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More on modules

Sums of modules

Suppose that $R$ is a ring and $M$ is an $R$-module. Let $N_1,\ldots, N_{k}$ be submodules of $M$. Then the sum $N_1+\ldots+N_{k}$ is the collection

\[N_1+\ldots+N_{k}=\lbrace n_1+\cdots+n_{k} : n_{i}\in N_{i}\rbrace\]

It is a submodule of $M$ and the smallest submodule containing all the $N_{i}$.

One can also consider infinite collections of submodules:

\[\sum_{i\in I} N_{i} = \lbrace \sum_{j\in J} n_{j} : n_{j}\in N_{j},\ J\subset I\mathrm{\ finite\ }\rbrace\]

Generating submodules (compare vector spaces)

Suppose $A\subset M$. Then the submodule $RA$ of $M$ generated by $A$ is the smallest submodule of $M$ containing $A$. In practice it is the collection

\[RA = \lbrace r_1 a_1 + \cdots + r_k a_k : r_1,\ldots, r_k\in R, a_1,\ldots, a_k\in A, k\in\Z, k\ge 0\rbrace\]

In linear algebra, we would say that $RA$ is the submodule of $M$ that is spanned by $A$ and this terminology can be used here as well.

We can also say that $RA$ is the set of (finite) $R$-linear combinations of elements of $A$.

Generating sets - an example

Suppose that $V$ is a $\Q$-vector space of dimension $n$ and $w_1,\ldots, w_k$ are a set of vectors in $V$.

Since $V$ is also a $\Z$ module (by “restriction of scalars”) we can consider the sub-$\Z$-module of $V$ generated by the $w_{i}$. This is all $\Z$-linear combinations of the $w_i$.

For example if $V=\Q^{2}$ and $A=\lbrace w_1, w_2\rbrace$ are the standard basis elements then $\Z A$ is the subset of $V$ of vectors with integer coefficients in the standard basis.

Finite generation

Definition: An $R$-module $M$ is finitely generated if there is a finite subset $A\subset M$ such that $RA=M$.

Note that $\Q$ is finitely generated as a $\Q$-module (in fact it’s generated by one element) but not as a $\Z$-module.

For vector spaces, finitely generated means finite dimensional. A generating set is the same as a spanning set.

Comparison with vector spaces

A set $m_1,\ldots, m_k$ in an $R$-module $M$ is linearly independent if, whenever $\sum r_i m_i=0$, all $r_i=0$.

For vector spaces, a maximal linearly independent set (meaning a linearly independent set which becomes dependent when any nonzero element is added to it) automatically spans the vector space, and we call this a basis.

For modules, this fails. Consider $\Z^{2}$ and let $e_1=[2,0]$ and $e_2=[0,2]$. If $e=[a,b]$ then \(2e-ae_1-be_2=0\) so $e_1, e_2$ is a maximal linearly independent set. But they don’t generate all of $\Z^{2}$.

Cyclic modules

Definition: An $R$ module $M$ is cyclic if it is generated by one element: $M=Ra$ for some $a\in M$.

  • Cyclic groups are cyclic $\Z$-modules.
  • If $R$ is a ring with unity and $I$ is a left ideal, then $R/I$ is a cyclic $R$-module generated by $1+I$.
  • If $R$ is a ring with unity, an ideal $I$ is a cyclic module if and only if it is a principal ideal.
  • If $R=M_{n}(F)$ for a field $F$ and $M=F^{n}$ is the space of column vectors viewed as an $R$-module, then $M$ is cyclic.

If $R=\Z[i]$, then $(1+i)R$ is a cyclic module for $R$ generated by $(1+i)$. But if we view $(1+i)R$ as a $\Z$-module inside the $\Z$-module $R=\Z+\Z i$ then $(1+i)R$ is generated over $\Z$ by $1+i$ and $(1+i)i=i-1$; it is not cyclic as a $\Z$-module.

Characterization of cyclic modules

Proposition: Let $M$ be a cyclic $R$-module. Then $M$ is isomorphic to $R/I$ where $I$ is a left ideal of $R$.

Proof: Let $m\in M$ generate $M$. Consider the map $f: R\to M$ defined by $f(r)=rm$. This is a module homomorphism since

\[f(r_1 r_2)=r_1 r_2 m = r_1 (r_2 m) = r_1 f(r_2c).\]

(Remember that we are thinking of $R$ here as an $R$-module, not a ring.)

Characterization of cyclic modules cont’d

The kernel of the map $f(r)=rm$ is the set $I=\lbrace r\in R : rm=0\rbrace$.

This is a left ideal since if $rm=0$ then $srm=0$ for all $s\in R$.

Since $M$ is cyclic, the map $f$ is surjective.

Therefore by the isomorphism theorem $M$ is isomorphic to $R/I$.

More on cyclic modules

Recall that a module $M$ for $F[x]$ is the same as an $F$-vector space $V$ together with a linear map $T:V\to V$.

If $M$ is cyclic then there is an $m\in M$ so that every $m’\in M$ is given by $p(x)m$ for some $p(x)\in F[x].$

This means that that there is a vector $v\in V$ so that every vector $v’\in V$ is of the form $p(T)v$. In other words, the set $v, Tv, T^2v,\ldots, T^{n}v,\ldots$ spans $V$.

If $V=F^{2}$ and $T$ satisfies $Te_{1}=0$ and $Te_{2}=e_{2}$ then $V$ is not cyclic.

If $Te_{1}=0$ and $Te_{2}=e_{1}$ then $V$ is cyclic and generated by $e_{2}$. Also $T^2e_2=0$ and so as an $R$-module $V$ is isomorphic to $F[x]/(x^2)$.

Direct Sums and Direct Products

Direct Products (definition)

Suppose that $M_{1},\ldots, M_{k}$ are $R$ modules. The direct product $M_1\times\cdots\times M_{k}$ of the $M_{i}$ is the set of “vectors” $(m_1,\ldots, m_k)$ with $m_{i}\in M_{i}$. Addition and multiplication by $R$ are done componentwise.

Internal direct sums

Suppose that $M$ is an $R$-module and $N_1,\ldots, N_{k}$ are submodules of $M$. There is a module homomorphism

\[N_1\times\cdots\times N_{k}\to N_1+\cdots N_{k}\subset M\]

defined by sending $(n_1,\ldots, n_k)\to n_1+\cdots n_k$.

Internal direct sums (continued)

Definition: The sum map above is an isomorphism if and only if either of the following two conditions are satisfied:

  • $N_{j}\cap (N_1+\cdots N_{j-1}+N_{j+1}+\cdots N_{k})=0$ for all $j=1,2,\ldots, k$
  • Any $x\in N_1+N_2+\ldots+N_{k}$ can be written uniquely as a sum $x=n_1+n_2+\ldots +n_k$ with $n_{i}\in N_{i}$.

If $M$ is isomorphic to $N_1\times \cdots \times N_{k}$ via the sum map, we say that

\[M = N_1\oplus N_2\oplus \cdots \oplus N_{k}\]

and say that $M$ is the internal direct sum of the $N_{i}$.

Direct Sums vs Direct Products


Suppose that $I$ is a set and $M_{i}$ is an $R$-module for each $i\in I$.

The direct product $\prod_{I} M_{i}$ is the collection of all functions $f:I\to \cup_{i\in I} M_{i}$ such that $f(i)\in M_{i}$. It is an $R$-module: $(f+g)(i)=f(i)+g(i)$ and $(rf)(i)=r(f(i))$.

The direct sum $\oplus_{I} M_{i}$ is the submodule of $\prod_{I} M_{i}$ consistsing of functions $f$ with the additional property that there is a finite subset $J\subset I$ such that $f(i)=0$ unless $i\in J$.

Notice that if $I$ is finite then these two things are the same.

Countable sums and products

Suppose that $I=\N$, the natural numbers, and $M_{i}$ is a family of $R$-modules indexed by $I$. Then:

  • $\prod_{i\in I} M_{i}$ consists of sequences $(m_1,m_2,\ldots,m_k,\ldots)$ where $m_{i}\in M_{i}$.
  • $\oplus_{i\in I} M_{i}$ consists of sequences $(m_1,m_2,\ldots, m_k,\ldots)$ where $m_{i}\in M_{i}$ and there is an $N$ such that $m_{i}=0$ for all $i\ge N$.

Notice that, if each $M_{i}$ is countable, then so is $\oplus_{i\in I}M_{i}$, but $\prod_{i\in I}M_{i}$ is not.

Free Modules


Definition: A module $M$ is free on a set $A$ of generators if, for every nonzero element $m$ of $M$, there are unique nonzero $r_1,\ldots, r_k$ in $R$ and elements $a_1,\ldots, a_k$ in $A$ such that

\[m = r_1 a_1 + \cdots + r_k a_k.\]

Such a set $A$ is called a basis of $M$, so a module $M$ is free if it has a basis.

Examples and non-examples

If $A=\lbrace a_1,\ldots,a_n\rbrace$ is finite, then $M$ is free on $A$ if the map

\[\oplus_{i=1}^{n} R \to M\]

defined by $(r_1,\ldots, r_n)\mapsto r_1 a_1+\cdots+r_n a_n$ is an isomorphism. So basically $M$ is free on a set $A$ with $n$ elements if and only if it is isomorphic to $R^{n}$.

If $R=\Z$, then $M=\Z/m\Z\oplus \Z/m\Z$ is not free on $(1,0)$ and $(0,1)$. Every $m\in M$ is a linear combination $r_1(1,0)+r_2(0,1)$ for $r_1,r_2\in\Z$, but $r_1$ and $r_2$ are not uniquely determined. In fact $M$ is not free on any set of generators.

Any vector space over $F$ is a free $F$-module.

Rings with nonprincipal ideals.

A principal ideal in a (commutative) ring is a free module, but a non-principal ideal is not. Consider $I=(2,1+\sqrt{-5})\subset R=\Z[\sqrt{-5}]$. Choose any two elements of this ideal, say $x$ and $y$. Then $-y\cdot x + x\cdot y=0$ which shows that the map $R\oplus R\to I$ is not injective. On the other hand we know that the ideal is not principal.

Mapping property

Let $A$ be a set. There exists a module $F(A)$, called the free module on $A$, which contains $A$ as a subset.
It satisfies the following property.

Let $M$ be any module and let $f:A\to M$ be any map of sets. Then there is a unique module homomorphism $\Phi:F(A)\to M$ such that the following diagram commutes:

\[\begin{xy} \xymatrix { A\ar[r]^{\subset}\ar[rd]^{f} & F(A)\ar[d]^{\Phi}\\ & M\\ } \end{xy}\]

Examples of mapping property

  • If $V$ is a vector space and $B$ is a basis, then $V$ is free on $B$. A linear map from $V\to W$ is determined by where you send $B$. In this situation, $f:B\to W$ is the map of sets sending the basis of $V$ to a subset of $W$, and $\Phi$ is the resulting linear map.

  • If $A$ is any set, then $F(A)$ is the $R$-module of “formal linear combinations of elements of $A$”: the set of sums $\sum r_{i}a_{i}$ over finite collections $\lbrace a_1,\ldots,a_n\rbrace$ of elements of $A$.

  • Alternatively it is the set of functions $f:A\to R$ that are zero for all but a finite subset of $A$ with pointwise addition and scalar multiplication.


Any two free modules on the same set are isomorphic via the module map induced by the identity map on $A$.


Let $R$ be an integral domain.

Definition: The rank of an $R$-module is the maximum number of $R$-linear independent elements of $M$.

Proposition: Let $M$ be a free $R$ module of rank $n$. Then any $n+1$ elements of $M$ are linearly dependent. Thus any submodule of $M$ has rank at most $n$.

Proof: Let $m_1,\ldots, m_{n+1}$ be elements of $M$ and let $e_1,\ldots, e_n$ be a basis of $M$. Each $m_{i}$ is an $R$-linear combination of the $e_i$. We can view the $m_i$ as vectors in $F^{n}$ where $F$ is the fraction field of $R$. They are linearly dependent in $F^{n}$, meaning there is a relation \(\sum f_i m_{i}=0\) where the $f_i$ are in $F$. Clearing denominators gives a relation over $R$.


Torsion Definition

Suppose that $R$ is a ring with unity.

Definition: Let $M$ be an $R$-module. An element $m\in M$ is a torsion element if $rm=0$ for some nonzero $r\in R$. The set of torsion elements in $M$ is called $\Tor(M)$.

  • Any finite abelian group is a torsion $\Z$-module.
  • Any cyclic $R$-module is torsion.
  • Any finite dimensional vector space $V$ over a field $F$ with a linear map $T:V\to V$ is a torsion $F[x]$-module.

Lemma: If $R$ is an integral domain and $M$ is an $R$-module, then the set of torsion elements is a submodule.

Proof: If $m_1$ and $m_2$ are torsion, $r_1 m_1=0$ and $r_2 m_2=0$, with both $r_1$ and $r_2$ nonzero, then $r_1 r_2 (m_1+m2)=0$ and $r_1 r_2 (m_1 m_2)=0$, and $r_1 r_2$ is nonzero since $R$ is an integral domain.

Torsion-free modules

If $R$ is an integral domain, an $R$-module $M$ is called torsion-free if $\Tor(M)=0$.

Any free module is torsion-free, but the converse is false. For example, non-principal ideals in integral domains are not free. This follows from the following lemma.

Lemma: An ideal of $R$ is free if and only if it is principal.

Proof: $R$ is a free module of rank $1$, so a submodule has rank at most $1$; if it has rank $1$, it is a principal ideal.

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