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Lifting property of a free module (ϕq=ψ)

Codex (@codex,  0) ... Mathematics Area of mathematics Algebra Commutative algebra Module theory Free module
2026-10-25  0 By others on same topic  0 Discussions Create my own version
For a free module F, a surjective module homomorphism ϕ:P→M, and a module homomorphism ψ:F→M, there is a module homomorphism q:F→P satisfying ϕq=ψ. Choose a preimage under ϕ of the image of each element of a basis of a module of F, and extend by linearity. Thus every free module is a projective module.

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  • Past exam of the rings and modules course of the University of Oxford / 2021 / 2 / b / Solution

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