So Im having some trouble in solving projective module problems. I will give an example which extends to the other problems I also wasn´t also to solve.

Let A be a ring, $P$ a left $A$-module. We call a dual basis for $P$ to the family $\{(x_i,f_i)\}_{i \in I}$ where $(x_i,f_i)\in P \times P^*$ for every $i\in I$ such that

i) for every $x\in P$, the set $\{i \in I:f_i(x) \neq 0\}$ is finite

ii) for every $x\in P$, $x=\sum_{i \in I}f_i(x)x_i$

Show that $P$ is projective if and only if it has a dual basis.

So I started by trying to prove that if $P$ has a dual basis, then $P$ is projective. So let $f:T \to N$ be a surjective $A$-module homomorphism and $g: P \to N$ an $A$-module homomorphism. I want to see that there exist an $A$-module homomorphism $\overline{g}:P \to T$ such that $f \circ \overline{g} = g$.

So Im thinking about the following, consider, for every $x \in P$, an element $t_x \in T$ such that $g(x) = f(t_x)$. This element exists since $f$ is surjective. Now the problem in defining $\overline{g}:P \to T$ as $\overline{g}(x)=t_x$ is the good definition, since there can be different $t_x$ such that $f(t_x)=g(x)$. My problem is that im not being able to deal with this, and further more, I don't understand why in some cases (for example when prooving that $\Bbb{Z}$ is a projective $\Bbb{Z}$-module, the uniqueness of the homomorphism leaving the module implies good definition).

Can anyone help me get me out of this confution?

  • $\begingroup$ It's helpful to use other criteria for a module to be projective, such as the one in terms of exact sequences splitting, or being a direct summand of a free module. $\endgroup$ – Ben West Jun 25 '17 at 23:59

(Sorry, I wrote this with $\varphi_i$ instead of $f_i$, and $a_i$ instead of $x_i$.)

First, suppose $P$ is projective. As with any module, $P$ is a quotient of a free $A$-module $F$, so there is a surjection $$ 0\to{\ker\psi}\to F\xrightarrow{\psi}{P}\to 0 $$ and since $P$ is projective, this sequence is split, with splitting $\varphi\colon P\to F$ such that $\psi\varphi=1_P$. Let $\{e_i\}$ a basis of $F$, and denote $\psi(e_i)=a_i$. Also, for $x\in P$, we have a unique expression $\varphi(x)=\sum_i r_ie_i$ for some $r_i\in A$. Let $\varphi_i$ be the coordinate maps $\varphi_i(x)=r_i$, which are $A$-maps. It follows $\varphi_i(x)=0$ for almost all $i$, since $\varphi(x)$ is a finite linear combination by the definition of a basis. For the second property, $$ x=1_P(x)=\psi\varphi(x)=\psi\left(\sum_i r_ie_i\right)=\sum_i r_i\psi(e_i)=\sum_i\varphi_i(x)a_i. $$ Hence the set $\{a_i\}$ is a generating set for $P$.

Conversely, suppose the elements $a_i$ and maps $\varphi_i$ exist. To show $P$ is projective, prove it is a direct summand of a free module. By property (ii), $\{a_i\}_{i\in I}$ is a generating set of $P$. Let $F$ be a free module with basis $\{e_i\}_{i\in I}$, and define a surjection $\psi\colon F\to P$ via $\psi(e_i)=a_i$. Define $\varphi\colon P\to F$ by $\varphi(x)=\sum_i \varphi_i(x)e_i$. This sum is finite, hence $\varphi$ is well-defined, by property (i), and is easily checked to be an $A$-map. Compute $$ \psi\varphi(x)=\psi\left(\sum_i\varphi_i(x)e_i\right)=\sum_i\varphi_i(x)a_i=x $$ so that $\psi\varphi=1_P$, and hence $P$ is a direct summand of $F$.


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