

Remark. Modules over a field F and vector spaces over F are identical.

Definition An R-submodule is a subset$N \subseteq M$ which is closed under the action taken forall r \in R. That is, given r \in R, n \in N, then rn \in N. Every module has at least two submodules: itself and the trivial (empty) submodule.

Remark. If F is a field, submodules are equivalent to subspaces.

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Example. Let F be a field and F[x] a polynomial ring. Then, let V be a vector space of F, and T be a linear transformation from V to itself. That is, V: T \rightarrow T. We know that V is an $F$-module. We will want to show that V can be written as an $F[x]$-module for some choice of T. That is, we want an action F[x] \times V \rightarrow V.

Now, for a given linear transformation T, consider some polynomial p(x) = a_n x^n + \ldots + a_0 and some v \in V. We define p(x) \times v by$

p(x) \times v = a_n T^n(v) + a_{n-1} T^{n-1}(v) + \ldots + a_0 v

with T^n being defined as applying T a total of n times.

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Proposition. Let R be a ring and M an $R$-module. Then, a subset N of M is a submodule of M if and only if

1. N \neq \emptyset
2. For all r \in R, x, y \in N, then rx - y \in N

Definition. Let R be a commutative ring with identity. An $R$-algebra is a ring A together with a ring homomorphism f: R \rightarrow A such that \varphi(1_R) = 1_A. Thus, the subring f(R) \subseteq A is contained in the center of A.

Recall. The center of a ring A is the subring A' such that for all x, y \in R', then xy = yx. In other words, it is the commutative subring of A.

Definition. Given two $R$-algebras A, B, an *$R$-algebra homomorphism$ is a ring homomorphism \varphi: A \rightarrow B that maps 1_A \rightarrow 1_B such that \varphi(ra) = r\varphi(a).

Section 10.2 - Quotient Modules and Module Homomorphisms

Definition. Let R be a ring and M, N be $R$-modules. then a ring homomorphism \varphi: M \rightarrow N is an $R$-module homomorphism if for all r \in R, \varphi(rx) = r\varphi(x).

Theorem. An $R$-module homomorphism is an isomorphism if it is 1-1 and onto, and said modules are isomorphic.

Definition. Let M, N be $R$-modules. The set \text{Hom}_R(M, N) is the set of all homomorphisms from M to N.

Proposition. Let M, N, and L be $R$-modules. Then,

1. A function \varphi: M \rightarrow N is an $R$-module homomorphism if and only if \varphi(rx + y) = r\varphi(x) + \varphi(y) for all x, y \in M and r \in R.
2. Let \varphi, \psi \in \text{Hom}_R(M, N). Then, define \varphi + \psi as

(\varphi + \psi)(m) = \varphi(m) + \psi(m)

Then, \varphi + \psi \in \text{Hom}_R(M, N). Additionally, if R is commutative, with (r\varphi)(m) = r(\varphi(m)), then r\varphi \in \text{Hom}_R(M,N) 3. If \varphi \in \text{Hom}_R(L, M) and \psi \in \text{Hom}_R(M, N), then \psi \circ \varphi \in \text{Hom}_R(L, N) 4. \text{Hom}_R(M, M) is a ring with identity. With R being commutative, \text{Hom}_R(M, M) is an $R$-algebra.

Proposition. Let R be a ring, M an $R$-module, and N \subseteq M an $R$-submodule. then, M/N can be made into an $R$-module by defining addition. With r \in R and x + N \in M/N,

r(x + N) = (rx) + N

That is,

r \overline{x} = \overline{rx}

Definition. Let A, B be submodules of the $R$-module M. Then, the sum of A and B is defined as

A + B = {a + b | a \in A, b \in B}

This is the smallest submodule that contains both A and B.

Theorem. First Isomorphism Theorem. Let M, N be $R$-modules, and \varphi: M \rightarrow N be an $R$-module homomorphism. Then, \ker \varphi is a submodule of M, and M / \ker \varphi \cong \varphi(M).

Theorem. Second Isomorphism Theorem. Let A, B be submodules of the $R$-module M. Then, (A + B)/B \cong A/(A \cap B).

Theorem. Third Isomorphism Theorem. Let M be an $R$-module, and A \subseteq B be submodules of M. Then, \frac{M/A}{B/A} \cong M/B.

Theorem. Lattice Isomorphism Theorem. Let N be a submodule of the $R$-module M. Then, there is a bijection between submodules of M containing N and submodules of M/N. This is given by A \leftrightarrow A/N, for A \supseteq N.

Section 10.3 - Generation of Modules, Direct Sums, and Free Modules

Definition. Let M be an $R$-module and N_1, \ldots, N_n be submodules of M.

1. The sum of N_1, \ldots, N_n is the set of all finite sums of elements from the sets N_i. That is, N_1, \ldots, N_n := \{a_1 + a_2 + \ldots + a_n | a_i \in N_i\}
2. For any subset A of M, let RA = \{r_1 a_1 + r_2 a_2 + \ldots + r_m a_m | r_i \in R, a_i \in A\}. If N is a submodule of M such that N = RA, then A is called the generating set for N.
3. A submodule N of M is finitely generated if there is some finite subset A of M such that N = RA. That is, N is generated by some finite subset.
4. A submodule of M (up to equality) is cyclic if there exists some element a \in M such that N = Ra = \{ra | r \in R\}.

Definition. Let M_1, \ldots, M_k be a collection of $R$-modules. Then, the direct product is defined as

M_1 \otimes \ldots M_k = (m_1, \ldots, m_k), m_i \in M_i

This direct product is in itself an $R$-module.

Proposition. Let N_1, \ldots, N_n be submodules of the $R$-module M. Then, the following are equivalent:

1. The map \pi: N_1 \otimes \ldots \otimes N_k \rightarrow N_1 + \ldots + N_k defined by \pi(a_1, \ldots, a_n) = a_1 + \ldots + a_n is an isomorphism
2. N_j \cup (N+1 + \ldots + N_{j-1} + N{j+1} + \ldots + N_n) = 0 for all j \in \{1, 2, \ldots, k\}
3. Every x \in N_1 + \ldots + N_n can be written uniquely in the form a_1 + \ldots + a_n, with a_i \in N_i

Definition. An $R$-module F is said to be free on the subset A of F if for every nonzero x \in F, there exists nonzero elements r_1, \ldots, r_n of R and unique a_1, \ldots, a_n such that x = r_1 a_1 + \ldots + r_n a_n for some n \in \mathbb{Z}^+. That is, A is a basis or set of free generators of F.

Theorem. For any set A, there is a free $R$-module F(A) on A such that F(A) satisfies the universal property: if M is any $R$-module, and \varphi: A \rightarrow M is a map of sets, there exists a unique $R$-module homomorphism: \Phi: F(A) \rightarrow M such that \Phi(a) = \varphi(a) for all a \in A.

Corollary. If F_1 and F_2 are free modules on A, then there is a unique isomorphism between F_1 and F_2, which is the identity map on A.

Corollary. If F is a free $R$-module with basis A, then F \cong F(A).

Definition For a free module F with basis A, if R is commutative, then the rank of F is the cardinality of A.

Section 10.4 - Tensor Products of Modules

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Section 10.5 - Exact Sequences - Projective, Injective, and Flat Modules

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