# Being direct summand of free module implies having dual basis.

I need to provet equality of two definitions of projective module: being direct summand of free module (or equally: having embedding into free module) and having dual basis.

Lets use Wikipedia notation: https://en.wikipedia.org/wiki/Projective_module#Direct_summands_of_free_modules (points 1.3 and 1.5)

I understand that map which takes x to direct sum of $$f_{i}(x)$$ is an embedding of P into free module which is direct sum or R through indexes $$i \in I$$, therefore having dual basis implies being direct summand.

I dont understand how second implication from direct summan to dual basis works. https://ncatlab.org/nlab/show/dual+basis On NCatlab these generators $${x_{i}}$$ are taken from nowhere (as it often happens on ncatlab).

Okay, as P is direct summand, we have emebedding into free module F = direct sum of R through indexes in I. Okey, let (f_{i}(x)) be embedding of element x.

Of course, for given one specific x we can find such $$x_{i}'s$$ in P such that $$x = sum_{i \in I} f_{i}(x)x_{i}$$. But how do i know, that for other element $$y \in P$$, $$y = \sum_{i \in I} f_{i}(y)x_{i}$$? I dont get that. Should that dual basis be related with basis of free module F?

• "On NCatlab these generators $x_i$ are taken from nowhere (as it often happens on ncatlab)." No, a free module has associated with it a set of generators (how else do you know it's free?); those generators are the $x_i$. Dec 14, 2021 at 20:20

Not every submodule of a free module is a direct summand! A submodule being a direct summand is equivalent to the natural inclusion map having a retraction. The retraction is key in establishing the dual basis result.

By definition, a submodule $$P$$ of a module $$F$$ being a direct summand means that there is also a submodule $$Q$$ such that every $$f\in F$$ can be written uniquely as $$p+q$$ for $$p\in P,q\in Q$$. In particular, we can define $$r\colon F\to P$$ by $$r(f)=p\in P$$ where $$p$$ is the element such that $$f=p+q$$. Moreover, uniqueness guarantees this is a linear map, and that if $$s\colon P\hookrightarrow F$$ is the natural inclusion map, then $$r\circ s=\mathrm{id}_P$$, i.e. that $$r\colon F\to P$$ is a retraction of $$s\colon P\to F$$.

To see that not every submodule of a free module is a direct summand, note that $$e=r\circ s\colon P\to P$$ is idempotent, i.e. satisfies $$e\circ e=(s\circ r)\circ(s\circ r)=s\circ(r\circ s)\circ r=s\circ\mathrm{id}_N\circ r=s\circ r=e$$. Moreover, we have that $$P$$, as a direct summand of $$F$$, is the image of $$e$$, which is also given by $$\{f\in F:e(f)=f\}$$. Thus direct summands determine idempotents on $$F$$.

Suppose now that $$F$$ is a free module of rank $$1$$. Then submodules of $$F$$ are ideals of the ring $$R$$, endomorphisms of $$F$$ simply multiplications by elements of the ring $$R$$, and the idempotents are those of $$R$$. Therefore any ideal that is not a principal ideal generated by an idempotent is not a direct summand.

Exercise: show that every map with a retraction determines a direct summand.

Now, to understand the relationship with dual basis, let $$FB$$ be a free module with basis $$B$$. Then $$P$$ being a direct summand implies we have $$s\colon P\to FS$$ with a retraction $$r\colon FI\to P$$ such that $$r\circ s=\mathrm{id}_P$$. In particular, the retraction is surjective, and so $$r(I)=\{x_i\}\subseteq P$$ is a generating family for $$P$$ because $$I$$ is a generating set for $$FI$$. Moreover, the coordinate functions $$f_i\colon FB\to R$$ such that $$x=(f_i(x))$$ then have to satisfy $$r(x)=\sum f_i(x)x_i$$, and we obtain coordinate function $$f_i\circ s\colon P\to R$$ such that $$p=r\circ s(p)=\sum f_i\circ s(p)x_i$$.

• It seems like I proved only that a module with dual basis is a submodule of a free module, but not that is direct summand. I dont know how to prove that. Dec 19, 2021 at 19:20