I have a question related to this: Projective modules

I'm trying to understand the "philosophy" of the statement, because it seems too similar to the statement "a module is free iff every element can be written uniquely as a finite linear combination of elements of a basis".

Is this "projective basis" property saying this:

a module P is projective iff every element in P can be written as a finite linear combination of some elements of P?

We lose uniqueness in the expression as a sum: in the elements of P, in the elements of R, and in the number of terms (so the concept of "rank" wouldn't make sense). Is this all, or am I misunderstanding the statement?

Any other intuition related to that property is also appreciated.


The statement "every element in P can be written as a finite linear combination of some elements of P.", where "some" means a finite set, just says that the module is finitely generated.

This has nothing to do with being projective.

Take for instance the $\mathbb Z$-module $\mathbb Z/2$. Here every element can be written as a multiple of $[1]$. So $\mathbb Z/2$ is finitely generated but not projective.

On the other hand the infinite direct sum $\bigoplus_{\mathbb N}\mathbb Z$ is a projective $\mathbb Z$-module which is not finitely generated.

The crucial thing in the definition of a projective basis is really that you have homomorphisms out of $P$ into the ring (regarded as module over itself).

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    $\begingroup$ But when I say "some" I do not mean a finite set. Of course, in the expression as a finite sum, they are finite, but they could be chosen from an infinite subset of P. Also, why is it crucial to have homomorphisms? in the case of free module, the homomorphisms are of the form $f_i(x)=r_i$, not in the case of projective modules? I'm confused. $\endgroup$ – Bruno Stonek Feb 16 '11 at 19:29
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    $\begingroup$ As Mariano points out, if you just mean any fixed generating set, you can always take the whole module. It's crucial to have homomorphisms because that's were the condition becomes non-trivial. $\endgroup$ – Rasmus Feb 16 '11 at 19:53
  • $\begingroup$ I understand it now. I was thinking of the $f_i$ merely as functions, and this won't do. Asking for them to be homomorphisms is what renders the condition non-trivial. Thank you. $\endgroup$ – Bruno Stonek Feb 16 '11 at 20:11
  • $\begingroup$ @Bruno: It was a pleasure to help. $\endgroup$ – Rasmus Feb 16 '11 at 20:45

The statement you're linking to is: A module $P$ is projective if and only if there is a family $\{x_{i}\}_{i \in I} \subset P$ and morphisms $f_{i}: P \to R$ such that for each $x \in P$ we have $x = \sum_{i \in I} f_{i}(x) x_{i}$. The last statement says three things:

  1. In order for the sum to make sense we must have that for all $x$ the set $\{i\,:\,f_{i}(x) \neq 0 \}$ is finite. Or, as stated there: for all $x$ we have $f_{i}(x) = 0$ for almost all $i$.
  2. The set $(x_{i})_{i \in I}$ generates $P$. In other words, the map $g: \bigoplus_{i \in I} R \to P$ sending $(r_{i})$ to $\sum_{i \in I} r_{i} x_{i}$ is an epimorphism (it suffices to take $r_{i} = f_{i}(x)$ to see that this map is onto).
  3. The epimorphism $g$ splits: there is a right inverse $f: P \to \bigoplus_{i \in I} R$ of $g$, i.e. $gf = \operatorname{id}_{P}$ (this morphism $f$ is of the form $(f_{i})_{i \in I}$ with morphisms $f_{i}: P \to R$ and by the definition of a direct sum we have $f_{i}(x) \neq 0$ only for finitely many $i$).

In my opinion its just an extremely explicit way of phrasing the much more catchy "a module is projective if and only if it is a direct summand of a free module".

  • $\begingroup$ hey, but sometimes you need extremely explicit descriptions! Catchy will only get you as far as it'll get you :) $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '11 at 20:58
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    $\begingroup$ @Mariano: Right, I second that. Probably the important thing to achieve is that you understand the situation well enough that you can pass back and forth between catchy and extremely explicit. $\endgroup$ – t.b. Feb 16 '11 at 21:10

The phrasing of your statement

a module P is projective iff every element in P can be written as a finite linear combination of some elements of P

is pretty bad, because it is not at all evident from that those some elements are fixed: what you mean, but not what you wrote, probably is

a module is projective iff there is a set $X\subseteq P$ such that every element of $P$ is a finite combination of elements of $X$.

Now, this last statement is false: you can always pick $X=P$, so its truth would imply that all modules are projective---and there are rings which have non-projective modules!

  • $\begingroup$ Yes, that is what I meant, and yes, I see it can't be true. $\endgroup$ – Bruno Stonek Feb 16 '11 at 20:09

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