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A dual basis for a left $R$-module $M$ consists of elements $m_1, \dots, m_r \in M$ and homomorphisms $f_1, \dots, f_r: M \to R$ such that every $m \in M$ satisfies
\[
m = \sum_{i=1}^r f_i(m) \, m_i.
\]
This reconstructs any element of $M$ using "coordinates" given by the $f_i$. (Note: For possibly infinite index sets with the finite support condition—i.e., $f_i(m) \neq 0$ for only finitely many $i$—the property characterizes projective modules that are not necessarily finitely generated.) If $M$ is not projective, like $\mathbb{Z}/2\mathbb{Z}$ over $\mathbb{Z}$, no such dual basis exists.
Why This Characterizes Finitely Generated Projective Modules
Projective modules are those that lift homomorphisms through surjections, equivalently direct summands of free modules. The finite dual basis encodes this structure explicitly.
Proof: Existence of Finite Dual Basis Implies $M$ Is Finitely Generated Projective
Suppose such a dual basis exists.
Let $F = R^r$ be the free left $R$-module of rank $r$, with standard basis $\{e_1, \dots, e_r\}$.
Define the surjective homomorphism $p: F \to M$ by $p\left(\sum_{i=1}^r r_i e_i\right) = \sum_{i=1}^r r_i m_i$. (This is onto because the $m_i$ generate $M$, as every $m$ is an $R$-linear combination of them.)
Define the homomorphism $s: M \to F$ by $s(m) = \sum_{i=1}^r f_i(m) e_i$.
Then $p \circ s = \text{id}_M$, since $p(s(m)) = \sum_{i=1}^r f_i(m) m_i = m$.
This shows $0 \to \ker p \to F \xrightarrow{p} M \to 0$ is a split exact sequence, so $M$ is a direct summand of the free module $F$ (specifically, $F \cong M \oplus \ker p$).
Thus, $M$ is projective. Since $r$ is finite, $M$ is finitely generated (generated by $\{m_1, \dots, m_r\}$).
Proof: Finitely Generated Projective Implies Existence of Finite Dual Basis
Suppose $M$ is a finitely generated projective left $R$-module.
Since $M$ is finitely generated, there exists a surjective homomorphism $p: F \to M$ where $F = R^n$ is free of finite rank $n$ (e.g., take generators of $M$ and map the basis of $F$ to them).
Since $M$ is projective, the sequence $0 \to \ker p \to F \xrightarrow{p} M \to 0$ splits: there exists a homomorphism $s: M \to F$ such that $p \circ s = \text{id}_M$.
Let $\{e_1, \dots, e_n\}$ be the standard basis of $F$.
Set $m_i = p(e_i) \in M$ for $i = 1, \dots, n$.
For each $m \in M$, we have $s(m) \in F$, so $s(m) = \sum_{i=1}^n r_i e_i$ for unique $r_i \in R$. Define $f_i: M \to R$ by $f_i(m) = r_i$; these are $R$-linear (as $s$ is $R$-linear, and the coefficients depend linearly on $s(m)$).
Then $m = p(s(m)) = p\left(\sum_{i=1}^n f_i(m) e_i\right) = \sum_{i=1}^n f_i(m) p(e_i) = \sum_{i=1}^n f_i(m) m_i$.
Thus, $\{(m_i, f_i)\}_{i=1}^n$ is the desired finite dual basis (with $r = n$). |
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