Suppose $R$ is a ring in which every left invertible element is invertible. Does this condition imply that every left invertible matrix in $\mathrm{M}_{n\times n}(R)$ is necessarily invertible?

  • 2
    $\begingroup$ No. This is Exercise 1.18 in Lam, Exercises in Modules and Rings, and has a solution there. (Btw, the "rings with some special property" are called Dedekind-finite. See also here.) $\endgroup$
    – user26857
    Jan 6 '15 at 11:38
  • $\begingroup$ I think Lam even points out that erroneous proofs that Dedekind finiteness passes to matrix rings exist in the literature, so be careful, @mostafa. $\endgroup$
    – rschwieb
    Jan 6 '15 at 11:55

(As user26857 brought up in his comment) a ring for which $xy=1$ implies $yx=1$ is called Dedekind finite or directly finite. This is readily confirmed to be equivalent to saying "every left invertible element is invertible."

Exercise 1.18 in Lam's Lectures on modules and rings is to show there exists a Dedekind finite ring $R$ such that $M_2(R)$ is not Dedekind finite. A solution for it appears in the solution book Exercises in modules and rings on page 10.

Without verifying all the details, the construction goes like this. Let $w,x,y,z,s,t,u,v$ be symbols, and let $\begin{bmatrix}s&u\\t&v\end{bmatrix}\begin{bmatrix}x&y\\z&w\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ under formal matrix multiplication. This provides us with four relations between the symbols, and then we look at the ring $R$ which is the quotient of the free $k$-algebra on these eight symbols subject to those relations.

To finish, it's shown that the product of the matrices in the reverse order does not make the identity (so $M_2(R)$ isn't Dedekind finite), and that $R$ is a domain (so it is Dedekind finite.)

If you'd rather learn directly from the original source, Lam credits it to this article:

Shepherdson, J. C. Inverses and zero divisors in matrix rings. Proc. London Math. Soc. (3) 1, (1951). 71--85. MR0041831 (13,7i)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.