Rings with only finitely many nilpotent elements are Dedekind-finite

Posted: February 14, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , ,

For the definition and some examples of Dedekind-finite rings see this post. We will assume that R is a ring with 1.

Notation. For a,b \in R and integers i,j \geq 1 we let e_{ij}=b^{i-1}a^{j-1} - b^ia^j.

The following shows that if ab=1, then e_{ij} behave like matrix units!

Problem 1. If ab = 1, then e_{ij}e_{k \ell} = \delta_{jk}e_{i \ell}, for all i,j,k, \ell \geq 1.

Solution. Since ab=1, an easy induction shows that for every integers r,s \geq 1 we have

a^rb^s = \begin{cases} a^{r-s} & \text{if} \ r \geq s \\ b^{s-r} & \text{if} \ s> r. \end{cases} \ \ \ \ \ \ \ \ \ \ \ \ (1)

We also have

e_{ij}e_{k \ell} = b^{i-1}(a^{j-1}b^{k-1})a^{\ell -1} - b^{i-1}(a^{j-1}b^k)a^{\ell} - b^i(a^jb^{k-1})a^{\ell - 1} + b^i(a^jb^k)a^{\ell}. \ \ \ \ \ \ \ \ \ \ \ \ (2)

Now apply (1) to the terms in brackets on the right hand side of (2) to finish the proof. You will see very quickly that the right hand side of (2) is 0 if j<k or j > k and it is e_{i \ell} if j=k. \ \Box

Problem 2. Prove that a ring R which is not Dedekind-finite has infinitely many nilpotent elements.

Solution. Since R is not Dedekind-finite, there exist a,b \in R such that ab=1 and ba \neq 1. Now consider the set A = \{e_{1n} : \ \ n=2,3,4, \cdots \}. By Problem 1, we have e_{1n}^2 = 0. So every element of A is nilpotent. We now need to show that these elements are pairwise distinct and so A is an infinite set of nilpotent elements of R. So suppose, to the contrary, that e_{1n}=e_{1m}, for some distinct integers n,m \geq 2. Then by Problem 1,

1-ba=e_{11}=e_{1m}e_{m1}=e_{1n}e_{m1}=0,

contradicting ba \neq 1. \ \Box

Example. A trivial result of Problem 2 is that every finite ring is Dedekind-finite. See also Example 3 in here.

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