LEC ALG4 2

Recall that $R$-modules, along with $R$-module homomorphisms, from an abelian category $R\text{-Mod}$. If $\phi :M\to N$ is a homomorphism, then $\phi$ is an isomorphism $⟺$ $\phi$ is bijective. The proof goes exactly as seen here.

Definition 373.1(Definition).

1. A generating set of $M$ is a subset $X\subseteq M$ such that for all $y\in M$, there exist $r_1,\dots ,r_n\in R$ and $x_1,\dots ,x_n\in X$ such that ${\sum }_{i=1}^n{r}_i{x}_i=y$. If finite $X$ exists, $M$ is said to be finitely generated.
2. A set $X\subseteq M$ is said to be linearly independent if ${\sum }_{i=1}^n{r}_i{x}_{i_n}=0$ $⟹$ $r_i=0\forall i$, i.e, every finite trivial linear combination has trivial coefficients.
3. A set $X\subseteq M$ is said to be a basis if $X$ is generating and linearly independent.

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Free modules

$R$ is a (commutative) ring, $M$ is a $R$-module.

Definition 373.2(Free module).

Given a set $A$, the free $R$-module on $A$, denoted $F^R(A)$ (together with a set function $j:A\to F^R(A)$) is an $R$-module such that for all $R$-modules $M$ and set functions $f:A\to M$ there exists a unique $R$-module homomorphism $\phi :F^R(A)\to M$ such that the diagram

commutes.

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Note that $j$ must be necessarily injective!

The construction of such a module generalizes directly the case of abelian groups ¹, replacing $\mathbb{Z}$ with an arbitrary ring $R$. We will denote the coproduct ${⨁}_{a\in A}M$ by $M^{\oplus A}$.

By the virtue of being defined by a universal property, we know that $F^R(A)$, if it exists, is unique up to isomorphism for any given set $A$.

We now provide an explicit realization of $F^R(A)$.

Proposition 373.3(Proposition).

$R^{\oplus A}$ is a free $R$-module on $A$.

Proof.

Recall that $R^{\oplus A}$ consists of all tuples $(r_a{)}_{a\in A}$ with finite support. It is easily seen that $E:=\{e_a:a\in A\}$, where $e_a:A\to R$ are defined by

$$e_a(a^{\prime })=\{\begin{array}{ll}{1}_R& a=a^{\prime }\\ 0_R& a\ne a^{\prime }\end{array}$$

is a basis for $R^{\oplus A}$. Commutativity of the diagram from Def 2 fixes the values of $\phi$ on $E$; Since we require $\phi$ to be a homomorphism, this determines $\phi$ on all of $R^{\oplus A}$. It remains to check that the determined map is indeed an $R$-module homomorphism, which is easily done ².□

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Some preliminary propositions

A simple application of Zorn’s lemma shows that every module has maximal linearly independent subsets. In fact, we can prove something stronger:

Proposition 373.4(Proposition).

Let $M$ be an $R$-module, and let $S\subseteq M$ be a linearly independent subset. Then there exists a maximal linearly independent subset of $M$ containing $S$.

Proof.

Consider the family $\mathcal{I}$ of linearly independent subsets of $M$ containing $S$, ordered by inclusion. Since $S$ is linearly independent, $\mathcal{I}\ne 0$. By Zorn’s lemma, it suffices to verify that every chain in $\mathcal{I}$ has an upper bound. Indeed, the union of a chain of linearly independent subsets containing $S$ is also linearly independent: because any relation of linear dependence only involves finitely many elements and these elements would all belong to one subset in the chain.□

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See Addendum A to 373 for results analogous results on minimal generating sets.

We also see that being free is equivalent to having a basis.

Proposition 373.5(Proposition).

$M$ is a free $R$-module on some set $A$ $⟺$ $M$ has a basis.

Proof.

If $B\subseteq M$ is linearly independent and generates $M$, then the corresponding homomorphism $R^{\oplus B}\to M$ is injective and surjective. Conversely, if $\phi :R^{\oplus B}\to M$ is an isomorphism, then $B$ is identified with a subset of $M$ which generates it (because $\phi$ is surjective) and is linearly independent (because $\phi$ is injective).□

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Prp 5 was used by Kummini as the definition of free module.

Vector spaces

Bases are necessarily maximal linearly independent subsets and minimal generating subsets; this holds over every ring. What makes vector spaces special is that the converse also holds.

We have previously seen that every vector space has a basis (Thm 348.4), from which it follows that every module over a field is free. Here’s another proof using Prp 4: Let $M$ be a $k$-module. Let $A$ be a maximal linearly independent subset of $M$. We will show that $A$ is a basis of $M$. Suppose $v\in M$, $v\notin A$. Then $A\cup \{v\}$ is not linearly independent, by the maximality of $A$; therefore, there exist distinct $a_1,\dots ,a_t\in A$ such that $c_{0}v+c_1{a}_1+\dots c_t{a}_t=0$ for some $c_0,\dots ,c_t\in k$ not all zero. Now, $c_0\ne 0$, since otherwise we would get a linear dependence relation among the elements of $A$. Since $k$ is a field, we can write

$$v=\sum _{i=1}^t\frac{c_i}{c_0}{a}_i,$$

proving that $v\in ⟨A⟩$.

In summary, using the results of this section and Addendum A to 373, we have

Proposition 373.6(Proposition).

Let $M$ be a $k$-module. For $A\subseteq M$, TFAE:

1. $A$ is a basis of $M$;
2. $A$ is a maximal linearly independent subset of $M$;
3. $A$ is a minimal generating set of $M$.

Further, we have the following:

4. $M$ is free.
5. If $S$ is a linearly independent subset of $M$, there exists a basis $S^{\prime }$ of $M$ containing $S$.
6. If $S$ is a generating subset of $M$, there exists a basis $S^{\prime }$ of $M$ contained in $S$.

When working with modules over integral domains, you can easily reduce the problem to the case of a vector space over a field by passing to the field of fractions:

Proposition 373.7(Aluffi (2009) Exr VI.1.7 ✦).

Let $R$ be an integral domain, and let $M=R^{\oplus A}$ be a free $R$-module. Let $k$ be the field of fractions of $R$, and view $M$ as a subset of $V=k^{\oplus A}$ in the evident way. A subset $S\subseteq M$ is linearly independent in $M$ (over $R$) iff it is linearly independent in $V$ (over $k$). Conclude that the rank of $M$ (as an $R$-module) equals the dimension of $V$ (as a $k$-vector space). Prove that if $S$ generates $M$ over $R$, then it generates $V$ over $K$. The converse is not true.

The rank of a free module

We now perform the familiar³ ritual of showing $|X_1|=|X_2|⟺F^R(X_1)\cong F^R(X_2)$, making the notion of rank well-defined and in particular allowing us to call $R^{\oplus n}$ (also denoted by just $R^n$) the free $R$-module of rank $n$. Kummini does this later. Here’s how Aluffi does it:

Lemma 373.8(Aluffi (2009) VI.1.9 ✦).

Let $R$ be an integral domain, and let $M$ be a free $R$-module. Let $B$ be a maximal linearly independent subset of $M$, and let $S$ be a linearly independent subset. Then, $|S|⩽|B|$. In particular, any two maximal linearly independent subsets of a free module over an integral domain have the same cardinality.

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Since every basis for $M$ is a maximal linearly independent subset, it follows that every basis of $M$ has the same cardinality.

Theorem 373.9(Theorem).

Let $R$ be an integral domain, and let $X_1,X_2$ be sets. Then,

$$F^R(X_1)\cong F^R(X_2)⟺X_1\cong X_2.$$

Proof.

$(⟹)$ Let $i_1:X_1\to F^R(X_1)$ and $i_2:X_2\to F^R(X_2)$ be the (injective) set maps associated with $F^R(X_1)$ and $F^R(X_2)$ respectively. Then, $i_1(X_1)$ is a basis for $F^R(X_1)$. If $\phi :F^R(X_1)\to F^R(X_2)$ is an isomorphism, then $\phi (i_1(X_1))$ is a basis for $F^R(X_2)$. Since $i_2(X_2)$ is also a basis for $F^R(X_2)$ and bases are maximal linearly independent subsets, it follows from Lem 8 that $|X_1|=|X_2|$.

$(⟸)$ See Prp 102.6.□

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Corollary 373.10(Corollary).

If $R$ is an integral domain, then

$$R^m\cong R^n⟺m=n.$$

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We can now make the definition

Definition 373.11(Rank and dimension).

Let $R$ be an integral domain. The rank of a free $R$-module $M$, denoted ${\text{rk}}_{R}M$, is the cardinality of a maximal linearly independent subset of $M$. The rank of a vector space is called its dimension, denoted ${\dim }_{k}V$.

Remark 373.12(Remark).

Lem 8 tells us that every linearly independent subset $S$ of a free $R$-module $M$ must have cardinality lower than or equal to ${\text{rk}}_{R}M$. Similarly, every generating set must have cardinality greater than or equal to the rank.

Misc

Proposition 373.13(Proposition).

Let $M$ be an $R$-module. Then there exists a free module $F$ and a surjective $R$-module homomorphism $F\twoheadrightarrow M$.

Proof.

Let $F$ be the free module with basis $\{e_m:m\in M\}$. Define $\phi :F\to M$ by $e_m\mapsto m$.□

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If we have a generating set for $M$, we can do better:

Proposition 373.14(Proposition).

Let $M$ be an $R$-module. Let $X\subseteq M$ be a generating set. Let $F$ be the free module with basis $X$. Then there exists an $R$-linear surjective map $F\to M$ defined by $\sum r_x{e}_x\mapsto \sum r_{x}x$.

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Proposition 373.15(Macdonald & Atiyah (n.d.) Ex. 2.11 ✦).

Let $R$ be commutative.

1. If $\phi :R^m\to R^n$ is surjective, then $m⩾n$.
2. If $\phi :R^m\to R^n$ is injective, then $m⩽n$.

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Analogue of Lem 74.5:

Proposition 373.16(Proposition).

Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. Let $X$ be a finite linearly independent subset of $M$, and $Y$ be a finite generating set of $M$. Then, $|X|⩽|Y|$.

Proof.

Let $X=\{x_1,\dots ,x_s\}$ and $Y=\{y_1,\dots ,y_t\}$. Let $\sigma :R^{\oplus X}\hookrightarrow M$ and $\pi :R^{\oplus Y}\twoheadrightarrow M$ be defined in the obvious ways. Now write $x_i={\sum }_{j=1}^t{a}_{ij}{y}_j$ and define $\psi :R^{\oplus X}\to R^{\oplus Y}$ by $\psi (e_i)={\sum }_{j=1}^t{a}_{ij}{e}_j$. Clearly, $\pi \psi =\sigma$, so $\psi$ is injective. By Prp 15, $s⩽t$.□

Footnotes

1. Which, remember, are just $\mathbb{Z}$-modules! ↩

2. This proof looks very similar to that of Prp 31.1 - noting that every element of the coproduct can be uniquely expressed via a “basis”, using this to define a map, and then verifying that this map is actually a group / module homomorphism. The proofs of Prp 103.2 / Prp 103.3 bypass this by invoking Prp 31.1 / Rmk 31.2.2 respectively and using the universal property of coproducts. You can see a Prp 31.1-style proof of Prp 103.2 at Aluffi (2009) II.5.4. Indeed, you can prove Prp 3 by invoking Rmk 31.2.2 and the universal property of coproducts instead. ↩

3. Prp 103.4, Prp 102.6, Prp 102.7 ↩

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References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.

Macdonald, G., & Atiyah, M. F. (n.d.). Introduction to Commutative Algebra.