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Hi. I have read that stably free modules not finitely generated are free; this is proved in M.R. Gabel, stably free projectives over commutative rings, Thesis, Brandeis Univ., Waltham, MA 1972.

But I can't find a proof of that, can anyone of you help me?

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How do you define "stably free"? By $P\oplus R^m\cong R^{\left(X\right)}$ for some (possibly infinite) set $X$ but finite $m$ ? Then see exercise 1.9 of . – darij grinberg Jan 27 '11 at 0:28
(Note that Weibel considers only the case $\left|X\right|=\Aleph_0$, but the proof works for all cases. I am rather unhappy with the hint given in Weibel, however; it could be written up in a better way.) – darij grinberg Jan 27 '11 at 0:34
Ah, this is better: Thm. A.1. – darij grinberg Jan 27 '11 at 0:36
thanks, however the definition I know is: a stably free module is a finitely generated module $P$ such that there exist finitely generated free modules $F$ and $F^\prime$ such that $P\oplus F\cong F^\prime$ – Aaron Bennet Jan 27 '11 at 1:09
With this definition all stably free modules are finitely generated. – Torsten Ekedahl Jan 27 '11 at 5:21

Let $P\oplus R^k$ be free. By induction it's enough to take $k=1$, so the assertion comes down to saying that an infinitely long unimodular row can be completed to an infinite invertible matrix. But an infinitely long unimodular row contains (after rearrangement) a finitely long unimodular sub-row. So it suffices to show that any unimodular row with a non-trivial initial unimodular sub-row can be completed.

Write the row as (s,x,t) where s is is unimodular and x has length 1. Let p be a column such that s.p = 1. Consider the matrix

$$\matrix{s&x&t\cr I&p&0\cr 0&0&I\cr}$$

where the $I$'s and $0$'s are identity and 0 matrices of approrpiate sizes. This is easily seen to be invertible.

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(Of course these "rows" need not be countable.) – Steven Landsburg Jan 27 '11 at 7:50
"so the assertion comes down to saying that an infinitely long unimodular row can be completed to an infinite invertible matrix." I don't understand this very first step. Can you explain it? – Martin Brandenburg Jan 27 '11 at 8:07
Martin: We are given an isomorphism $R^M\rightarrow R\oplus P$ (with $M$ possibly infinite). The restriction $R^M\rightarrow R$ is represented by a unimodular row. If $P$ is free, then the isomporphism itself is represented by an invertible matrix with that row as its first row. Is there a problem with this that I'm failing to see? – Steven Landsburg Jan 27 '11 at 14:13

The proper definition of 'stably free' is as follows: let $\Lambda$ be a ring. Then a $\Lambda$-module $S$ is stably free when $S\oplus\Lambda^n$ is a free module of unspecified rank, finite or infinite, where $n$ is an integer. The reason that $n$ must be finite is to avoid Eilenberg's trick which shows that $P\oplus \Lambda^\infty \cong \Lambda^\infty$ for any countably generated projective module $P$. Here $\Lambda^\infty$ denotes the direct limit ${\lim}_{\to}\Lambda^n$ where $\Lambda^n\subset\Lambda^n\oplus\Lambda\cong\Lambda^{n+1}$ under the inclusion $x\mapsto (x,0)$.

Gabel's theorem actually holds for non-commutative rings also. There is a proof in T.Y. Lam, Serre's conjecture. Although Gabel's theorem does not hold for projective modules, a theorem of Kaplansky [Projective modules, Ann. of math, 68:2 (1958) 372 - 377] says that every projective module is a direct sum of countably generated projective modules, and so even infinitely generated projectives cannot be 'too bad'.

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I guess Eilenburg is a typo. – HenrikRüping Jan 27 '11 at 13:04
Ah yes. Not strictly a typo; I'm just a terrible speller. – Seamus Jan 27 '11 at 13:14
The Eilenberg Swindle shows that for any projective module $P$, there is a free module $F$ with $P \oplus F \cong F$. – Pete L. Clark Apr 13 '13 at 0:30
up vote 1 down vote accepted

I think that the proof contained in (link given by darij grinberg) it's the best proof. Thanks to all

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