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?

  • 1
    $\begingroup$ "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 $x_i$. $\endgroup$ Dec 14, 2021 at 20:20

1 Answer 1


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 $p=r\circ s(p)=\sum f_i\circ s(p)x_i$.

  • $\begingroup$ 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. $\endgroup$
    – robin3210
    Dec 19, 2021 at 19:20

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