Generated Modules and Linear Independence

Section 13.1 Generated Modules and Linear Independence

Subsection Generated Modules

“You need to learn independence. You have to be independent - it builds character.”
―Mike Tyson

Where We Came From, Where We’re Going.

In Chapter 4, we learned about free groups and presentations, ways of representing groups in more digestible ways. Recall that a Free Group is essentially a way of creating a generic group from a specific set. These sets were the generators of the free group, which was called free due to the lack of relations, or combinations of elements that canceled out to the identity.

In the realm of modules, there are many similarites, such as the notion of a free module. Free modules are, essentially, ways of creating generic modules from specific sets. These sets are called the generators of the free module,

Though, in the world of linear algebra and vector spaces, such sets are referred to as spanning sets.

which is called free due to the lack of relations, or linear combinations of elements that cancel out to zero.

Every group has a Presentation, a set of generators and relations that express the structure of the group. This presentation can be obtained by starting with a free group and taking quotients corresponding to the relations between the generators. Thus, every group can be expressed as the quotient of a free group.

Similarly, we shall see that every module can be expressed as the quotient of a free module. Additionally, the concept of a presentation for a module exists as well, which we will explore in greater depth in Chapter 14.

Definition 13.1. Linear Combination.

Let \(M\) be an \(R\)-module and \(m_1, \dots, m_n \in M\text{.}\) An \(R\)-linear combination of \(m_1, \dots, m_n\) is an element of \(M\) of the form \(r_1m_1 + \cdots + r_nm_n\) for some \(n \geq 0\) and \(r_1,\dots,r_n \in R\text{.}\)

Notice that there are no exponents here, hence the use of the term “linear”.

(If \(n = 0\text{,}\) this gives the empty sum which is interpreted to give \(0_M\text{.}\))

Definition 13.2. Generated Submodule.

Let \(R\) be a ring with \(1 \ne 0\) and let \(M\) be an \(R\)-module. For a subset \(A\) of \(M\) , the submodule of \(M\) generated by \(A\) is

\begin{equation*} RA := \{r_1a_1 + · · · + r_na_n | n ≥ 0, r_i \in R, a_i \in A\}. \end{equation*}

We say \(M\) is generated by \(A\) if \(M = RA\text{.}\)

To use the language of Definition 13.1, an \(R\)-module \(M\) is generated by \(A\) if every element in \(M\) can be written as a linear combination of elements in \(A\text{.}\)

Lemma 13.3. \(R\cdot A\) Smallest Submodule of \(M\) Containing \(A\).

For any subset \(A\) of a \(R\)-module \(M\text{,}\) the subset \(R \cdot A\) is indeed a submodule of \(M\text{,}\) and it is the smallest submodule of \(M\) that contains \(A\) as a subset. In fact, we have

\begin{equation*} RA = \bigcap\limits_{A\subseteq N, N \text{ submodule of }M} N. \end{equation*}

Definition 13.4. Cyclic Module.

If \(M = Ra\) for some single element \(a \in M\text{,}\) we say that \(M\) is cyclic.

Example 13.5. Cyclic Modules.

If \(R = \Z\text{,}\) then (recalling that a \(\Z\)-module is the same thing as an abelian group) we see that \(M\) is a cyclic \(\Z\)-module if and only if \(M\) is a cyclic group.
\(R\) is cyclic as a module over itself, since \(R = R \cdot 1\text{.}\)

More generally, cylic modules are those that can be expressed as a quotient of \(R\) by some ideal.

Exercise 13.6. Cyclic Modules and Ideals.

A left \(R\)-module \(M\) is cyclic if and only if there exists a left ideal \(I\) of \(R\) such that \(M \cong R/I\text{.}\)

Solution.

Say \(M\) is cyclic and \(m \in M\) is a generator of \(M\text{,}\) so that \(M = \{rm \mid r \in R\}\text{.}\) Define \(f: R \to M\) to be the unique \(R\)-map with \(f(1) = m\text{.}\) Here I am applying the UMP for bases, using that \(\{1\}\) is a basis of \(R\) as a left \(R\)-modules. More explicitly,

\begin{equation*} f(r) = f(r\cdot 1) = r f(1) = rm \end{equation*}

for all \(r \in R\text{.}\) Then \(f\) is onto, since \(m\) generates \(M\text{.}\) Its kernel is a left ideal \(I\) of \(R\text{,}\) since submodules of \(R\) are the same thing as left ideals. By the Item 2, there is an isomorphism \(R/I \xra{\cong} M\) sending \(r + I\) to \(rm\text{.}\)

Definition 13.7. Finitely Generated Module.

A module \(M\) is finitely generated if there exists some finite subset \(A\) of \(M\) such that \(M = R \cdot A\text{.}\)

Remark 13.8.

For any \(R\)-module \(M\text{,}\) we have \(R \cdot \emptyset = \{0_M\}\text{.}\) This is because the empty sum is interpreted as giving \(0_M\text{.}\)

Example 13.9. Standard Free Module Finitely Generated.

Let \(R\) be a ring. The standard free \(R\)-module CITEX of rank \(n\text{,}\) \(R^n\text{,}\) is finitely generated, since it is generated by \(e_1, \dots, e_n\) where \(e_i := \begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}\text{,}\) with a \(1\) in the \(i\)-th position. This holds since given any element \(v = \begin{bmatrix} r_1 \\ \vdots \\ r_n\end{bmatrix}\) of \(R^n\) we have \(v = \sum_i r_i e_i\text{.}\)

In particular, taking \(n = 1\text{,}\) \(R\) is cyclic as a module over itself, since \(R = R \cdot 1\text{.}\) More generally, for any (two sided) ideal \(I\text{,}\) \(R/I\) is a cyclic left \(R\)-module, generated by \(\ov{1}\text{.}\)

Lemma 13.10. Finite Generation and Quotient Modules.

Let \(R\) be a ring with \(1 \neq 0\text{,}\) let \(M\) be an \(R\)-module, and let \(N\) be an \(R\) submodule of \(M\text{.}\)

If \(M\) is finitely generated as an \(R\)-module, then so is \(M / N\text{.}\)
If \(N\) and \(M / N\) are finitely generated as \(R\)-modules, then so is \(M\text{.}\)

Proof.

Note that if \(M=R A\) then \(M / N=R \bar{A}\text{,}\) where \(\bar{A}=\{a+N \mid a \in A\}\text{.}\)
Coming soon!

Subsection Linear Independence

Definition 13.11. Linearly Independent.

Let \(M\) be an \(R\)-module and let \(A\) be a subset of \(M\text{.}\) The set \(A\) is linearly independent if whenever \(r_1,\ldots,r_n \in R\) and \(a_1,\ldots ,a_n\) are distinct elements of \(A\) satisfying \(r_1a_1 + \dots + r_na_n = 0\text{,}\) then \(r_1 = \dots = r_n = 0\text{.}\) Otherwise \(A\) is linearly dependent.

Remark 13.12.

The empty subset of any module is linearly independent (vacuously).

Example 13.13. One Element Subsets of \(R\)-Modules.

A one element subset \(\{m\}\) of an \(R\)-module \(M\) is linearly independent if and only if whenever \(rm =0\text{,}\) we have \(r = 0\text{.}\)

But it is possible for one elements subsets to be linearly dependent: For example, let \(R\) be any ring and \(I\) and (two-sided) ideal such that \(I \ne 0\text{.}\) Then I claim that every non-empty subset of \(M = R/I\) is linearly dependent. For say \(A\) is a such a nonempty subset. For any \(a \in A\text{,}\) pick any \(r \in I\) such that \(r \ne 0\text{.}\) Then \(ra = 0\) (since \(a = y + I\) for some \(y \in R\) and hence \(ra = ry + I = I = 0_M\)) and this shows \(A\) is linearly dependent. In particular, even a one-element subset of \(M\) is linearly dependent.

Example 13.14. \(\{3\}\) Linearly Independent in \(\Z\).

The singleton \(\{3\}\) is a linearly independent subset of the \(\Z\)-module \(M = \Z\text{.}\) But it does not generate all of \(M\text{.}\) The subset \(\{3, 5\}\) does generate all of \(M\text{,}\) but it is not linearly independent, since \(5 \cdot 3 + (-3) \cdot 5 = 0\text{.}\) More on this later.

Exploration 13.1. Maximally Linearly Independent.

Let \(R\) be a commutative integral domain and \(M\) an \(R\)-module. Recall that a subset \(S\) of \(M\) is called a maximal linearly independent set of \(M\) if \(S\) is linearly independent and any subset of \(M\) properly containing \(S\) is linearly dependent.

Let \(T\) be a linearly independent subset of \(M\text{.}\) Prove that \(T\) is contained in some maximal linearly independent subset of \(M\text{.}\)
Let \(T\) be a linearly independent subset of \(M\) and \(N\) the \(R\)-submodule of \(M\) generated by \(T\text{.}\) Prove that \(T\) is a maximal linearly independent subset if and only if \(M/N\) is torsion. (Recall that an \(R\)-module \(P\) is called torsion if for each \(p \in P\text{,}\) there is a \(r \in R\) such that \(r \ne 0\) and \(rp = 0\text{.}\))

Solution.

Let \(A\) be the set of all linearly independent subsets of \(M\) that contain \(T\text{.}\) We can order \(A\) with respect to inclusion. Let \(X\) be a totally ordered subset of \(A\text{,}\) and let \(U\) be the union of all elements in \(X\text{.}\) Let \(\{u_i\}\) be a set of elements in \(U\) such that \(\sum_{i=0}^nr_iu_i=0\) for some \(r_i\in R\text{,}\) where \(i\leq n\) for some \(n\in\N\text{.}\) As \(U\) is the union of all elements in \(X\text{,}\) there exists some \(B_1\) such that \(u_1\in B_1\text{.}\) However, as \(X\) is totally ordered, there exists some \(B_2\) such that \(B_2\) contains \(B_1\) and \(u_2\in B_2\text{.}\) Continuing in this way, we see that there exists some \(B_n\in X\) such that \(\{u_i\}\subseteq B_n\text{.}\) As \(B_n\) is linearly independent, we know that \(\sum_{i=0}^nr_iu_i=0\) means that \(r_i=0\) for all \(i\text{.}\) Thus \(U\) is indeed linearly independent, making it an upper bound for \(X\text{.}\) Thus by Zorn’s Lemma there exists a maximal element of \(A\text{,}\) which we denote \(S\text{.}\) Thus \(S\) is linearly independent, contains \(T\text{,}\) and is maximal.
\((\Rightarrow)\) Suppose \(T\) is maximal linearly independent, and suppose by way of contradiction that \(M/N\) is not torsion. Thus there exists some \(xN\in M/N\) such that for all \(r\in R\text{,}\) we see that \(rx\neq0\text{.}\) However, as \(x\neq 0\) and \(x\not\in N\text{,}\) this means that \(x\not\in T\text{.}\) Consider \(T\cup\{x\}\text{.}\) This set is linearly independent, contradicting the assumption that \(T\) was maximal.

\((\Leftarrow)\) Suppose \(M/N\) is torsion. Let \(x\neq0\in M\) and consider \(T\cup\{x\}\text{.}\) Consider \(x+N\in M/N\text{.}\) As \(M/N\) is torsion, there exists an \(r\neq0\in R\) such that \(rxN=0\text{.}\) Thus \(rx\in N\text{.}\) (Note, if \(x\in N\text{,}\) then \(r=1\)). As \(rx\in N\) and \(N\) is generated by \(T\) (\(N=RT\)), \(rx=\sum r_it_i\text{.}\) Subtracting over we see that \(rx-\sum r_it_i=0\text{.}\) But as \(r\neq 0\text{,}\) we see that each \(t_i\) and \(x\) are in \(T\cup\{x\}\text{,}\) but the sum is \(0\text{.}\) Thus \(T\cup\{x\}\) is linearly dependent.

Qual Watch.

Proving Exploration 13.1 was [provisional cross-reference: may-2018-5] on the [provisional cross-reference: may-2018] qualifying exam.

Summary

An \(R\)-module \(M\) is generated by a set \(A\) if every element of \(M\) can be written as an \(R\)-Linear Combination of elements of \(A\text{.}\) The module \(M\) is finitely generated if the set \(A\) is finite.
Every ring \(R\) can viewed as a cyclic \(R\)-module generated by \(\{1\}\text{.}\)
A set \(A\) is Linearly Independent if there are no \(R\)-linear combinations of elements in \(A\) that equal \(0\text{.}\)

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