How does one prove that Q is flat but not projective over Z?

Flatness: Q⊗_Z - is localisation at Z0, an exact functor. Non-projectivity: projective Z-modules are free, and Q is not free — any two rationals are linearly dependent over Z, so a basis would have one element, but Q is not cyclic.

Flat Modules and Their Basic Theory

Q: Why is only injectivity part of the definition of flatness?

Because M ⊗_R - is right exact for every module M: it always preserves cokernels and surjections. The only possible failure of exactness is at the left-hand end, so flatness is precisely the demand that injections are preserved.

Q: What is the point of (24.22) if (24.23) is what gets used?

(24.22) is the mechanism that lets one *transfer* exactness between the two variables of the tensor product. (24.23) uses it twice, once in each direction, and the pair of hypotheses — F' flat for one part, F flat for the other — is exactly what makes the two transfers available.

Q: Why does (24.24) produce a chain of left ideals when the module is a right module?

The relations e_i = e_i+1a_i+1 have coefficients acting on the right of the basis vectors, so composing them accumulates the a_i on the left: e_0 = e_n a_n … a_1. Any condition extracted from the a_i therefore concerns products a_n … a_1 and the left ideals they generate. This is the origin of the side switch in Bass's Theorem P.

Executive Summary

Tensoring is right exact for free: $M \otimes_{R} A \to M \otimes_{R} B \to M \otimes_{R} C \to 0$ is exact for any short exact sequence of left modules. Flatness is the demand that it also be left exact — that $A ↪ B$ give $M \otimes_{R} A ↪ M \otimes_{R} B$ .

Free $\Rightarrow$ projective $\Rightarrow$ flat, with the last implication strict in general: $ℚ$ is flat but not projective over $ℤ$ . Bass's theorem $(24.25)$ says the implication is an equivalence precisely over right perfect rings, and the module built in $(24.24)$ from a sequence $a_{1}, a_{2}, \dots$ is the device that proves it.

$- \otimes$ exactDefinition

$ℚ$ Flat, not projective over

ℤ

Torsion-freeFlatness over

ℤ

(24.20)–(24.24)Lam's numbering

Overview

Flatness is a right-module property tested against left modules, so it is intrinsically two-sided in its bookkeeping even though it is a property of one module. Throughout, $M$ is a right $R$ -module, primed letters denote left modules, and $\otimes$ means $\otimes_{R}$ .

M flat : ⟺ (A ↪ B of left R -modules ⟹ M \otimes_{R} A ↪ M \otimes_{R} B)

(24.20)

Right exactness of $M \otimes_{R} -$ is automatic; flatness adds injectivity on the left.

The elementary permanence properties are immediate from the fact that tensor products commute with direct sums: $R_{R}$ is flat because $R \otimes_{R} A ≅ A$ ; direct sums of flat modules are flat; direct summands of flat modules are flat. Hence free $\Rightarrow$ projective $\Rightarrow$ flat.

The two technical results on this page — the lemma $(24.22)$ comparing two presentations and the theorem $(24.23)$ — exist to make flatness checkable on a single chosen presentation rather than against all short exact sequences, and $(24.24)$ then applies that machinery to a deliberately constructed module.

Learning Objectives

State $(24.20)$ and explain why only injectivity needs checking.
Prove that direct sums and direct summands of flat modules are flat, and deduce projective implies flat.
Verify that $ℚ$ is $ℤ$ -flat and that $ℤ / 2 ℤ$ is not.
State $(24.22)$ with the correct flatness hypothesis on each side.
Use $(24.23)$ to test flatness against one fixed presentation of $M$ .
Build the module $F / K$ of $(24.24)$ and prove it is flat.

Definitions

Definition(24.20)Flat module

A right $R$ -module $M$ is flat if the functor $M \otimes_{R} -$ is exact on the category of left $R$ -modules: for every short exact sequence $0 \to A \to B \to C \to 0$ of left $R$ -modules, the induced sequence $0 \to M \otimes_{R} A \to M \otimes_{R} B \to M \otimes_{R} C \to 0$ is exact. Since the tail $M \otimes A \to M \otimes B \to M \otimes C \to 0$ is always exact, the content is that $M \otimes_{R} -$ preserves injectivity.

$M \otimes_{R} -$: The tensor functor from left $R$ -modules to abelian groups. It is always right exact and commutes with arbitrary direct sums and direct limits.
Faithfully flat: Flat, and in addition $M \otimes_{R} N = 0$ forces $N = 0$ . A stronger condition used in descent theory; flatness alone does not detect vanishing.
${Tor}_{1}^{R} (M, N)$: The first derived functor of the tensor product; $M$ is flat exactly when ${Tor}_{1}^{R} (M, N) = 0$ for all left modules $N$ .
Torsion-free: For abelian groups: $n x = 0$ with $n \neq 0$ implies $x = 0$ . Over $ℤ$ this is equivalent to flatness, by $(24.21)$ .
Presentation: An exact sequence $0 \to K \to F \to M \to 0$ with $F$ free or flat. $(24.23)$ makes flatness of $M$ testable on any one such sequence.

Modules on the right are the ones tested for flatness; left modules are the test objects. Over a commutative ring the distinction evaporates, and much of the intuition comes from that case.

Core Concepts

The hierarchy and where it is strict

freeprojectiveflattorsion-free (over a domain)

Each arrow can be strict. Over $ℤ$ , projective and free coincide but flat is strictly weaker: $ℚ$ is flat and not projective. Over a general commutative domain, flat implies torsion-free but not conversely — the ideal $(x, y)$ in $k [x, y]$ is torsion-free and not flat.

Why localisations are flat

For a commutative ring $R$ and a multiplicatively closed $S \subseteq R$ , the functor $S^{- 1} R \otimes_{R} -$ is naturally isomorphic to localisation $S^{- 1} (-)$ , which is exact because a fraction is zero only if some element of $S$ kills its numerator. So $S^{- 1} R$ is a flat $R$ -module for every $S$ ; it is projective only in special cases, and for $R = ℤ$ , $S = ℤ ∖ {0}$ it is $ℚ$ , which is not.

Detecting non-flatness

To show $M$ is not flat, exhibit one injection that it destroys. Two standard patterns: a torsion module over a domain kills the multiplication map, and a module annihilated by $n$ kills the inclusion of $n B$ into $B$ . Both are visible in the $ℤ$ -module $ℤ / 2 ℤ$ .

Key Results

Proposition—Elementary permanence properties

Let $R$ be a ring. (i) $R_{R}$ is flat. (ii) An arbitrary direct sum $⨁_{i} M_{i}$ of right $R$ -modules is flat if and only if each $M_{i}$ is flat. (iii) A direct summand of a flat module is flat. Consequently every free right $R$ -module is flat, and every projective right $R$ -module is flat.

Proof

(i) $R \otimes_{R} A ≅ A$ naturally, so $R \otimes_{R} -$ is the identity functor up to isomorphism and is exact. (ii) Tensor products commute with direct sums, so for an injection $A ↪ B$ the map $(⨁_{i} M_{i}) \otimes A \to (⨁_{i} M_{i}) \otimes B$ is the direct sum of the maps $M_{i} \otimes A \to M_{i} \otimes B$ ; a direct sum of maps is injective exactly when each summand is. (iii) is the only if half of (ii). Free modules are direct sums of copies of $R_{R}$ , projective modules are their direct summands.

Example—Flat but not projective; not flat

For a commutative ring $R$ and a multiplicatively closed set $S$ , $S^{- 1} R$ is a flat $R$ -module, because $S^{- 1} R \otimes_{R} -$ is the exact localisation functor. Taking $R = ℤ$ and $S = ℤ ∖ {0}$ shows $ℚ$ is $ℤ$ -flat; it is not projective, since projective $ℤ$ -modules are free and $ℚ$ is not free. In the other direction, $ℤ / 2 ℤ$ and $ℚ / ℤ$ are not $ℤ$ -flat.

Proposition(24.21)Flatness over the integers

An abelian group $M$ is flat as a $ℤ$ -module if and only if $M$ is torsion-free. (Lam records this without proof; it follows from the criterion that flatness need only be tested on finitely generated ideals, which over $ℤ$ are the $n ℤ$ .) Flat modules may therefore be thought of as a generalisation of torsion-free abelian groups.

Lemma(24.22)Comparing two presentations

Let $ε : 0 \to K \to F \to M \to 0$ be an exact sequence of right $R$ -modules and $ε' : 0 \to K' \to F' \to M' \to 0$ an exact sequence of left $R$ -modules.

If $F'$ is flat, then exactness of $M \otimes_{R} ε'$ implies exactness of $ε \otimes_{R} M'$ ;
if $F$ is flat, then exactness of $ε \otimes_{R} M'$ implies exactness of $M \otimes_{R} ε'$ .

Here *exactness of $M \otimes_{R} ε'$ * means injectivity of $M \otimes K' \to M \otimes F'$ , and *exactness of $ε \otimes_{R} M'$ * means injectivity of $K \otimes M' \to F \otimes M'$ . The two statements are exchanged by passing to $R^{op}$ , so it suffices to prove one.

Proof

We prove (1) by a diagram chase in the $3 \times 3$ array with entries $X \otimes Y'$ for $X \in {K, F, M}$ and $Y' \in {K', F', M'}$ . All rows and columns are right exact. Because $F'$ is flat, the row $0 \to K \otimes F' \to F \otimes F' \to M \otimes F' \to 0$ is exact; by hypothesis the column $0 \to M \otimes K' \to M \otimes F' \to M \otimes M' \to 0$ is exact.

Let $x \in K \otimes M'$ have image $0$ in $F \otimes M'$ . Since $K \otimes F' \to K \otimes M'$ is onto, choose $y \in K \otimes F'$ mapping to $x$ . Its image $γ (y) \in F \otimes F'$ dies in $F \otimes M'$ , so by exactness of the column through $F$ there is $z \in F \otimes K'$ with image $γ (y)$ .

Push $z$ into $M \otimes K'$ . Its further image in $M \otimes F'$ equals the image of $γ (y)$ , which is $0$ because $γ (y)$ comes from $K \otimes F'$ . As $M \otimes K' \to M \otimes F'$ is injective, the image of $z$ in $M \otimes K'$ is $0$ , so by exactness of the row through $K'$ we may write $z$ as the image of some $w \in K \otimes K'$ .

Now the images of $w$ and of $y$ in $F \otimes F'$ agree, and $K \otimes F' \to F \otimes F'$ is injective because $F'$ is flat; hence $y$ is the image of $w$ . Finally the composite $K \otimes K' \to K \otimes F' \to K \otimes M'$ is zero, so $x = 0$ . This proves injectivity of $K \otimes M' \to F \otimes M'$ .

In the language of derived functors both parts read off from the long exact sequences: flatness of $F'$ gives ${Tor}_{1}^{R} (M, M') ≅ \ker (M \otimes K' \to M \otimes F')$ , while flatness of $F$ gives an injection ${Tor}_{1}^{R} (M, M') ↪ K \otimes M'$ whose image is the kernel of $K \otimes M' \to F \otimes M'$ .

Theorem(24.23)Flatness tested on one presentation

Let $ε : 0 \to K \to F \to M \to 0$ be an exact sequence of right $R$ -modules with $F$ flat. Then $M$ is flat if and only if $ε \otimes_{R} M'$ is exact for every left $R$ -module $M'$ — that is, $K \otimes_{R} M' \to F \otimes_{R} M'$ is injective for every $M'$ .

Proof

Necessity. Assume $M$ is flat and let $M'$ be a left $R$ -module. Choose an exact sequence $ε' : 0 \to K' \to F' \to M' \to 0$ with $F'$ free, hence flat. Flatness of $M$ makes $M \otimes_{R} ε'$ exact, so $(24.22) (1)$ — whose hypothesis * $F'$ flat* is satisfied — gives exactness of $ε \otimes_{R} M'$ .

Sufficiency. Assume $ε \otimes_{R} M'$ is exact for every left module $M'$ . Let $ε' : 0 \to K' \to F' \to M' \to 0$ be an arbitrary short exact sequence of left $R$ -modules. Applying $(24.22) (2)$ , legitimate because $F$ is flat by hypothesis, exactness of $ε \otimes_{R} M'$ yields exactness of $M \otimes_{R} ε'$ , i.e. injectivity of $M \otimes K' \to M \otimes F'$ . Since every injection of left modules occurs inside such a sequence, $M$ is flat.

Proposition(24.24)The flat module attached to a sequence (Bass)

Let $a_{1}, a_{2}, \dots$ be any sequence of elements of $R$ . Let $F = ⨁_{i \geq 0} e_{i} R$ be free of countable rank and let $K \subseteq F$ be the submodule generated by

f_{i} = e_{i} - e_{i + 1} a_{i + 1}, i \geq 0 .

(24.24a)

Then the right $R$ -module $M : = F / K$ is flat. Moreover, if $M$ is projective then the descending chain of principal left ideals $R a_{1} \supseteq R a_{2} a_{1} \supseteq R a_{3} a_{2} a_{1} \supseteq \dots$ is eventually stationary.

Proof

Flatness. The elements $f_{i}$ generate $K$ freely: a relation $\sum_{i} f_{i} r_{i} = 0$ expands, in the free basis ${e_{i}}$ , to $r_{0} = 0$ , then $r_{i} - a_{i} r_{i - 1} = 0$ for $i \geq 1$ , giving all $r_{i} = 0$ . So $K = ⨁_{i \geq 0} f_{i} R$ is free, and $F$ is free; by $(24.23)$ it suffices to show $K \otimes M' \to F \otimes M'$ is injective for every left module $M'$ .

Let $α = \sum_{i = 0}^{n} f_{i} \otimes x_{i} \in K \otimes M'$ map to $0$ . Expanding in $F \otimes M' = ⨁_{i} (e_{i} \otimes M')$ :

0 = e_{0} \otimes x_{0} + \sum_{i = 1}^{n} e_{i} \otimes (x_{i} - a_{i} x_{i - 1}) - e_{n + 1} \otimes a_{n + 1} x_{n} .

(P.1)

Each component must vanish separately, so $x_{0} = 0$ , then $x_{i} = a_{i} x_{i - 1} = 0$ successively for $i = 1, \dots, n$ . Hence $α = 0$ and $M$ is flat.

Projectivity forces stationarity. Suppose $M$ is projective. Then $0 \to K \to F \to M \to 0$ splits, so there is $π : F \to K$ restricting to the identity on $K$ . Write $π (e_{i}) = \sum_{j} f_{j} b_{i j}$ with $b_{i j} \in R$ , almost all zero for each fixed $i$ . Applying $π$ to $f_{i} = e_{i} - e_{i + 1} a_{i + 1}$ and comparing coefficients in the free basis ${f_{j}}$ of $K$ :

b_{i i} - b_{i + 1, i} a_{i + 1} = 1, b_{i j} - b_{i + 1, j} a_{i + 1} = 0 (i \neq j) .

(P.2)

Fix $j$ and iterate the second relation for $i = 0, 1, \dots, j - 1$ : $b_{0 j} = b_{1 j} a_{1} = b_{2 j} a_{2} a_{1} = \dots = b_{j j} a_{j} \dots a_{2} a_{1}$ . Since $π (e_{0})$ has finite support, $b_{0 j} = 0$ for all sufficiently large $j$ , so $b_{j j} a_{j} \dots a_{1} = 0$ for such $j$ . Using the first relation in the form $1 - b_{j j} = - b_{j + 1, j} a_{j + 1}$ ,

a_{j} \dots a_{1} = (1 - b_{j j}) a_{j} \dots a_{1} = - b_{j + 1, j} a_{j + 1} a_{j} \dots a_{1} \in R a_{j + 1} a_{j} \dots a_{1} .

(P.3)

Thus $R a_{j} \dots a_{1} \subseteq R a_{j + 1} a_{j} \dots a_{1}$ for all large $j$ , and the reverse inclusion always holds, so the chain $R a_{1} \supseteq R a_{2} a_{1} \supseteq \dots$ becomes stationary.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Reduce to one presentation

Testing flatness against every injection is impractical. $(24.23)$ fixes one exact sequence with flat middle term and tests only that; the price is the lemma $(24.22)$ , which is where the diagram chase lives.

Move 2

Expand in a free basis

For a free module, $F \otimes M' = ⨁_{i} (e_{i} \otimes M')$ , so an element vanishes iff all its coordinates do. Flatness verifications for explicitly presented modules reduce to solving a triangular system.

Move 3

Turn a splitting into equations

Projectivity gives a retraction $π : F \to K$ ; writing $π$ in the two bases turns an abstract splitting into the coefficient identities $(P . 2)$ , from which the chain condition falls out.

Move 3 is the pivotal one for the sequel. The module $M$ of $(24.24)$ is engineered so that its projectivity forces a specific chain of principal left ideals to stabilise, which is exactly the DCC appearing in Bass's Theorem P — and that is how the flatness criterion for perfect rings gets its chain condition.

Worked Example

$ℤ / 2 ℤ$ is not flat

Let $M = ℤ / 2 ℤ$ over $R = ℤ$ ; then $M \otimes_{ℤ} -$ is reduction modulo $2$ , that is $N \mapsto N / 2 N$ . Take $B = ℤ / 4 ℤ$ and its unique subgroup of order $2$ , $A = 2 ℤ / 4 ℤ$ , with the inclusion $A ↪ B$ .

A / 2 A ≅ ℤ / 2 ℤ \neq 0, A = 2 B ⟹ the induced map A / 2 A \to B / 2 B is 0 .

(E.1)

$2 A = 0$ because every element of $A$ has order dividing $2$ ; and $A = 2 B$ maps into $2 B / 2 B = 0$ .

So $M \otimes A \to M \otimes B$ is the zero map from a nonzero group: not injective, hence $M$ is not flat. Consistently with $(24.21)$ , $ℤ / 2 ℤ$ has torsion.

$ℚ / ℤ$ is not flat, by a different injection

Take $M = ℚ / ℤ$ and the injection $A = ℤ ↪ B = ℚ$ . Then $M \otimes_{ℤ} ℤ ≅ ℚ / ℤ \neq 0$ , while $M \otimes_{ℤ} ℚ = 0$ because every element of $ℚ / ℤ$ is divisible by every positive integer, so $x \otimes q = x \otimes n (q / n) = n x \otimes (q / n)$ can be made to vanish. A nonzero group mapping to zero is not an injection.

Bass's module for $R = ℤ$ and $a_{i} = 2$

Take $R = ℤ$ and $a_{i} = 2$ for all $i$ . Then $F = ⨁_{i \geq 0} e_{i} ℤ$ , $K$ is generated by $e_{i} - 2 e_{i + 1}$ , and $M = F / K$ identifies $e_{i}$ with $2 e_{i + 1}$ , so

M ≅ ℤ [\frac{1}{2}] = varinjlim (ℤ \overset{2}{⟶} ℤ \overset{2}{⟶} ℤ \to \dots),

(E.2)

The class of $e_{i}$ corresponds to $2^{- i}$ .

This is torsion-free, hence flat by $(24.21)$ , matching the general assertion of $(24.24)$ . It is not projective: the chain $2 ℤ \supseteq 4 ℤ \supseteq 8 ℤ \supseteq \dots$ of principal ideals is strictly descending, so the necessary condition in $(24.24)$ fails.

Comparison and Classification

Flat, projective and free over familiar rings
Ring	Flat but not projective	Projective but not free	Flat = projective?
A field or division ring	none	none	yes
$ℤ$	$ℚ$ , $ℤ [1 / 2]$	none	no
$ℤ_{(p)}$	$ℚ$	none	no
$T_{2} (k)$ , $k$ a field	none	$e_{1} R$	yes
A Dedekind domain, not a field	the fraction field	non-principal ideals	no
Any right perfect ring	none	possible	yes

Closure properties
Operation	Flat	Projective
Arbitrary direct sums	preserved	preserved
Direct summands	preserved	preserved
Direct limits	preserved	not preserved
Arbitrary direct products	not in general	not in general
Extensions	preserved	preserved (the sequence splits)
Base change $- \otimes_{R} S$	preserved	preserved

The direct limit row is the essential difference and explains everything else: every flat module is a direct limit of finitely generated free modules, by Lazard's theorem, and projectivity is not a limit-stable condition.

Relationship Map

All right $R$ -modulesno condition

Flat

M \otimes_{R} -

exact; equivalently

{Tor}_{1}^{R} (M, -) = 0

Projectivesummand of a free module; lifts along epimorphisms

Freehas a basis

Finitely generated free

R^{n}

for some

n

The middle containment collapses precisely over right perfect rings, by $(24.25)$ ; the outer one collapses over local rings and over $ℤ$ , where projective already implies free.

sequence

a_{1}, a_{2}, \dots

flat module

F / K

F / K

projective

R a_{1} \supseteq R a_{2} a_{1} \supseteq \dots

stationary

This is the bridge to the chain conditions of Bass's Theorem P: if every flat module is projective, then every such chain is stationary, which is DCC on principal left ideals.

Computational Notes

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

Over a commutative Noetherian ring, flatness of a finitely presented module is decidable: a finitely presented flat module is projective, and projectivity of a finitely presented module can be tested by computing Fitting ideals or by checking local freeness at the primes in the support.
For modules given by a presentation matrix over a polynomial ring, Gröbner basis packages (Macaulay2, Singular, Sage) compute ${Tor}_{1}$ and hence certify flatness; the cost is dominated by the syzygy computation, which is doubly exponential in the worst case.
Local criterion: over a commutative local ring, a finitely generated module is flat if and only if it is free, which converts flatness testing into a rank computation over the residue field.
For infinitely generated modules there is no algorithm; $ℚ$ and $ℤ [1 / 2]$ show that the interesting flat modules are direct limits and are not finitely presentable.
Lazard's theorem — every flat module is a direct limit of finitely generated free modules — is the structural statement behind all of the above, and explains why flatness is stable under limits while projectivity is not.

Failure Modes and Common Mistakes

Do not assume infinite direct products of flat modules are flat; that requires coherence of the ring, and fails for a general $R$ .
Do not confuse flat with faithfully flat: $ℚ$ is flat over $ℤ$ but $ℚ \otimes_{ℤ} ℤ / 2 ℤ = 0$ , so it does not detect vanishing.
Do not forget which side is being tested: $M_{R}$ flat is a statement about left modules $M'$ , and over a noncommutative ring a module can be flat on one side of a bimodule structure and not the other.
Do not expect $(24.24)$ to produce a projective module when the chain is stationary; stationarity is stated only as a necessary condition for projectivity, not a sufficient one.

Quick Reference

Definition

M_{R}

flat

iff M \otimes_{R} -

exact on left modules

What needs checkingOnly preservation of injections

Tor form

M

flat

iff {Tor}_{1}^{R} (M, N) = 0

for all

_{R} N

Ideal test

M \otimes_{R} 𝔞 \to M

injective for finitely generated left ideals

𝔞

Hierarchyfree

\Rightarrow

projective

\Rightarrow

flat, both strict in general

Over $ℤ$ flat

iff

torsion-free

(24.21)

Localisation

S^{- 1} R

is always

R

-flat

One-presentation test

(24.23)

: with

F

flat,

M

flat

iff K \otimes M' \to F \otimes M'

injective for all

M'

Bass's module

F / K

with

f_{i} = e_{i} - e_{i + 1} a_{i + 1}

is flat; projective forces DCC on

R a_{n} \dots a_{1}

The results of (24.20)–(24.24)
Item	Statement	Hypotheses
(24.20)	Definition of flatness	none
(24.21)	$ℤ$ -flat $iff$ torsion-free	$R = ℤ$
(24.22)(1)	$M \otimes ε'$ exact $\Rightarrow ε \otimes M'$ exact	$F'$ flat
(24.22)(2)	$ε \otimes M'$ exact $\Rightarrow M \otimes ε'$ exact	$F$ flat
(24.23)	Flatness testable on one presentation	$F$ flat in $0 \to K \to F \to M \to 0$
(24.24)	$F / K$ is flat; projective forces a stationary chain	any sequence $a_{1}, a_{2}, \dots \in R$

Frequently Asked Questions

Why is only injectivity part of the definition of flatness?

Because $M \otimes_{R} -$ is right exact for every module $M$ : it always preserves cokernels and surjections. The only possible failure of exactness is at the left-hand end, so flatness is precisely the demand that injections are preserved.

Is a flat module over a noncommutative ring flat on both sides?

The question is not well posed for a one-sided module: flatness of $M_{R}$ is tested against left $R$ -modules. For a bimodule $_{S} M_{R}$ one may ask about flatness over $S$ and over $R$ separately, and the two are independent conditions.

How does one prove that

ℚ

is flat but not projective over

ℤ

Flatness: $ℚ \otimes_{ℤ} -$ is localisation at $ℤ ∖ {0}$ , an exact functor. Non-projectivity: projective $ℤ$ -modules are free, and $ℚ$ is not free — any two rationals are linearly dependent over $ℤ$ , so a basis would have one element, but $ℚ$ is not cyclic.

What is the point of

(24.22)

(24.23)

is what gets used?

$(24.22)$ is the mechanism that lets one transfer exactness between the two variables of the tensor product. $(24.23)$ uses it twice, once in each direction, and the pair of hypotheses — $F'$ flat for one part, $F$ flat for the other — is exactly what makes the two transfers available.

Why does

(24.24)

produce a chain of left ideals when the module is a right module?

The relations $e_{i} = e_{i + 1} a_{i + 1}$ have coefficients acting on the right of the basis vectors, so composing them accumulates the $a_{i}$ on the left: $e_{0} = e_{n} a_{n} \dots a_{1}$ . Any condition extracted from the $a_{i}$ therefore concerns products $a_{n} \dots a_{1}$ and the left ideals they generate. This is the origin of the side switch in Bass's Theorem P.

Does every module have a flat cover?

Yes — over every ring. This is the flat cover conjecture, proved by Bican, El Bashir and Enochs in 2001. The contrast with projective covers, which exist only over perfect rings, is one of the more striking asymmetries in the subject.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §24, (24.20)–(24.24) (pp. 367–369).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
H. Cartan and S. Eilenberg, Homological Algebra, Princeton University Press, 1956, Chapters II and VI.
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, §4 (flat modules and the equational criterion).
L. Bican, R. El Bashir and E. Enochs, “All modules have flat covers”, Bulletin of the London Mathematical Society 33 (2001), 385–390.

AI Suggested Questions

Prove Lazard's theorem that every flat module is a direct limit of finitely generated free modules.
Show that a finitely presented flat module is projective, and identify where finite presentation is used.
Give a torsion-free module over $k [x, y]$ that is not flat, and compute the obstructing Tor group.
Work out the equational criterion for flatness and use it to reprove that localisations are flat.
For which rings is an arbitrary direct product of flat right modules flat?
Compute the module $F / K$ of $(24.24)$ for $R = ℤ$ and a sequence of distinct primes and identify it explicitly.
How do flat modules behave under change of rings, and what does faithful flatness add?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The hierarchy and where it is strict

Why localisations are flat

Detecting non-flatness

06Key Results

07Proof Techniques and Method

Reduce to one presentation

Expand in a free basis

Turn a splitting into equations

08Worked Example

ℤ/2ℤ is not flat

ℚ/ℤ is not flat, by a different injection

Bass's module for R=ℤ and ai=2

09Comparison and Classification

10Relationship Map

11Computational Notes

12Failure Modes and Common Mistakes

13Quick Reference

14Frequently Asked Questions

15Related KEVOS Topics

16References

17AI Suggested Questions

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Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

$ℤ / 2 ℤ$ is not flat

$ℚ / ℤ$ is not flat, by a different injection

Bass's module for $R = ℤ$ and $a_{i} = 2$

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions