D&F, Ch 10, Proposition 30 - Proof that (3) ===> (4)

[Math Amateur]

I am reading Dummit and Foote, Section 10.5 : Exact Sequences - Projective, Injective and Flat Modules.

I am studying Proposition 30 and its proof (D&F, pages 389-390)

I need some help in order to fully understand the proof of \(\displaystyle (3) \Longrightarrow (4)\).

Proposition 30 reads as follows:

View attachment 2507

The relevant part of the proof i.e. \(\displaystyle (3) \Longrightarrow (4) \) is as follows:

View attachment 2508
https://www.physicsforums.com/attachments/2509

In the exact sequence \(\displaystyle 0 \longrightarrow ker \ \phi \longrightarrow \mathcal{F} \stackrel{\phi}{\longrightarrow} P \longrightarrow 0 \) , the mapping between \(\displaystyle ker \ \phi \text{ and } \mathcal{F} \) is the inclusion map and the map \(\displaystyle \phi \) would be the unique map identity on the (finite) generators for P ...

BUT ... how do we know that P is finitely generated ...

Indeed, I am assuming that a free module is one with a finite basis ... is this the case?

Can someone please clarify the above issues?

Peter

 

[Deveno]

Well, no, a free module does NOT need to be finitely generated. The "easiest" counter-example I can think of is the free $\Bbb Z$-module, $\Bbb Z[x]$, which is easily seen to have a countably infinite basis:

$\{1,x,x^2,\dots\}$

The map that sends "generators of $\mathcal{F}$ to "generators of $P$" does not have to be defined on a finite set. Free modules (indeed free abelian groups, free groups, free monoids, and free semi-groups) can be constructed using ANY set (including infinite sets) as a set of generators.

When embedding these free objects in a "direct sum" (that is to say, a co-product in an abelian category) it is typical to require that any linear combination have finite basis support. For if not, then one can have trouble defining the canonical "evaluation" homomorphisms (skirting this problem is what the study of Hilbert spaces is all about...some mathematicians still feel a bit "uneasy" about evaluating "formal power series in an indeterminate").

Your understanding of the maps involved isn't quite correct, however.

Let me give a (somewhat sophisticated) example.

Suppose $R = \Bbb Q$, and suppose that $T$ is the matrix:

$T = \begin{bmatrix}1&0&1\\2&1&0\\3&1&1 \end{bmatrix}$

so that $T$ defines a mapping $\Bbb Q^3 \to \Bbb Q^3$.

It can be shown that: $T(\Bbb Q^3) = \{(a,b,a+b): a,b \in \Bbb Q\}$, which is generated by $\{(1,0,1),(0,1,1)\}$.

Define the mapping: $S:\Bbb Q \to \Bbb Q^3$ by $S(q) = (-q,2q,q)$. Then the sequence:

$\{0\} \to \Bbb Q \stackrel{S}{\to} \Bbb Q^3 \stackrel{T}{\to} \Bbb Q^3 \to \{(0,0,0)\}$

is a short exact sequence of $\Bbb Q$-modules, and we have it splits because of the map:

$R:\Bbb Q^3 \to \Bbb Q^3$ defined by the matrix:

$\begin{bmatrix}1&1&1\\-2&-1&-2\\0&-1&-1 \end{bmatrix}$

and $TR(a,b,a+b) = T(2a+2b,-4a-3b,-a-2b) = (a,b,a+b)$.

Proposition 30 then tells us that: $\Bbb Q^3 \cong \text{ker }T \oplus \text{im }T$, and we can even make this isomorphism explicit:

$(x,y,z) \leftrightarrow (x(-1,2,1),T(0,z,y))$

Which shouldn't be surprising since $\Bbb Q^3$ is naturally isomorphic to $\Bbb Q \oplus \Bbb Q^2$, and $T(\Bbb Q^3)$ is clearly a free $\Bbb Q$-module with 2 generators.

 

[Math Amateur]

Well, no, a free module does NOT need to be finitely generated. The "easiest" counter-example I can think of is the free $\Bbb Z$-module, $\Bbb Z[x]$, which is easily seen to have a countably infinite basis:

$\{1,x,x^2,\dots\}$

The map that sends "generators of $\mathcal{F}$ to "generators of $P$" does not have to be defined on a finite set. Free modules (indeed free abelian groups, free groups, free monoids, and free semi-groups) can be constructed using ANY set (including infinite sets) as a set of generators.

When embedding these free objects in a "direct sum" (that is to say, a co-product in an abelian category) it is typical to require that any linear combination have finite basis support. For if not, then one can have trouble defining the canonical "evaluation" homomorphisms (skirting this problem is what the study of Hilbert spaces is all about...some mathematicians still feel a bit "uneasy" about evaluating "formal power series in an indeterminate").

Your understanding of the maps involved isn't quite correct, however.

Let me give a (somewhat sophisticated) example.

Suppose $R = \Bbb Q$, and suppose that $T$ is the matrix:

$T = \begin{bmatrix}1&0&1\\2&1&0\\3&1&1 \end{bmatrix}$

so that $T$ defines a mapping $\Bbb Q^3 \to \Bbb Q^3$.

It can be shown that: $T(\Bbb Q^3) = \{(a,b,a+b): a,b \in \Bbb Q\}$, which is generated by $\{(1,0,1),(0,1,1)\}$.

Define the mapping: $S:\Bbb Q \to \Bbb Q^3$ by $S(q) = (-q,2q,q)$. Then the sequence:

$\{0\} \to \Bbb Q \stackrel{S}{\to} \Bbb Q^3 \stackrel{T}{\to} \Bbb Q^3 \to \{(0,0,0)\}$

is a short exact sequence of $\Bbb Q$-modules, and we have it splits because of the map:

$R:\Bbb Q^3 \to \Bbb Q^3$ defined by the matrix:

$\begin{bmatrix}1&1&1\\-2&-1&-2\\0&-1&-1 \end{bmatrix}$

and $TR(a,b,a+b) = T(2a+2b,-4a-3b,-a-2b) = (a,b,a+b)$.

Proposition 30 then tells us that: $\Bbb Q^3 \cong \text{ker }T \oplus \text{im }T$, and we can even make this isomorphism explicit:

$(x,y,z) \leftrightarrow (x(-1,2,1),T(0,z,y))$

Which shouldn't be surprising since $\Bbb Q^3$ is naturally isomorphic to $\Bbb Q \oplus \Bbb Q^2$, and $T(\Bbb Q^3)$ is clearly a free $\Bbb Q$-module with 2 generators.

Thanks Deveno ... most helpful ...

You write:

"Well, no, a free module does NOT need to be finitely generated. ... ... "

and give an example of a free module that is not finitely generated ... but I am still a little confused ...

In M.E. Keating's book " A First Course in Module Theory" we read the following:

https://www.physicsforums.com/attachments/2515

So Keating defines a free module as one with a basis ... and the subset B that he is talking about (maybe as an example) is finite ... but I think your point is that B does not have to be finite ... reading the passage from Keating he does not rule out an infinite basis ...

So then the basis could be an infinite set? Is that right?

On reading your post I went back to check D&F as well ... and they define a free module as follows:

View attachment 2516

Well ... D&F talk about an R-Module being free on A and, when you read it carefully it seems never require A to be a finite set - but they do (and I think that this was my point of confusion) talk about every non-zero element \(\displaystyle x \in F \) being able to be expressed as a unique (and finite - n elements!) sum of the form

\(\displaystyle x = r_1a_1 + r_2a_2 + ... \ ... r_na_n \)

Are my comments/analysis of the situation correct?

Peter

 

[Deveno]

Yes, a basis is a set that spans and is linearly independent. The set itself need not be finite. Let's look at what linear independence and spanning mean:

A set is said to be linearly independent if the only linear combination of its elements that sums to 0 is the 0-combination (all the $R$-coefficients (or as they are sometimes called, "co-ordinates") are 0).

A set is said to span another (set) if every element of the spanned set is a linear combination of the first set.

These notions usually presume we are talking about an $R$-module (linear combinations involve the notion of sum and scalar multiplication). Linear combinations are finite sums of scalar multiples of elements of the basis. The reason we have to choose finite sums, is because we, in general, cannot "evaluate" infinite sums. For example, if $x > 1$ it is impossible to calculate what:

$1 + x + x^2 +\cdots$

"is".

Polynomial rings are the best example of this: it is immediate that monomial terms may be of arbitrarily high degree, and that different monomial terms are all linearly independent (equality of polynomials is, by definition, equality of term coefficients), so any basis is infinite, yet any polynomial has but a finite number of terms.

In point of fact, though, finitely-generated modules are of more interest, because we can say more about them with greater certainty. And to further leverage the power of dvisibility in PID's, it is not uncommon to restrict study even MORE, and consider finitely-generated modules over a PID, whose structure we can catalog very "concretely".

With arbitrary rings, we usually seek to find "a better quotient", that is, some ideal that contains "all of the bad elements" that prevent our ring from being "nice". Or, to coin a nifty slogan: "when in doubt, mod it out!".

 

[blue_raver22]

Dear Peter,

Thank you for your questions regarding Proposition 30 and its proof. In order to fully understand the proof of (3) \Longrightarrow (4), we need to first understand the concept of exact sequences and projective modules.

An exact sequence is a sequence of modules and homomorphisms between them, where the image of one homomorphism is equal to the kernel of the next. In Proposition 30, the exact sequence is 0 \longrightarrow ker \ \phi \longrightarrow \mathcal{F} \stackrel{\phi}{\longrightarrow} P \longrightarrow 0. This means that the image of the homomorphism from ker \ \phi to \mathcal{F} is equal to the kernel of the homomorphism from \mathcal{F} to P.

A projective module is a module that satisfies a certain lifting property, which essentially means that any homomorphism from the module to a larger module can be extended to a homomorphism from the larger module. This is the case for P in Proposition 30, as the map \phi can be extended to a homomorphism from \mathcal{F} to P.

Now, in the proof of (3) \Longrightarrow (4), we are given that \mathcal{F} is a projective module, which means that it has a lifting property. This allows us to extend the inclusion map from ker \ \phi to \mathcal{F} to a homomorphism from \mathcal{F} to P. This extended map is the same as the unique map identity on the finite generators for P, which shows that P is indeed finitely generated.

To answer your question, a free module is indeed one with a finite basis. This means that it is generated by a finite set of elements, and therefore it is also finitely generated.

I hope this clarifies the issues you had with the proof. If you have any further questions, please don't hesitate to ask.

Best,