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Let $R$ be a commutative ring with unit, and let $A, B$ be $R$-modules. Into the book "A Course in Homological Algebra" of Peter J. Hilton and U. Stammbach, the authors present the functor $Ext_R(A, B)$, by starting from a projective presentation of $A$, i.e. a short exact sequence of the type $0 \rightarrow R \rightarrow P \rightarrow A \rightarrow 0$, where $P$ is a projective module. From this, it would seem that such an exact sequence always exists for every module $A$. Could someone explain to me why?

Thank you!

Rick88
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  • If $R$ does not need to satisfy any special properties, then this sequence can be constructed by just taking a surjection $\varphi \colon F \to A$, where $F$ is a free module $R$-module, and setting $R = \ker(\varphi)$. Such a surjection always exists: just take $F$ to be the free $R$-module on the set $A$, and let $\varphi \colon F \to A$ be the unique $R$-linear map sending the basis element $a \in F$ to $a \in A$. – Alex Wertheim Apr 26 '21 at 09:40
  • It is clear, thank you! – Rick88 Apr 26 '21 at 15:06

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