Proposition 8

Tag 0DU7, permanent. Cite it as [Bourbaki, Tag 0DU7].

Let $M$ be an $A$ -module, $P$ and $P'$ two minimal projective resolutions of $M$ , and $f:P\to P'$ a morphism of resolutions. Then $f$ is an isomorphism. In particular, two minimal projective resolutions of $M$ are isomorphic.

Put $\widetilde P_n=P_n$ for $n\ne-1$ and $\widetilde P_{-1}=M$ ; define $\widetilde P'_n$ analogously and put $f_{-1}=1_M$ . We shall prove by induction starting from $-1$ that $f_n:\widetilde P_n\to\widetilde P'_n$ is an isomorphism for every $n$ . This is evident for $n=-1$ ; suppose that $f_n$ and $f_{n-1}$ are isomorphisms. It follows from the commutativity of the diagram

\begin{array}{ccc} P_n & \xrightarrow{\ d_n\ } & P_{n-1}\\ {\scriptstyle f_n}\downarrow && \downarrow{\scriptstyle f_{n-1}}\\ P'_n & \xrightarrow{\ d'_n\ } & P'_{n-1} \end{array}

that $f_n$ induces an isomorphism $g_n$ of $\operatorname{Ker}d_n$ onto $\operatorname{Ker}d'_n$ . It then follows from the commutativity of the diagram

\begin{array}{ccc} P_{n+1} & \xrightarrow{\ \delta_{n+1}\ } & \operatorname{Ker}d_n\\ {\scriptstyle f_{n+1}}\downarrow && \downarrow{\scriptstyle g_n}\\ P'_{n+1} & \xrightarrow{\ \delta'_{n+1}\ } & \operatorname{Ker}d'_n \end{array}

and from VIII, loc. cit., that $f_{n+1}$ is an isomorphism.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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Cites

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