Projective Modules, Splitting and Semisimplicity

Q: Is every projective module free?

No. Z/3 is projective and not free over Z/6, and non-principal ideals of a Dedekind domain are projective and not free. It is true over local rings by Kaplansky's theorem, over principal ideal domains, and — for finitely generated modules — over polynomial rings over a field by the Quillen–Suslin theorem.

Q: What is the practical difference between (2.8) and (2.9)?

(2.8) gives a genuinely cheap test — restrict attention to cyclic modules. (2.9) as stated requires all modules, and the cyclic version, though true, rests on Osofsky's theorem. In practice one tests semisimplicity through projectivity, never through injectivity.

Q: Does a module have to be finitely generated to be projective?

No. Any direct sum of projective modules is projective, so R^(I) for infinite I is projective and typically not finitely generated. Finite generation matters when one wants to conclude freeness or to compute with a dual basis of finite size.

Executive Summary

A left $R$ -module $P$ is projective if every map out of $P$ into a quotient can be lifted to the thing being quotiented. Equivalently — and this is Lam's $(2.7)$ — $P$ is a direct summand of a free module, and equivalently every surjection onto $P$ splits.

That last form is what connects projectivity to semisimplicity. Semisimple rings are the rings over which all short exact sequences split, so over them everything is projective. Theorem $(2.8)$ shows the converse holds with a much weaker hypothesis: it is enough that every cyclic left module be projective, because that alone forces every left ideal to split off from $_{R} R$ .

Lifting

g = f \circ h

for some

h : P \to A

Structural form

P \oplus Q ≅ R^{(I)}

free

Splitting formevery

A ↠ P

splits

Ring testcyclic modules projective

\Rightarrow

semisimple

Dual testall modules injective

\Leftrightarrow

semisimple

Homological

gl.dim R = 0

Overview

Free modules are easy to map out of: define the map on a basis. Projective modules are the class for which this convenience survives, and they are exactly the retracts of free modules. Everything in homological algebra that begins with a resolution begins with this class.

f : A ↠ B, g : P \to B ⟹ \exists h : P \to A with f \circ h = g

(2.7a)

The lifting property. The map $g$ into the quotient is given; $h$ into the object being quotiented must be produced.

The injective notion is obtained by reversing every arrow: $I$ is injective if maps into $I$ extend along injections. The two theories are formally dual but behave very differently in practice — free modules give projectives for nothing, while producing injectives requires the theory of divisible modules and injective hulls.

For semisimple rings the asymmetry disappears: over such a ring every module is simultaneously projective and injective, and no resolution is ever longer than one step. That statement is the homological signature of the class defined on Semisimple Rings: Definition and Equivalent Characterisations.

Learning Objectives

State the lifting property and the extension property precisely, with all quantifiers.
Prove the cycle of implications in $(2.7)$ , including why free modules are projective.
Prove $(2.8)$ , isolating the step where cyclicity is used.
Prove $(2.9)$ and say exactly which half of the injective analogue of $(2.7)$ it needs.
Show that $S_{2}$ is a cyclic non-projective module over $T_{2} (k)$ .
Place free, projective and flat in the correct order and give separating examples.

Definitions

Definition(2.7·0)Projective module

A left $R$ -module $P$ is projective if for every surjective $R$ -homomorphism $f : A \to B$ of left $R$ -modules and every $R$ -homomorphism $g : P \to B$ , there exists an $R$ -homomorphism $h : P \to A$ with $f \circ h = g$ .

Definition(2.9·0)Injective module

A left $R$ -module $I$ is injective if for every injective $R$ -homomorphism $f : A \to B$ of left $R$ -modules and every $R$ -homomorphism $g : A \to I$ , there exists an $R$ -homomorphism $h : B \to I$ with $h \circ f = g$ .

Free module: $F ≅ R^{(I)} = ⨁_{i \in I} R$ ; equivalently $F$ has a basis, and $R$ -maps out of $F$ correspond to arbitrary set maps from the basis.
Retract: $P$ is a retract of $M$ if there are maps $P \to M \to P$ composing to ${id}_{P}$ ; equivalently $P$ is isomorphic to a direct summand of $M$ .
Split short exact sequence: $0 \to A \to B \to C \to 0$ with $B ≅ A \oplus C$ compatibly with the maps.
Flat module: $M$ is flat if $- \otimes_{R} M$ preserves injections. Projective implies flat; the converse holds for finitely presented modules.
Divisible module: For $ℤ$ -modules, $M$ with $n M = M$ for all $n \neq 0$ ; over $ℤ$ divisible and injective coincide.

Both definitions are stated for left modules; the right-handed versions are the same statements read in the opposite ring.

Core Concepts

Why free modules lift

Let $F$ be free on a basis ${x_{i}}_{i \in I}$ , let $f : A ↠ B$ and $g : F \to B$ . For each $i$ choose $a_{i} \in A$ with $f (a_{i}) = g (x_{i})$ — possible because $f$ is onto, and requiring the axiom of choice when $I$ is infinite. Defining $h (x_{i}) = a_{i}$ and extending $R$ -linearly gives $f \circ h = g$ on a basis, hence everywhere. Projectivity of free modules is therefore nothing but the universal property of a basis.

Direct summands inherit lifting

If $F = P \oplus Q$ and $g : P \to B$ is given, extend $g$ to $F$ by declaring it zero on $Q$ , lift the extension to $\tilde{h} : F \to A$ , then restrict to $P$ . This one-line argument is why the class of projectives is closed under direct summands and under arbitrary direct sums.

The dual basis description

Projectivity can be written without quantifying over all modules. $P$ is projective iff there exist elements ${x_{i}}_{i \in I}$ of $P$ and homomorphisms ${f_{i}} \subseteq {Hom}_{R} (P, R)$ such that for every $x \in P$ only finitely many $f_{i} (x)$ are nonzero and

x = \sum_{i \in I} f_{i} (x) x_{i} .

(dual basis)

A basis without linear independence: the coordinates exist but need not be unique.

Where projectives sit

**Free $\Rightarrow$ projective $\Rightarrow$ flat**, and both implications are strict in general.
Over a local ring every projective module is free (Kaplansky); over a principal ideal domain every submodule of a free module is free, so finitely generated projective equals free.
Over a polynomial ring $k [x_{1}, \dots, x_{n}]$ finitely generated projective modules are free — the Quillen–Suslin theorem, answering Serre's problem.
Over a Dedekind domain the finitely generated projectives are the direct sums of ideals; the failure of freeness is measured exactly by the class group.
$P$ is projective iff ${Ext}_{R}^{1} (P, M) = 0$ for every $M$ ; the class of projectives is what makes $Ext$ and $Tor$ computable.

Key Results

Proposition(2.7)Three descriptions of projectivity

For a left $R$ -module $P$ the following are equivalent.

$P$ is projective.
$P$ is isomorphic to a direct summand of a free left $R$ -module.
Every surjective $R$ -homomorphism from a left $R$ -module onto $P$ splits.

Proof

**(1) $\Rightarrow$ (3).** Let $f : A ↠ P$ be surjective. Apply the lifting property to $f$ and to $g = {id}_{P} : P \to P$ : there is $h : P \to A$ with $f \circ h = {id}_{P}$ , which is precisely a splitting.

**(3) $\Rightarrow$ (2).** Choose any generating set of $P$ and let $F$ be free on it, with the resulting surjection $π : F ↠ P$ . By (3) there is $h$ with $π \circ h = {id}_{P}$ , so $F = h (P) \oplus \ker π$ and $h (P) ≅ P$ .

**(2) $\Rightarrow$ (1).** Write $F = P' \oplus Q$ with $F$ free and $P' ≅ P$ ; it suffices to treat $P'$ . Given $f : A ↠ B$ and $g : P' \to B$ , extend $g$ to $\tilde{g} : F \to B$ by $\tilde{g} |_{Q} = 0$ . Since $F$ is free, choose for each basis element $x_{i}$ an element $a_{i} \in A$ with $f (a_{i}) = \tilde{g} (x_{i})$ and let $\tilde{h} : F \to A$ be the induced map; then $f \circ \tilde{h} = \tilde{g}$ . Restricting $\tilde{h}$ to $P'$ gives the required lift.

Theorem(2.8)Homological characterisation of left semisimplicity

For a ring $R$ with identity the following are equivalent.

$R$ is left semisimple.
Every left $R$ -module is projective.
Every finitely generated left $R$ -module is projective.
Every cyclic left $R$ -module is projective.

Proof

**(1) $\Leftrightarrow$ (2).** If $R$ is left semisimple then every short exact sequence of left modules splits by $(2.5) (1)$ ; in particular every surjection onto a module $P$ splits, so $P$ is projective by $(2.7)$ . Conversely, if every left module is projective, then given any surjection $B ↠ C$ the module $C$ is projective, so the surjection splits by $(2.7)$ ; hence every short exact sequence splits and $R$ is left semisimple by $(2.5) (1)$ .

**(2) $\Rightarrow$ (3) $\Rightarrow$ (4)** are specialisations.

**(4) $\Rightarrow$ (1).** We verify $(2.5) (5)$ , that $_{R} R$ is semisimple. Let $𝔄 \subseteq R$ be any left ideal. The module $R / 𝔄$ is cyclic, hence projective by hypothesis, so by $(2.7)$ the short exact sequence

0 ⟶ 𝔄 ⟶_{R} R ⟶ R / 𝔄 ⟶ 0

(2.8a)

splits. A splitting exhibits $𝔄$ as a direct summand of $_{R} R$ . Since $𝔄$ was arbitrary, every submodule of $_{R} R$ is a direct summand, i.e. $_{R} R$ is semisimple, and $(2.5)$ gives (1).

Theorem(2.9)The injective analogue

For a ring $R$ with identity the following are equivalent: (1) $R$ is left semisimple; (2) every left $R$ -module is injective.

Proof

**(1) $\Rightarrow$ (2).** Let $I$ be a left $R$ -module, $f : A ↪ B$ injective and $g : A \to I$ given. By $(2.5) (2)$ the module $B$ is semisimple, so $f (A)$ is a direct summand: $B = f (A) \oplus B'$ . Define $h : B \to I$ by $h (f (a) + b') = g (a)$ , which is well defined because the sum is direct and because $f$ is injective. Then $h \circ f = g$ .

**(2) $\Rightarrow$ (1).** Let $M$ be a left $R$ -module and $N \subseteq M$ a submodule. By hypothesis $N$ is injective, so applying the extension property to the inclusion $N ↪ M$ and to ${id}_{N}$ produces $h : M \to N$ with $h |_{N} = {id}_{N}$ . Then $M = N \oplus \ker h$ , so $M$ is semisimple; by $(2.5) (2)$ , $R$ is left semisimple.

Remark—Cyclic injectives: Osofsky's theorem

The list in $(2.9)$ can in fact be extended by (3) every finitely generated left $R$ -module is injective, and (4) every cyclic left $R$ -module is injective. The implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) are immediate; the converse (4) $\Rightarrow$ (1) is a theorem of B. Osofsky (1964) and is substantially harder than anything on this page. The asymmetry with the projective case is real: $(2.8) (4) \Rightarrow (1)$ is three lines, its injective mirror image is a research paper.

Corollary—Global dimension zero

$R$ is left semisimple iff $gl.dim R = 0$ , iff ${Ext}_{R}^{1} (M, N) = 0$ for all left $R$ -modules $M, N$ . Indeed ${Ext}_{R}^{1} (M, -) = 0$ says precisely that $M$ is projective, and $(2.8)$ says that all modules being projective is semisimplicity.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Test against the identity. Almost every proof about projectivity begins by applying the lifting property to $g = {id}_{P}$ . That single substitution converts an abstract universal property into a concrete splitting map.
Present, then split. Every module is a quotient of a free one. If the resulting surjection splits, the module is a summand of a free; the structure theory follows from choosing the presentation and then splitting it.
Extend by zero. To move a map defined on a direct summand up to the whole module, extend it by zero on the complement. This is what makes summands of projectives projective and is the only place the direct sum decomposition is used.
Test on cyclic objects. $(2.8) (4)$ and Baer's criterion are the same trick: a property that must be checked against all modules can often be checked against the quotients $R / 𝔄$ alone, because $𝔄 \subseteq_{R} R$ carries the whole obstruction.

Worked Example

A cyclic non-projective module over $T_{2} (k)$

Let $k$ be a field and $R = T_{2} (k) = {(\begin{matrix} a & b \\ 0 & c \end{matrix}) : a, b, c \in k}$ , of dimension $3$ . Put $e_{1} = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ and $e_{2} = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ : orthogonal idempotents with $e_{1} + e_{2} = 1$ , so $_{R} R = R e_{1} \oplus R e_{2}$ .

R e_{1} = (\begin{matrix} k & 0 \\ 0 & 0 \end{matrix}), R e_{2} = (\begin{matrix} 0 & k \\ 0 & k \end{matrix}), \dim_{k} R e_{1} = 1, \dim_{k} R e_{2} = 2 .

(E.1)

There are exactly two simple left $R$ -modules, both one-dimensional over $k$ : on $S_{1}$ the matrix acts by its entry $a$ , on $S_{2}$ by its entry $c$ . Direct computation gives $R e_{1} ≅ S_{1}$ , and $R e_{2}$ has the submodule $N = (\begin{matrix} 0 & k \\ 0 & 0 \end{matrix}) ≅ S_{1}$ with $R e_{2} / N ≅ S_{2}$ .

Step 1: $S_{2}$ is cyclic

The set $𝔫 = (\begin{matrix} k & k \\ 0 & 0 \end{matrix})$ is a left ideal, since $(\begin{matrix} a & b \\ 0 & c \end{matrix}) (\begin{matrix} a' & b' \\ 0 & 0 \end{matrix}) = (\begin{matrix} a a' & a b' \\ 0 & 0 \end{matrix})$ , and $R / 𝔫 ≅ S_{2}$ . So $S_{2}$ is a cyclic left $R$ -module.

Step 2: $R e_{2}$ is indecomposable

${End}_{R} (R e_{2}) ≅ (e_{2} R e_{2})^{op}$ , and $e_{2} R e_{2} = (\begin{matrix} 0 & 0 \\ 0 & k \end{matrix}) ≅ k$ . An endomorphism ring with no idempotents other than $0$ and $1$ forces indecomposability, so $R e_{2}$ is not a direct sum of two nonzero submodules.

Step 3: $S_{2}$ is not projective

Consider the surjection $R e_{2} ↠ R e_{2} / N ≅ S_{2}$ . If $S_{2}$ were projective this would split by $(2.7)$ , giving $R e_{2} ≅ N \oplus S_{2}$ with both summands nonzero — contradicting Step 2. Hence $S_{2}$ is a cyclic left module that is not projective, condition $(2.8) (4)$ fails, and $T_{2} (k)$ is not left semisimple.

Projective but not free

Over $R = ℤ / 6$ , the ideal $⟨ 2 ⟩ = {0, 2, 4} ≅ ℤ / 3$ is a direct summand — indeed $ℤ / 6 = ⟨ 2 ⟩ \oplus ⟨ 3 ⟩$ — so it is projective. It is not free, because every free $ℤ / 6$ -module has $6^{n}$ elements and $| ⟨ 2 ⟩ | = 3$ . Over $R = ℤ / 4$ the picture changes: the only subgroup of order $2$ is $⟨ 2 ⟩$ itself, which is the kernel of $ℤ / 4 ↠ ℤ / 2$ , so that surjection does not split and $ℤ / 2$ is not projective over $ℤ / 4$ .

A commutative arithmetic instance: in $ℤ [\sqrt{- 5}]$ the ideal $𝔭 = (2, 1 + \sqrt{- 5})$ satisfies $𝔭^{2} = (2)$ and $𝔭 \oplus 𝔭 ≅ R \oplus 𝔭^{2} ≅ R^{2}$ , so $𝔭$ is projective of rank one; it is not principal, hence not free.

Process and Workflow

Is the left $R$ -module $M$ projective?

$R$ is semisimpleYes, unconditionally, by

(2.8)

. Stop here — this is the cheapest possible test and it also gives injectivity, by

(2.9)

$R$ is localProjective is equivalent to free by Kaplansky's theorem, so compare

M

with

R^{n}

; a finitely generated

M

is projective iff a minimal generating set is a basis.

$M$ is finitely presentedProjective is equivalent to flat. Over a commutative ring this becomes a local condition: check that

M_{𝔭}

is free of constant rank at every prime.

$R$ is a finite-dimensional algebraDecompose

_{R} R

into indecomposable projectives

R e_{i}

and compare

M

with direct sums of them; a module is projective iff it is such a sum.

OtherwiseTry to split a surjection

F ↠ M

from a free module, or refute projectivity by exhibiting a non-split surjection onto

M

— the route used for

S_{2}

over

T_{2} (k)

Comparison and Classification

Free, projective and flat on concrete modules
	Free	Projective	Flat
$ℤ$ over $ℤ$	yes	yes	yes
$ℚ$ over $ℤ$	no	no	yes
$ℤ / 2$ over $ℤ$	no	no	no
$⟨ 2 ⟩ ≅ ℤ / 3$ over $ℤ / 6$	no	yes	yes
$ℤ / 2$ over $ℤ / 4$	no	no	no
$𝔭 = (2, 1 + \sqrt{- 5})$ over $ℤ [\sqrt{- 5}]$	no	yes	yes
$k^{n}$ over $M_{n} (k)$ , $n \geq 2$	no	yes	yes
$S_{1}$ over $T_{2} (k)$	no	yes	yes
$S_{2}$ over $T_{2} (k)$	no	no	no

Free, projective and flat on concrete modules

Ring conditions expressed through projectivity
Condition on $R$	Module-theoretic form	Example
Left semisimple	every left module is projective	$M_{n} (D)$ , $ℤ / 6$ , $ℚ S_{3}$
Left hereditary	every left ideal is projective	$ℤ$ , any Dedekind domain, $T_{2} (k)$
Local	every projective is free (Kaplansky)	$ℤ / 4$ , $k [[x]]$ , $ℤ_{(p)}$
Principal ideal domain	finitely generated projective equals free	$ℤ$ , $k [x]$
Polynomial ring over a field	finitely generated projective equals free	$k [x_{1}, \dots, x_{n}]$ , by Quillen–Suslin
Quasi-Frobenius	projective and injective coincide	$ℤ / 4$ , $k G$ for $G$ finite

Note that $T_{2} (k)$ is hereditary but not semisimple: every left ideal is projective, yet not every cyclic module is. The gap between those two statements is exactly the content of $(2.8) (4)$ .

Relationship Map

$P$ projective — equivalently a retract of a free module
- always implies
  - every surjection onto $P$ splits
  - $P$ is flat
  - ${Ext}_{R}^{1} (P, -) = 0$
  - $P$ has a dual basis
- is implied by
  - $P$ free
  - $P$ a direct summand of a projective
  - $P$ any module, when $R$ is semisimple
  - $P$ flat and finitely presented
- does not imply
  - $P$ free, unless $R$ is local or a PID
  - $P$ injective, unless $R$ is quasi-Frobenius
  - $P$ finitely generated

FreeProjectiveFlatTorsion-free (over a domain)

All three implications are strict: $ℤ / 3$ over $ℤ / 6$ separates the first, $ℚ$ over $ℤ$ separates the second, and over $ℤ$ the third is an equivalence but over a general domain it is not.

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Topology

Serre–Swan correspondence

For a compact Hausdorff space $X$ , finitely generated projective modules over $C (X)$ correspond exactly to vector bundles on $X$ . Projectivity is the algebraic shadow of local triviality, and this is the starting point of topological K-theory and of noncommutative geometry.

Systems and control

Multidimensional filter banks

Perfect-reconstruction filter banks in several variables correspond to unimodular completion problems over $k [x_{1}, \dots, x_{n}]$ . Quillen–Suslin guarantees a solution exists, and constructive versions of the theorem produce the filters.

Algebraic K-theory

$K_{0}$ and finiteness obstructions

The Grothendieck group of finitely generated projectives is the primary invariant of a ring; Wall's finiteness obstruction and the class group of a number field are both instances.

Computer algebra

Resolutions and homology

Free and projective resolutions are how $Ext$ and $Tor$ are computed, and hence how syzygies, Betti numbers and group cohomology are computed in Macaulay2, Singular and GAP.

The honest position: projectivity itself is machinery. Its value is that it converts a lifting problem, which is hard to check, into a splitting problem, which is concrete — and the applications above are all instances of that conversion.

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

Over a PID. Smith normal form of a presentation matrix decides projectivity and produces a basis when the module is free; cost is polynomial in the matrix size, with coefficient growth the practical bottleneck over $ℤ$ .
Over a finite-dimensional algebra. Compute a complete set of primitive orthogonal idempotents, giving the indecomposable projectives $A e_{i}$ ; a module is projective iff its multiplicity vector matches a direct sum of these, which is a comparison of dimension vectors.
Over a polynomial ring. Deciding projectivity reduces to a Gröbner basis computation; constructive Quillen–Suslin algorithms (Logar–Sturmfels, and later refinements) then produce an explicit basis. Implementations exist in Macaulay2 and Singular, and worst-case Gröbner cost is doubly exponential in the number of variables.
Injectivity. Baer's criterion turns injectivity into a finite check when $R$ is left noetherian: it suffices to test extension along finitely many generators of each left ideal.
Limits. For a general finitely presented ring, deciding whether a finitely presented module is projective is not algorithmically solvable, because the word problem already is not.

Failure Modes and Common Mistakes

Do not confuse the two uses of the word injective: an injective homomorphism is a monomorphism, an injective module is one with the extension property. Lam's $(2.9)$ uses both in the same sentence.
Do not assume flat implies projective. It does for finitely presented modules, and $ℚ$ over $ℤ$ shows what happens without that hypothesis.
Do not expect projectivity to be preserved by restriction of scalars. $ℚ$ is projective over $ℚ$ and not projective over $ℤ$ .
Do not forget that lifting along an infinite basis uses the axiom of choice; the statement that all free modules are projective is not choice-free.

Quick Reference

Projectivemaps out of

P

lift along surjections

Equivalently

P

is a direct summand of a free module

Equivalentlyevery surjection onto

P

splits

Equivalently

{Ext}_{R}^{1} (P, -) = 0

Injectivemaps into

I

extend along injections

Baertest injectivity on left ideals only

Semisimple testall cyclic left modules projective

Dual testall left modules injective

Which result to quote
You want	Use	Reference
A splitting of $A ↠ P$	$P$ projective	(2.7)(1) $\Rightarrow$ (3)
A free module containing $P$ as a summand	$P$ projective	(2.7)(3) $\Rightarrow$ (2)
Projectivity of a summand of a free module	extend by zero	(2.7)(2) $\Rightarrow$ (1)
Semisimplicity from projectivity	cyclic modules suffice	(2.8)(4) $\Rightarrow$ (1)
Semisimplicity from injectivity	all modules injective	(2.9)
Semisimplicity from cyclic injectivity	Osofsky's theorem	cite, do not reprove

Frequently Asked Questions

Why is projectivity defined by a lifting property rather than as a summand of a free module?

Because the lifting property is what gets used and is stated purely in terms of maps, so it transfers to any abelian category. The description as a summand of a free module is a theorem about module categories specifically — it depends on there being enough free objects, which not every abelian category has.

Is every projective module free?

No. $ℤ / 3$ is projective and not free over $ℤ / 6$ , and non-principal ideals of a Dedekind domain are projective and not free. It is true over local rings by Kaplansky's theorem, over principal ideal domains, and — for finitely generated modules — over polynomial rings over a field by the Quillen–Suslin theorem.

Why does

(2.8)

only need cyclic modules?

Because the obstruction to semisimplicity lives entirely inside $_{R} R$ . Applying projectivity of the cyclic module $R / 𝔄$ to the canonical surjection $R ↠ R / 𝔄$ splits off $𝔄$ as a direct summand, and doing this for every left ideal is precisely the statement that $_{R} R$ is semisimple.

Are projective and injective modules the same over any ring?

They coincide over quasi-Frobenius rings, which include semisimple rings, $ℤ / n$ , and group algebras of finite groups over a field. Over $ℤ$ they are almost disjoint: $ℤ$ is projective and not injective, $ℚ$ is injective and not projective, and the only module that is both is zero.

What is the practical difference between

(2.8)

and

(2.9)

$(2.8)$ gives a genuinely cheap test — restrict attention to cyclic modules. $(2.9)$ as stated requires all modules, and the cyclic version, though true, rests on Osofsky's theorem. In practice one tests semisimplicity through projectivity, never through injectivity.

Does a module have to be finitely generated to be projective?

No. Any direct sum of projective modules is projective, so $R^{(I)}$ for infinite $I$ is projective and typically not finitely generated. Finite generation matters when one wants to conclude freeness or to compute with a dual basis of finite size.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §2, results (2.7)–(2.9), pp. 28–30.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §16–§18 (projective and injective modules).
B. L. Osofsky, “Rings all of whose finitely generated modules are injective”, Pacific Journal of Mathematics 14 (1964), 645–650.
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, §2 and §3.
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
I. Kaplansky, “Projective modules”, Annals of Mathematics 68 (1958), 372–377.

AI Suggested Questions

Prove the dual basis lemma and use it to show that a direct summand of a projective module is projective.
Give a full proof that every module embeds in an injective module, and explain why this has no projective analogue.
Outline Osofsky's proof that a ring whose cyclic modules are all injective is semisimple.
Show that a finitely presented flat module is projective, and find a flat module that is not.
Explain how the Quillen–Suslin theorem is used constructively in multidimensional filter bank design.
Compute the indecomposable projective modules of the path algebra of a quiver with two vertices and one arrow, and match them against the $T_{2} (k)$ example.
Characterise the rings over which every finitely generated projective module is free, and give an example that is neither local nor a polynomial ring.

Projective Modules and Splitting

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why free modules lift

Direct summands inherit lifting

The dual basis description

Where projectives sit

Key Results

Proof Techniques and Method

Worked Example

A cyclic non-projective module over $T_{2} (k)$

Step 1: $S_{2}$ is cyclic

Step 2: $R e_{2}$ is indecomposable

Step 3: $S_{2}$ is not projective

Projective but not free

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Serre–Swan correspondence

Multidimensional filter banks

$K_{0}$ and finiteness obstructions

Resolutions and homology

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why free modules lift

Direct summands inherit lifting

The dual basis description

Where projectives sit

06Key Results

07Proof Techniques and Method

08Worked Example

A cyclic non-projective module over T2(k)

Step 1: S2 is cyclic

Step 2: Re2 is indecomposable

Step 3: S2 is not projective

Projective but not free

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Serre–Swan correspondence

Multidimensional filter banks

K0 and finiteness obstructions

Resolutions and homology

13Computational Notes

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

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Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

A cyclic non-projective module over $T_{2} (k)$

Step 1: $S_{2}$ is cyclic

Step 2: $R e_{2}$ is indecomposable

Step 3: $S_{2}$ is not projective

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

$K_{0}$ and finiteness obstructions

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions