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Hi everybody,

recall that a ring $A$ with unit has the invariant basis property if all bases of a given f.g. free $A$-module have the same number of elements. It is equivalent to say that every invertible matrix with entries in $A$ is a square matrix.

It is known that if for all $n\geq 1,$ every surjective endomorphism of $A^n$ is bijective, then $A$ has the invariant basis property.

Q1: is the converse true? Another way to put this is: if $A$ has the invariant basis property, and if $e_1,...,e_n$ span $A^n$, is it a basis of $A^n$?

Q2: Assume that $A$ has the invariant basis property. Is it true that if $e_1,....,e_m\in A^n$ are linearly independent, then $m\leq n$ ?

Q3: Assume that $A$ has the invariant basis property. Is it true that if $e_1,....,e_m\in A^n$ span $A^n$, then $m\geq n$ ?

Thanks!

greg

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  • $\begingroup$ I am pretty sure that the answer to Q1 is no, according to en.wikipedia.org/wiki/Stably_finite_ring . This is not a counterexample, though... $\endgroup$
    – darij grinberg
    Commented May 18, 2011 at 16:22
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    $\begingroup$ A counterexample can apparently be found in: P. M. Cohn, "Some remarks on the invariant basis property", Topology 5 (1966), 215-228. I do not have this paper. $\endgroup$
    – darij grinberg
    Commented May 18, 2011 at 16:29
  • $\begingroup$ Thanks for the reference. Apparently, all my questions are answered (by the negative) on this paper. Thanks a lot! $\endgroup$
    – GreginGre
    Commented May 18, 2011 at 17:25
  • $\begingroup$ Mmmmh, finally, I think that the counterexamples in this paper do NOT satisfy IBN, but I will have to read it more carefully. $\endgroup$
    – GreginGre
    Commented May 18, 2011 at 17:47
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    $\begingroup$ Counterexample to Q2: Let $A$ be a noncommutative polynomial ring over $\mathbb Z$ in $X$ and $Y$, i.e. the "group ring" of a free monoid on two generators. The left ideal (submodule of $A$) generated by $X$ and $Y$ is isomorphic to $A^2$, but $A$ has the IBP because it admits a unital homomorphism to $\mathbb Z$. $\endgroup$
    – Tom Goodwillie
    Commented May 18, 2011 at 22:25

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