Modules

Core idea

A module is a vector space whose scalars come from a ring instead of a field. That single change produces a theory that contains linear algebra, the theory of abelian groups, and the theory of ideals as special cases, and it is the language of most of modern algebra.

Contents

• 0. Background from ring theory
• 1. Modules
• 2. Submodules, homomorphisms, and quotients
• 3. Generators, direct sums, and free modules
• 4. Torsion
• 5. Exact sequences
• 6. Hom and duality
• 7. Projective and injective modules
• 8. Tensor products
• 9. Algebras
• 10. Modules over a principal ideal domain
• 11. Canonical forms
• 12. Chain conditions and composition series
• 13. Semisimple modules and rings
• 14. Tools over commutative rings
• 15. Looking ahead: resolutions, Ext, and Tor

Conventions

Throughout, every ring has an identity element $1$, ring homomorphisms send $1$ to $1$, and every module is unital: $1m=m$ for all $m$. Knowlpedia's general definition of a ring does not require an identity, so several boxes below say "ring" where these notes always mean "ring with identity".

An $R$-module is a left $R$-module unless right modules are named explicitly. Right modules are needed only for tensor products in Section 8. When $R$ is commutative, left and right modules are the same thing.

Some constructions need extra hypotheses on $R$. The module $\operatorname{Hom}_R(M,N)$, bilinear maps, the tensor product of two left modules, and algebras need $R$ commutative. Torsion needs an integral domain, and Sections 10 and 11 need a principal ideal domain. Each section states its hypotheses.

The words homomorphism, $R$-linear map, and $R$-map mean the same thing. The set of homomorphisms from $M$ to $N$ is written $\operatorname{Hom}_R(M,N)$, and $\operatorname{End}_R(M)=\operatorname{Hom}_R(M,M)$.

0. Background from ring theory

The module definitions depend directly on the following ring-theoretic notions. Further background, such as integral domains, polynomial rings, and principal ideal domains, appears where it is first used.

Background 0.1. Ring

Background 0.2. Unital ring

Background 0.3. Commutative ring

Background 0.4. Ring homomorphism

Background 0.5. Ideal

1. Modules

A vector space is an abelian group on which a field acts by scalar multiplication. A module is the same structure with the field replaced by an arbitrary ring. Three familiar theories become special cases: vector spaces (modules over a field), abelian groups (modules over $\mathbb Z$), and ideals (submodules of a ring over itself).

Definition 1.1. Module

Elementary consequences. The axioms give $0_R\,m=0_M$, $r\,0_M=0_M$ and $(-r)m=-(rm)=r(-m)$ for all $r\in R$ and $m\in M$. For instance, $0m=(0+0)m=0m+0m$, and cancelling in the group $(M,+)$ gives $0m=0$. Unlike in a vector space, $rm=0$ does not force $r=0$ or $m=0$: in the $\mathbb Z$-module $\mathbb Z/6\mathbb Z$, $2\cdot 3=0$.

Modules as ring homomorphisms. For each $r\in R$, the map $\lambda_r\colon M\to M$, $m\mapsto rm$, is an endomorphism of the abelian group $(M,+)$, because $r(m+n)=rm+rn$. The remaining axioms say that

is a ring homomorphism with $\lambda_1=\mathrm{id}_M$. Here $\operatorname{End}_{\mathbb Z}(M)$ is the ring of group endomorphisms of $M$ under pointwise addition and composition, and $(rs)m=r(sm)$ is the statement $\lambda_{rs}=\lambda_r\circ\lambda_s$. Conversely, a ring homomorphism $\varphi\colon R\to\operatorname{End}_{\mathbb Z}(M)$ with $\varphi(1)=\mathrm{id}_M$ defines a module structure by $rm=\varphi(r)(m)$. A module structure on an abelian group is therefore the same thing as a ring homomorphism into its endomorphism ring, just as a group action on a set is a homomorphism into its symmetric group.

Examples. The following modules recur throughout the notes.

1. Every ring $R$ is a left module over itself by left multiplication. This is the left regular module, also written ${}_RR$.
2. Every abelian group is a $\mathbb Z$-module in exactly one way, with $n\cdot m=m+\cdots+m$ ($n$ terms) for $n>0$. Abelian groups and $\mathbb Z$-modules are the same thing.
3. A module over a field $F$ is an $F$-vector space.
4. The set $R^n$ of column vectors is an $R$-module under coordinatewise operations, and it is also a left module over the matrix ring $M_n(R)$ by matrix multiplication.
5. A vector space together with a linear operator is a module over a polynomial ring, as the next two items explain. This example connects module theory to linear algebra, and Section 11 is built on it.

Background 1.2. Polynomial ring

Definition 1.3. Polynomial module of an operator

Write $V_T$ for this $F[x]$-module. The construction can be reversed. An $F[x]$-module $V$ is an $F$-vector space through the action of the constants, and $T(v)=x\cdot v$ is an $F$-linear operator on $V$. The two constructions are inverse to each other, so an $F[x]$-module is the same thing as a vector space with a chosen operator. Submodules of $V_T$ are the $T$-invariant subspaces, and homomorphisms $V_T\to W_S$ are the linear maps $u$ with $uT=Su$.

Example. Let $T$ be the operator on $F^2$ with $Te_1=e_2$ and $Te_2=0$. Then $x\cdot e_1=e_2$ and $x^2\cdot e_1=0$, so every element of $V_T$ has the form $(a+bx)\cdot e_1$. The module $V_T$ is generated by the single element $e_1$, and the polynomials that kill $e_1$ are exactly the multiples of $x^2$.

Construction 1.4. Restriction of scalars

The box states this for commutative rings, but the construction works for any ring homomorphism $\varphi\colon R\to S$ with $\varphi(1)=1$. Every $F[x]$-module becomes an $F$-vector space by restriction along $F\hookrightarrow F[x]$. Every $R$-module becomes an abelian group by restriction along the unique ring homomorphism $\mathbb Z\to R$. A ring $S$ containing $R$ as a subring is an $R$-module, for example $\mathbb C$ as a real vector space or $F[x]$ as an $F$-vector space with basis $1,x,x^2,\dots$. An $R$-module $M$ with $IM=0$ for a two-sided ideal $I$ is the restriction of a unique $R/I$-module along the quotient map $R\to R/I$.

Exercises for Section 1

1. Show that $r\,0=0$ and $(-1)m=-m$ in every module, and give an example of a module in which $rm=0$ with $r\neq0$ and $m\neq0$.
2. Show that an abelian group has exactly one structure of a $\mathbb Z$-module.
3. Let $n\ge1$. Show that a $\mathbb Z/n\mathbb Z$-module is the same thing as an abelian group $M$ with $nM=0$.
4. Let $T$ be the operator on $F^3$ with $Te_1=e_2$, $Te_2=e_3$ and $Te_3=0$. Show that the $F[x]$-module $V_T$ is generated by $e_1$, and find all of its submodules.
5. Let $R^{\mathrm{op}}$ be the ring with the same addition as $R$ and multiplication $a\ast b=ba$. Show that a right $R$-module is the same thing as a left $R^{\mathrm{op}}$-module.

2. Submodules, homomorphisms, and quotients

The constructions of this section work exactly as for vector spaces or groups: subobjects, structure-preserving maps, kernels and images, quotients, and the isomorphism theorems. The only new requirement is closure under multiplication by every ring element.

Definition 2.1. Submodule

Proposition 2.2. Submodule criterion

In practice a one-step test is faster: a subset $N\subseteq M$ is a submodule if and only if $N\neq\varnothing$ and $x+ry\in N$ for all $x,y\in N$ and $r\in R$. Conversely, suppose $N$ is nonempty and closed under $x+ry$, and pick $y\in N$. Then $y+(-1)y=0\in N$, so $ry=0+ry\in N$ and $x-y=x+(-1)y\in N$.

Operations on submodules. The intersection of any family of submodules is a submodule. The union of two submodules usually is not. The sum $N_1+N_2=\{n_1+n_2:n_i\in N_i\}$ is the smallest submodule containing both. For a subset $S\subseteq M$, the submodule generated by $S$ is

the set of finite $R$-linear combinations of elements of $S$. It is the intersection of all submodules containing $S$, since $1\in R$ gives $S\subseteq RS$. For a single element, $Rm=\{rm:r\in R\}$.

In the left regular module ${}_RR$, the submodules are exactly the left ideals. For $R=\mathbb Z$ the submodules are exactly the subgroups, and in $V_T$ they are the $T$-invariant subspaces.

Definition 2.3. Module homomorphism

An isomorphism is a bijective homomorphism, and its inverse map is automatically a homomorphism. For $\mathbb Z$-modules the homomorphisms are the group homomorphisms, and for vector spaces they are the linear maps. If $r$ is central in $R$, for instance when $R$ is commutative, then $m\mapsto rm$ is an endomorphism of every module. For noncommutative $R$, right multiplication $x\mapsto xa$ is an endomorphism of ${}_RR$ for each $a\in R$, while left multiplication $x\mapsto ax$ need not be.

Definition 2.4. Kernel

Proposition 2.5. Kernels are submodules

Definition 2.6. Image

More generally, if $f\colon M\to N$ is a homomorphism, then $f(M')$ is a submodule of $N$ for every submodule $M'\subseteq M$, and $f^{-1}(N')$ is a submodule of $M$ for every submodule $N'\subseteq N$.

Definition 2.7. Quotient module

Closure under scalars is what makes the action on cosets well defined: if $m-m'\in N$, then $rm-rm'=r(m-m')\in N$. The quotient map $\pi\colon M\to M/N$, $m\mapsto m+N$, is a surjective homomorphism with kernel $N$, so every submodule is a kernel.

Example. For a left ideal $I$ of $R$, the quotient $R/I$ is a left $R$-module generated by $1+I$. For a monic polynomial $f\in F[x]$ of degree $d$, the $F[x]$-module $F[x]/(f)$ has $F$-basis $1,x,\dots,x^{d-1}$, and multiplication by $x$ is a linear operator on this $d$-dimensional space. Section 11 returns to this operator.

Proposition 2.8. Universal property of quotient modules

Theorem 2.9. First isomorphism theorem

Proof sketch. Apply Proposition 2.8 to the submodule $\ker f\subseteq M$ to obtain $\bar f\colon M/\ker f\to N$ with $\bar f(m+\ker f)=f(m)$. Its image is $\operatorname{im} f$, and it is injective because $\bar f(m+\ker f)=0$ means $m\in\ker f$.

Example. Let $c\in F$, and let $F_c$ be the field $F$ made into an $F[x]$-module by letting $x$ act as multiplication by $c$. Evaluation $p\mapsto p(c)$ is a surjective $F[x]$-linear map $F[x]\to F_c$, because $(xp)(c)=c\,p(c)$. Its kernel is the ideal $(x-c)$, so $F[x]/(x-c)\cong F_c$ as $F[x]$-modules. The module structure matters here: as $F[x]$-modules, $F_c$ and $F_d$ are not isomorphic when $c\neq d$, although both are the same one-dimensional vector space.

Theorem 2.10. Second isomorphism theorem

Theorem 2.11. Third isomorphism theorem

Theorem 2.12. Correspondence theorem

Example. Let $n\ge1$. The submodules of the $\mathbb Z$-module $\mathbb Z/n\mathbb Z$ correspond to the subgroups $d\mathbb Z$ of $\mathbb Z$ that contain $n\mathbb Z$, that is, to the positive divisors $d$ of $n$. So $\mathbb Z/n\mathbb Z$ has exactly one submodule for each positive divisor of $n$.

Exercises for Section 2

1. Show that the union of two submodules is a submodule if and only if one of them contains the other.
2. (Modular law.) Let $A\subseteq C$ and $B$ be submodules of $M$. Show that $A+(B\cap C)=(A+B)\cap C$.
3. Use the correspondence theorem to find all submodules of the $F[x]$-module $F[x]/(x^3)$.
4. Let $c,d\in F$. Show that $F[x]/(x-c)\cong F[x]/(x-d)$ as $F[x]$-modules if and only if $c=d$.
5. Let $f\colon M\to N$ be a homomorphism and $N'\subseteq N$ a submodule. Show that $f$ induces an injective homomorphism $M/f^{-1}(N')\to N/N'$.

3. Generators, direct sums, and free modules

A vector space has a basis, any two bases have the same size, and every vector space is a direct sum of copies of the field. Over a ring each of these statements fails in general. This section introduces the language for describing how a module is built from smaller pieces, and identifies the modules that do behave like vector spaces: the free modules.

Definition 3.1. Annihilator of an element

Definition 3.2. Cyclic module

The map $R\to Rm$, $r\mapsto rm$, is a surjective homomorphism with kernel $\operatorname{ann}(m)$, so $Rm\cong R/\operatorname{ann}(m)$ by the first isomorphism theorem. The cyclic modules are thus exactly the quotients of ${}_RR$ by left ideals, and a cyclic module is determined up to isomorphism by the annihilator of a generator.

Definition 3.3. Annihilator of a module

Definition 3.4. Finitely generated module

Definition 3.5. Direct product of modules

Definition 3.6. Direct sum of modules

Proposition 3.7. Direct sum universal property

Internal direct sums. Let $N_1,\dots,N_k$ be submodules of $M$. The map $N_1\oplus\cdots\oplus N_k\to M$, $(n_1,\dots,n_k)\mapsto n_1+\cdots+n_k$, is a homomorphism. It is an isomorphism if and only if $N_1+\cdots+N_k=M$ and each $N_i$ meets the sum of the others only in $0$. In that case one writes $M=N_1\oplus\cdots\oplus N_k$ and calls $M$ the internal direct sum of the $N_i$. For two submodules the condition is $N_1+N_2=M$ and $N_1\cap N_2=0$. A submodule $N$ is a direct summand of $M$ if $M=N\oplus N'$ for some submodule $N'$, called a complement of $N$.

Unlike subspaces of a vector space, submodules need not have complements. The only proper nonzero submodule of the $\mathbb Z$-module $\mathbb Z/4\mathbb Z$ is $2\mathbb Z/4\mathbb Z$, and it has no complement.

Definition 3.8. Basis of a free module

Definition 3.9. Free module

Proposition 3.10. Free module universal property

Every module is a quotient of a free module. Map the free module on any generating set $S$ of $M$ onto $M$ by sending each basis element to the corresponding element of $S$. A module generated by $n$ elements is a quotient of $R^n$.

Definition 3.11. Rank of a free module

Free modules are not vector spaces. Even over $\mathbb Z$, several familiar facts about bases fail.

1. A minimal generating set need not be a basis. The set $\{2,3\}$ generates $\mathbb Z$ and no proper subset does, but it is not a basis since $3\cdot2-2\cdot3=0$.
2. A linearly independent set need not extend to a basis. The set $\{2\}$ in $\mathbb Z$ is an example.
3. A submodule of a free module need not be free. Over $R=\mathbb Z/6\mathbb Z$, the submodule $3R=\{0,3\}$ of the free module $R$ has two elements, while every nonzero free $R$-module has at least six.
4. A torsion-free module need not be free. In $\mathbb Q$, any two nonzero elements $a/b$ and $c/d$ satisfy $bc\cdot(a/b)-ad\cdot(c/d)=0$, so a basis would have one element. But $\mathbb Q$ is not cyclic.
5. Over a noncommutative ring, rank need not be defined. If $V$ is a vector space of countably infinite dimension and $R=\operatorname{End}_F(V)$, then $R\cong R^2$ as left $R$-modules.

Over a principal ideal domain, items 3 and 4 are repaired for finitely generated modules (Section 10).

Exercises for Section 3

1. Show that a left $R$-module is cyclic if and only if it is isomorphic to $R/I$ for some left ideal $I$.
2. Show that $\mathbb Q$ is not a finitely generated $\mathbb Z$-module.
3. Show that if $M=N_1\oplus N_2$ internally, then $M/N_1\cong N_2$.
4. Let $I$ be a two-sided ideal. Show that $\operatorname{ann}_R(R/I)=I$, and deduce that $\mathbb Z/m\mathbb Z\cong\mathbb Z/n\mathbb Z$ for $m,n\ge0$ only if $m=n$.
5. Show that a direct summand of a finitely generated module is finitely generated.

4. Torsion

In this section $R$ is an integral domain. Torsion measures how far multiplication by nonzero scalars is from being injective. In $\mathbb Z$, $nm=0$ with $n\neq0$ forces $m=0$, while in $\mathbb Z/n\mathbb Z$ every element is killed by $n$. Over a principal ideal domain, torsion splits every finitely generated module into a free part and a torsion part (Section 10).

Background 4.1. Integral domain

Definition 4.2. Torsion element

Definition 4.3. Torsion module

Definition 4.4. Torsion-free module

The torsion submodule. For a module $M$ over an integral domain, the set of torsion elements

is a submodule, called the torsion submodule of $M$. The module $M$ is torsion exactly when $\operatorname{Tor}(M)=M$ and torsion-free exactly when $\operatorname{Tor}(M)=0$. The quotient $M/\operatorname{Tor}(M)$ is always torsion-free, and homomorphisms carry torsion elements to torsion elements. These facts are standard, and the domain hypothesis is essential, as the next warning shows.

Example. Let $V$ be a vector space of finite dimension $n$ and $T$ an operator on $V$. The $n^2+1$ operators $1,T,\dots,T^{n^2}$ lie in the $n^2$-dimensional space of operators, so they are linearly dependent. A dependence relation is a nonzero polynomial $p$ with $p(T)=0$, and $p$ kills every element of $V_T$. So $V_T$ is a torsion $F[x]$-module.

Exercises for Section 4

1. Show that a submodule of a torsion-free module is torsion-free, and that a direct sum of torsion-free modules is torsion-free.
2. Show that a free module over an integral domain is torsion-free.
3. Let $M=\bigoplus_{p}\mathbb Z/p\mathbb Z$, the sum over all primes $p$. Show that $M$ is a torsion $\mathbb Z$-module with $\operatorname{ann}_{\mathbb Z}(M)=0$. Then show that a finitely generated module over an integral domain is torsion if and only if its annihilator is nonzero.
4. Let $M$ be torsion-free and $r\neq0$. Show that $m\mapsto rm$ is injective. Is it surjective for $M=\mathbb Z$? For $M=\mathbb Q$?

5. Exact sequences

Exact sequences record how a module is assembled from a submodule and a quotient. They are the bookkeeping device of the rest of the course: projective, injective, and flat modules are defined by how they interact with exact sequences.

Definition 5.1. Exact sequence of modules

Definition 5.2. Cokernel

Reading exact sequences. The sequence $0\to A\xrightarrow{i}B$ is exact if and only if $i$ is injective, and $B\xrightarrow{p}C\to0$ is exact if and only if $p$ is surjective. Every homomorphism $f\colon M\to N$ sits in an exact sequence

Definition 5.3. Short exact sequence

The end terms of a short exact sequence do not determine the middle term. The sequences

the second built from the inclusion of the first summand and the projection onto the second, have the same ends. Their middle terms are not isomorphic, since $\mathbb Z$ has no nonzero element of finite order and $\mathbb Z\oplus\mathbb Z/2$ does. Classifying the possible middle terms is the extension problem, which Section 15 returns to. The simplest middle term is the direct sum of the ends, and the next two items characterize when this happens.

Definition 5.4. Split exact sequence

Lemma 5.5. Splitting lemma

Proof sketch. Given a retraction $r$ with $ri=\mathrm{id}_A$, every $b\in B$ decomposes as $b=i(r(b))+(b-i(r(b)))$. The first term lies in $\operatorname{im} i$ and the second in $\ker r$. If $i(a)\in\ker r$, then $a=r(i(a))=0$, so $B=\operatorname{im} i\oplus\ker r$ internally, and $p$ restricts to an isomorphism $\ker r\to C$. Given a section $s$, one argues similarly with $B=\operatorname{im} i\oplus\operatorname{im} s$.

The two lemmas below are the standard diagram-chasing tools, and both are proved by chasing elements around the diagram. The five lemma is stated in an abelian category. For modules, read its objects as $R$-modules, its epimorphisms as surjections, and its monomorphisms as injections.

Lemma 5.6. Snake lemma

Lemma 5.7. Five lemma

The case used most often is the short five lemma: in a commutative diagram of two short exact sequences, if the two outer vertical maps are isomorphisms, so is the middle one. In the diagram below the rows are short exact and both squares commute:

Exercises for Section 5

1. Show that every short exact sequence $0\to A\to B\to F\to0$ with $F$ free splits.
2. Show that $0\to\mathbb Z\xrightarrow{n}\mathbb Z\to\mathbb Z/n\to0$ does not split for $n\ge2$. More generally, show that a short exact sequence of abelian groups $0\to A\to B\to C\to0$ does not split if $B$ has no nonzero elements of finite order and $C$ is finite and nonzero.
3. Let $0\to A\to B\to C\to0$ be a short exact sequence of abelian groups, let $n\ge1$, and write $X[n]=\{x\in X:nx=0\}$. Apply the snake lemma to multiplication by $n$. Show that the result joins $0\to A[n]\to B[n]\to C[n]$ and $A/nA\to B/nB\to C/nC\to0$ by a connecting map $C[n]\to A/nA$ into one exact sequence.
4. Prove the short five lemma directly by a diagram chase.
5. Show that a short exact sequence $0\to A\xrightarrow{i}B\to C\to0$ splits if and only if $i(A)$ is a direct summand of $B$.

6. Hom and duality

Homomorphisms between two modules form an abelian group, and an $R$-module when $R$ is commutative. Fixing one argument gives a functor, and this section records how these functors interact with direct sums and exact sequences. In this section, $R$ is commutative whenever $\operatorname{Hom}_R(M,N)$ is treated as an $R$-module.

Definition 6.1. Hom module

Commutativity is what makes $rf$ linear: $(rf)(sm)=r\,f(sm)=rs\,f(m)$, while linearity requires $(rf)(sm)=s\,(rf)(m)=sr\,f(m)$. A homomorphism $u\colon N\to N'$ induces $u_\ast\colon\operatorname{Hom}_R(M,N)\to\operatorname{Hom}_R(M,N')$, $f\mapsto u\circ f$. A homomorphism $v\colon M'\to M$ induces $v^\ast\colon\operatorname{Hom}_R(M,N)\to\operatorname{Hom}_R(M',N)$, $f\mapsto f\circ v$, and this reverses the direction of arrows.

Examples. The isomorphism $\operatorname{Hom}_R(R,M)\cong M$ is given by $f\mapsto f(1)$. For abelian groups:

Proposition 6.2. Hom turns sums into products

Dually, $\operatorname{Hom}_R\bigl(M,\prod_i N_i\bigr)\cong\prod_i\operatorname{Hom}_R(M,N_i)$; this is the universal property of the product. In particular $\operatorname{Hom}_R(R^n,M)\cong M^n$.

Proposition 6.3. Hom is left exact

The functor $\operatorname{Hom}_R(M,-)$ need not preserve surjections. Apply $\operatorname{Hom}_{\mathbb Z}(\mathbb Z/2,-)$ to $0\to\mathbb Z\xrightarrow{2}\mathbb Z\to\mathbb Z/2\to0$. The result is $0\to0\to0\to\mathbb Z/2$, and the last map is not surjective. Section 7 studies the modules for which these functors are exact.

Definition 6.4. Dual module

For a free module of finite rank, the evaluation map $M\to M^{\vee\vee}$, $m\mapsto(\varphi\mapsto\varphi(m))$, is an isomorphism. In general it can fail to be injective or surjective: $(\mathbb Z/n)^\vee=0$ over $\mathbb Z$, and for an infinite-dimensional vector space the evaluation map is injective but not surjective.

Exercises for Section 6

1. Show that $\operatorname{Hom}_{\mathbb Z}(\mathbb Z/m,\mathbb Z/n)\cong\mathbb Z/\gcd(m,n)$.
2. Show that $\operatorname{Hom}_{\mathbb Z}(\mathbb Q,\mathbb Z)=0$ and $\operatorname{Hom}_{\mathbb Z}(\mathbb Q,\mathbb Q)\cong\mathbb Q$.
3. For a commutative ring $R$, an ideal $I$, and a module $M$, show that $\operatorname{Hom}_R(R/I,M)\cong\{m\in M:Im=0\}$.
4. Prove the contravariant form of Proposition 6.3 directly: if $A\xrightarrow{u}B\xrightarrow{v}C\to0$ is exact, show that the induced maps $v^\ast$ and $u^\ast$ on $\operatorname{Hom}_R(-,N)$ satisfy $\ker v^\ast=0$ and $\ker u^\ast=\operatorname{im} v^\ast$.
5. Show that $M\to M^{\vee\vee}$ is an isomorphism when $M$ is free of finite rank.

7. Projective and injective modules

Free modules are easy to map out of: a homomorphism from a free module is any choice of images of a basis. Projective modules keep the lifting property that this gives, and injective modules have the dual extension property. In terms of Section 6, $P$ is projective exactly when $\operatorname{Hom}_R(P,-)$ is exact, and $E$ is injective exactly when $\operatorname{Hom}_R(-,E)$ is exact.

Definition 7.1. Projective module

Proposition 7.2. Projective modules are direct summands of free modules

Proof sketch. Choose a surjection $\pi\colon F\to P$ from a free module. Projectivity lifts $\mathrm{id}_P$ to $s\colon P\to F$ with $\pi s=\mathrm{id}_P$, so $F=s(P)\oplus\ker\pi$ by the splitting lemma. Conversely, lift a map from a summand by extending it by zero on the complement and using Proposition 3.10 on the free module.

Proposition 7.3. Projective short exact sequence criterion

Examples. Free modules are projective, and over a field every module is free and hence projective. A projective module need not be free. For $R=F\times F$, the ideal $P=F\times0$ is a direct summand of $R$, hence projective. It is not free, because $\dim_F P=1$ while a free $R$-module of rank $n$ has $F$-dimension $2n$. Over $\mathbb Z$, and more generally over a principal ideal domain, finitely generated projective modules are free (Section 10), and $\mathbb Z/n$ is not projective for $n\ge2$.

Definition 7.4. Injective module

Theorem 7.5. Baer's criterion

Injective modules over a principal ideal domain. An $R$-module $E$ is divisible if for every $m\in E$ and every $r\neq0$ there is $m'\in E$ with $rm'=m$. Over a principal ideal domain, Baer's criterion shows that injective is the same as divisible. A nonzero ideal is $(r)\cong R$, so a homomorphism $f\colon(r)\to E$ is determined by $m=f(r)$, which can be any element of $E$. If $F\colon R\to E$ extends $f$, then $m=F(r)=rF(1)$. Conversely, if $m=rm'$, then $F(x)=xm'$ extends $f$. So every such $f$ extends exactly when $E$ is divisible. Thus $\mathbb Q$ and $\mathbb Q/\mathbb Z$ are injective $\mathbb Z$-modules and $\mathbb Z$ is not. Every abelian group embeds in an injective one. Write $A\cong F/K$ with $F=\bigoplus_{i\in I}\mathbb Z$ free. Then $A$ embeds in $\bigl(\bigoplus_{i\in I}\mathbb Q\bigr)/K$, which is divisible because it is a quotient of a divisible group. The same holds over every ring: every module embeds in an injective module.

Exercises for Section 7

1. Show that a direct sum of modules is projective if and only if every summand is projective.
2. Use Proposition 7.3 to show that $\mathbb Z/n$ is not a projective $\mathbb Z$-module for $n\ge2$.
3. Show that $\mathbb Q$ is not a projective $\mathbb Z$-module. (Hint: a projective module embeds in a free module, and no nonzero element of a free abelian group is divisible by every positive integer.)
4. Show that a quotient of a divisible abelian group is divisible, and that $\mathbb Z$ is not an injective $\mathbb Z$-module.
5. Show that $\operatorname{Hom}_R(P,-)$ is exact when $P$ is projective.

8. Tensor products

A bilinear map $M\times N\to P$ is linear in each variable separately, but it is not a linear map out of the product module. The tensor product is the module that converts bilinear maps into linear ones. The general construction pairs a right $R$-module $M$ with a left $R$-module $N$ and produces an abelian group $M\otimes_RN$. When $R$ is commutative, $M\otimes_RN$ is again an $R$-module.

Definition 8.1. Bilinear and balanced maps

Definition 8.2. Universal property of the tensor product

Definition 8.3. Tensor product of modules

Definition 8.4. Bimodule

Bimodules supply the module structure on a tensor product. If $M$ is an $(S,R)$-bimodule and $N$ a left $R$-module, then $M\otimes_RN$ is a left $S$-module with $s(m\otimes n)=(sm)\otimes n$. When $R$ is commutative, every module is an $(R,R)$-bimodule with $rm=mr$, and $M\otimes_RN$ is an $R$-module with $r(m\otimes n)=(rm)\otimes n=m\otimes(rn)$.

Working with tensors. Three points cause most mistakes.

1. Every element of $M\otimes_RN$ is a finite sum $\sum_i m_i\otimes n_i$, but in general not a single pure tensor $m\otimes n$. In $F^2\otimes_FF^2\cong M_2(F)$, the pure tensors correspond to matrices of rank at most one, and $e_1\otimes e_1+e_2\otimes e_2$ corresponds to the identity matrix.
2. To define a homomorphism out of $M\otimes_RN$, write down a balanced map on $M\times N$ and apply the universal property. A formula on pure tensors defines a homomorphism only after this check.
3. Whether $m\otimes n$ is zero depends on the modules, not only on $m$ and $n$. In $\mathbb Z\otimes_{\mathbb Z}\mathbb Z/2$, the element $2\otimes1=1\otimes2=0$. In $2\mathbb Z\otimes_{\mathbb Z}\mathbb Z/2\cong\mathbb Z/2$, the element $2\otimes1$ is a generator. So tensoring the injection $2\mathbb Z\hookrightarrow\mathbb Z$ with $\mathbb Z/2$ gives the zero map $\mathbb Z/2\to\mathbb Z/2$, which is not injective.

Computations. $R^m\otimes_RR^n\cong R^{mn}$ and $\mathbb Z/m\otimes_{\mathbb Z}\mathbb Z/n\cong\mathbb Z/\gcd(m,n)$. For any abelian group $A$, $\mathbb Z/n\otimes_{\mathbb Z}A\cong A/nA$. Also $\mathbb Q\otimes_{\mathbb Z}\mathbb Z/n=0$, because

Proposition 8.5. Tensor product preserves direct sums

Proposition 8.6. Tensor product is right exact

Construction 8.7. Extension of scalars

Examples. Complexification $\mathbb C\otimes_{\mathbb R}V$ turns a real vector space into a complex one of the same dimension. For a finitely generated abelian group $A\cong\mathbb Z^r\oplus T$ with $T$ finite, $\mathbb Q\otimes_{\mathbb Z}A\cong\mathbb Q^r$: extending scalars to $\mathbb Q$ kills the torsion and remembers the rank.

Theorem 8.8. Tensor–Hom adjunction

Theorem 8.9. Tensor–Hom adjunction for bimodules

Proposition 8.10. Tensor commutes with direct limits and sums

The adjunction explains the exactness properties at once. A left adjoint preserves colimits, and cokernels and direct sums are colimits. This gives Proposition 8.5 and right exactness again. The statement about direct limits can be skipped on a first reading.

Definition 8.11. Flat module

Proposition 8.12. Projective implies flat

Examples. The $\mathbb Z$-module $\mathbb Z/n$ is not flat for $n\ge2$: tensoring the injection $\mathbb Z\xrightarrow{n}\mathbb Z$ with $\mathbb Z/n$ gives the zero map on $\mathbb Z/n$. The $\mathbb Z$-module $\mathbb Q$ is flat (Section 14) but not projective. Over a principal ideal domain, a module is flat exactly when it is torsion-free.

Proposition 8.13. Exactness properties of Hom and tensor

Exercises for Section 8

1. Compute $\mathbb Z/12\otimes_{\mathbb Z}\mathbb Z/18$ and $\mathbb Q\otimes_{\mathbb Z}(\mathbb Z^2\oplus\mathbb Z/5)$.
2. Show that $e_1\otimes e_1+e_2\otimes e_2\in F^2\otimes_FF^2$ is not a pure tensor.
3. Show directly from the universal property that $M\otimes_RR\cong M$ for a right $R$-module $M$.
4. For a commutative ring $R$ and ideals $I,J$, show that $R/I\otimes_RR/J\cong R/(I+J)$.
5. Show that the injection $2\mathbb Z\hookrightarrow\mathbb Z$ stays injective after tensoring with $\mathbb Q$, but not after tensoring with $\mathbb Z/2$.

9. Algebras

An algebra over a commutative ring $R$ is a ring that is also an $R$-module, with compatible multiplication. Polynomial rings, matrix rings, and group algebras are examples. Tensor products combine algebras, and three universal constructions build algebras from a module. This section can be read any time after Section 8.

Definition 9.1. Algebra over a commutative ring

Definition 9.2. Algebra homomorphism

Construction 9.3. Tensor product of algebras

Example. The map $a\otimes b\mapsto(ab,a\bar b)$ is an isomorphism of rings $\mathbb C\otimes_{\mathbb R}\mathbb C\cong\mathbb C\times\mathbb C$. Extending scalars can turn a field into a ring with zero divisors.

Construction 9.4. Tensor algebra

Construction 9.5. Symmetric algebra

Construction 9.6. Exterior algebra

Determinants. For $M=R^n$, the top exterior power $\bigwedge^nR^n$ is free of rank one with basis $e_1\wedge\cdots\wedge e_n$. An endomorphism $\varphi$ of $R^n$ induces an endomorphism of $\bigwedge^nR^n$, $v_1\wedge\cdots\wedge v_n\mapsto\varphi v_1\wedge\cdots\wedge\varphi v_n$, which is multiplication by a scalar. That scalar is $\det\varphi$. Multiplicativity $\det(\varphi\psi)=\det\varphi\det\psi$ follows from functoriality, with no computation.

Exercises for Section 9

1. In $\bigwedge^2_R(R^2)$, express $(ae_1+be_2)\wedge(ce_1+de_2)$ as a multiple of $e_1\wedge e_2$.
2. Use universal properties to show that $\operatorname{Sym}_R(R^n)\cong R[x_1,\dots,x_n]$.
3. Show that $F[x]\otimes_FF[y]\cong F[x,y]$ as $F$-algebras.
4. Show that $\mathbb C\otimes_{\mathbb R}\mathbb C\cong\mathbb C\times\mathbb C$ as rings.

10. Modules over a principal ideal domain

In this section $R$ is a principal ideal domain (PID). The main examples are $\mathbb Z$ and $F[x]$ for a field $F$. Over a PID, every finitely generated module is a direct sum of cyclic modules, and the cyclic summands can be chosen so that they are unique. For $R=\mathbb Z$ this classifies finitely generated abelian groups, and for $R=F[x]$ it gives the canonical forms of Section 11.

Background 10.1. Euclidean domain

Background 10.2. Principal ideal domain

Background 10.3. Prime element

Recall that in a PID every nonzero prime ideal is maximal, any two elements $a,b$ have a greatest common divisor $d$ with $(a,b)=(d)$ and $d=sa+tb$ for some $s,t\in R$, and every nonzero nonunit factors uniquely into primes.

Submodules of free modules. The first step toward the structure theorem repairs item 3 of the list in Section 3.

Theorem. If $R$ is a PID, then every submodule of $R^n$ is free of rank at most $n$.

Proof sketch. Induct on $n$; the case $n=0$ is trivial. Let $N\subseteq R^n$ and let $\pi\colon R^n\to R$ be the last coordinate. The image $\pi(N)$ is an ideal $(a)$. If $a=0$, then $N\subseteq R^{n-1}$ and induction applies. Otherwise $(a)\cong R$ is free, so $0\to N\cap R^{n-1}\to N\to(a)\to0$ splits (Exercise 1 of Section 5), and $N\cong(N\cap R^{n-1})\oplus R$. By induction $N\cap R^{n-1}$ is free of rank at most $n-1$.

Two consequences follow. A finitely generated projective module $P$ over a PID is free: a surjection $R^n\to P$ splits, so $P$ is isomorphic to a submodule of $R^n$. And every finitely generated module $M$ has a presentation. Choose generators to get a surjection $R^n\to M$. Its kernel $N$ is free with a basis of $m\le n$ elements. Writing that basis in coordinates gives an $n\times m$ matrix $A$ with