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A finitely generated projective module is stably free if $P \oplus R^m \cong R^n$ for some $m,n$. Show that every stably free $R$-module is free iff every unimodular row over $R$ can be completed to a non-singular matrix.

Now, what I know is that unimodular row is essentially row of some unimodular matrix, which is of determinant 1, which means the unimodular row can be completed to a non-singular matrix weather or not the given condition satisfies. So it will be great if anyone can explain it to me, am I reading the definitions of unimodular row wrong or the question is wrong. Thanks.

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  • $\begingroup$ See books by say Lang (Algebra), T.Y. Lam (old lecture notes and later text book) or K-theory by Bass, to cite a few, for definition of unimodular row (explained by Eric Wofsey) and much more. $\endgroup$
    – Mohan
    Oct 31 '19 at 14:13
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Your definition of unimodular row is totally wrong. Rather, a unimodular row is just a row whose entries generate the unit ideal in $R$. That is, $(a_1,\dots,a_n)\in R^n$ is unimodular if there exist $b_1,\dots,b_n\in R$ such that $\sum b_ia_i=1$.

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  • $\begingroup$ Ohhh okay okay. Can you send me any websites or books where I can read about this. I can't find about unimodular row in my algebra books that's why. $\endgroup$
    – user631697
    Oct 31 '19 at 6:19
  • $\begingroup$ I'm afraid I don't know much about it; I just learned the definition long ago in a course that used it to prove the Quillen-Suslin theorem. $\endgroup$ Oct 31 '19 at 6:26

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