I came across the following problem in Rotman's Advanced Modern Algebra:
11.69. If $\{M_\alpha\}_{\alpha\in A}$ is a family of left R-modules, prove that pd$(\sum\limits_{\alpha\in A}M_\alpha)=\sup\limits_{\alpha\in A}\{$pd$(M_\alpha)\}$, where pd denotes the projective dimension.
For now I am just looking at the case where pd$(\sum\limits_{\alpha\in A}M_\alpha)<\infty$. Here is my attempt:
Let
$0\rightarrow P_n\rightarrow ... \rightarrow P_1\rightarrow P_0\xrightarrow{\epsilon}\sum\limits_{\alpha\in A}M_\alpha\rightarrow 0$
be a projective resolution of minimal length. Then for each $\alpha$,
$0\rightarrow P_n\rightarrow ... \rightarrow P_1\rightarrow P_0\xrightarrow{p_\alpha\epsilon}M_\alpha\rightarrow 0$
is a projective resolution where $p_\alpha$ is the appropriate projection. Hence pd$M_\alpha\leq$ pd $(\sum\limits_{\alpha\in A}M_\alpha), \thinspace \forall\alpha\in A$. So pd $(\sum\limits_{\alpha\in A}M_\alpha)$ is an upper bound for $\{$pd$ M_\alpha\}_{\alpha\in A}$.
But I still need to show that it is the least upper bound. Any hints on how to show this last part as well as any corrections to my proof of the first part are appreciated.
