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Let $R$ be a (possibly non-commutative) ring. The left stable range of $R$ (denoted $sr_l(R)$) is the smallest $n$ such that every left unimodular row of length $>n$ is reducible. A similar definition exists for right stable range and it has been shown that the left stable range equals the right stable range so we refer simply to the stable range of $R$ (denoted $sr(R)$).

A ring is said to be Hermite (resp. n-Hermite) if any stably free module (resp. rank $n$ stably free module) is free. I've heard that if a ring $R$ has stable range 1, then it is Hermite (although I can't find a reference to this). I was wondering if the following is true for an arbitrary ring:

If $sr(R)=2$, then $R$ is $2$-Hermite.

Is this true? If so, why?

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    $\begingroup$ Exercise I.1.5 of Weibel's K-book shows that $sr(R)=n$ implies that the ring is $n$-Hermite. The essential idea is that reducibility of unimodular rows implies that they can be completed to invertible matrices and completeability of the unimodular row implies freeness of the stably free module associated to the unimodular row. $\endgroup$
    – Matthias Wendt
    Commented Dec 19, 2016 at 21:07
  • $\begingroup$ crossposted $\endgroup$
    – rschwieb
    Commented Dec 20, 2016 at 4:22

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