## 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 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.