Direct Sum of Projective and Free Modules In Kevin Ventullo's answer on Mathoverflow it is mentioned that if $F$ is a free module on infinitely many generators and $P$ is a projective module, then $P \oplus F$ is again free.
However, I cannot see why this should be true.
Could you please show me why this holds?
Thank you very much for your help!
 A: I wasn't aware of this result before, and I believe as you have stated it it's false, but for some specific pairs $P$, $F$ it makes sense using the fact that a projective is a summand of a free.  Specifically, let $F$ be free on $\kappa$ generators, where $\kappa$ is infinite, and suppose also that $P\oplus P' = F'$, where $F'$ is free on a set of generators with cardinality $\kappa' \leq \kappa$.  Then because $\kappa' \leq \kappa$ and $\kappa$ is infinite, we may write $F$ as an infinite sum of copies of $F'$:
\begin{align*}
 F = \bigoplus_iF' = \bigoplus_i(P\oplus P') = (\bigoplus_iP)\oplus(\bigoplus_i P')
\end{align*}
so that
\begin{align*}
 P\oplus F = P\oplus\left((\bigoplus_iP)\oplus(\bigoplus_i P')\right) = \left(P\oplus\bigoplus_iP\right)\oplus\bigoplus_i P' = (\bigoplus_iP)\oplus(\bigoplus_i P') = F.
\end{align*}
This trick doesn't work if $\kappa' > \kappa$, and I would guess that the result is actually false in that case.
A: I think the proof is not well written.
What the proof needs is a free module $F_0$ such that $P_i$ is a direct summand of $F_0$, for $i=0,\dots,n$: just take $Q_i$ with $P_i\oplus Q_i$ free and form $F_0=\bigoplus_{i=0}^n(P_i\oplus Q_i)$.
Then show that, if $F=F_0^{(\omega)}$ (a direct sum of a countable number of copies of $F_0$) and $P$ is a direct summand of $F_0$, then $P\oplus F$ is free: this is easy because, if $F_0=Q\oplus P$, we just need to formalize with suitable maps the idea that
$$
P\oplus (Q\oplus P)\oplus(Q\oplus P)\oplus\dotsb=
(P\oplus Q)\oplus(P\oplus Q)\oplus\dotsb
$$
Hence $P\oplus F\cong F$ is free.
