## Rings with only finitely many nilpotent elements are Dedekind-finite

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

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Definition 1. A ring $R$ is called Dedekind-finite if $\forall a,b \in R: \ ab=1 \Longrightarrow ba=1.$

Remark 1. Some trivial examples of Dedekind-finite rings: commutative rings, any direct product of Dedekind-finite rings, any subring of a Dedekind-finite ring.

Definition 2. A ring $R$ is called reversible if $\forall a,b \in R : \ ab = 0 \Longrightarrow ba = 0.$

Example 1. Every reversible ring $R$ is Dedekind-finite. In particular, reduced rings are Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Then $(ba-1)b=b(ab)-b=0$ and thus $b(ba-1)=0.$ So $b^2a=b$ and hence $ab^2a=ab=1.$ It follows that $ba=(ab^2a)ba=(ab^2)(ab)a=ab^2a=1.$ So $R$ is Dedekind-finite. Finally, note that every reduced ring is reversible because if $ab=0,$ for some $a,b \in R,$ then $(ba)^2=b(ab)a=0$ and thus $ba=0. \Box$

Example 2. Every (left or right) Noetherian ring $R$ is Dedekind-finite.

Proof. We will assume that $R$ is left Noetherian. Suppose that $ab=1$ for some $a,b \in R.$ Define the map $f: R \longrightarrow R$ by $f(r)=rb.$ Clearly $f$ is an $R$-module homomorphism and $f$ is onto because $f(ra)=(ra)b=r(ab)=r,$ for all $r \in R.$ Now we have an ascending chain of left ideals of $R$

$\ker f \subseteq \ker f^2 \subseteq \cdots.$

Since $R$ is left Noetherian, this chain stabilizes at some point, i.e. there exists some $n$ such that $\ker f^n = \ker f^{n+1}.$ Clearly $f^n$ is onto because $f$ is onto. Thus $f^n(c)=ba-1$ for some $c \in R.$ Then

$f^{n+1}(c)=f(ba-1)=(ba-1)b=b(ab)-b=0.$

Hence $c \in \ker f^{n+1}=\ker f^n$ and therefore $ba-1=f^n (c) = 0. \Box$

Example 3. Finite rings are obviously Noetherian and so Dedekind-finite by Example 2. More generally:

Example 4. If the number of nilpotent elements of a ring is finite, then the ring is Dedekind-finite. See here.

Note that Example 4 implies that every reduced ring is Dedekind-finite; a fact that we proved in Example 1.

Example 5. Let $k$ be a field and let $R$ be a finite dimensional $k$-algebra. Then $R$ is Dedekind-finite.

Proof. Every left ideal of $R$ is clearly a $k$-vector subspace of $R$ and thus, since $\dim_k R < \infty,$ any ascending chain of left ideals of $R$ will stop at some point. So $R$ is left Noetherian and thus, by Example 2, $R$ is Dedekind-finite. $\Box$

Remark 2. Two important cases of Example 5 are $M_n(R),$ the ring of $n \times n$ matrices over a field, and, in general, semisimple rings. As a trivial result, $M_n(R)$ is Dedekind-finite for any commutative domain $R$ because $M_n(R)$ is a subring of $M_n(Q(R))$, where $Q(R)$ is the quotient field of $R$.
So the ring of $n \times n$ matrices, where $n \geq 2,$ over a field is an example of a Dedekind-finite ring which is not reversible, i.e. the converse of Example 1 is not true. Now let $R_i = \mathbb{Z}, \ i \geq 1.$ Then $R= \prod_{i=1}^{\infty} R_i$ is obviously Dedekind-finite but not Noetherian. So the converse of Example 2 is not true.

Example 6 and Example 7 are two generalizations of Example 5.

Example 6. Every algebraic algebra $R$ over a field $k$ is Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Since $R$ is algebraic over $k,$ there exist integers $n \geq m \geq 0$ and some $\alpha_i \in k$ with $\alpha_n \alpha_m \neq 0$ such that $\sum_{i=m}^n \alpha_i b^i = 0.$ We will assume that $n$ is as small as possible. Suppose that $m \geq 1.$ Then, since $ab=1,$ we have

$\sum_{i=m}^n \alpha_i b^{i-1}=a \sum_{i=m}^n \alpha_i b^i = 0,$

which contradicts the minimality of $n.$ So $m = 0.$ Let $c = -\alpha_0^{-1}\sum_{i=1}^n \alpha_i b^{i-1}$ and see that $bc=cb=1.$ But then $a=a(bc)=(ab)c=c$ and therefore $ba=bc=1. \ \Box$

Remark 3. Regarding Examples 5 and 6, note that although any finite dimensional $k$-algebra $R$ is algebraic over $k,$ but $R$ being algebraic over $k$ does not necessarily imply that $R$ is finite dimensional over $k.$ For example, if $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ in $\mathbb{C},$ then it is easily seen that $\dim_{\mathbb{Q}} \overline{\mathbb{Q}}=\infty.$ Thus the matrix ring $R = M_n(\overline{\mathbb{Q}})$ is an algebraic $\mathbb{Q}$-algebra which is not finite dimensional over $\mathbb{Q}.$ So, as a $\mathbb{Q}$-algebra, $R$ is Dedekind-finite by  Example 6 not Example 5.

Example 7. Every PI-algebra $R$ is Dedekind-finite.

Proof. Let $J(R)$ be the Jacobson radical of $R.$ If $J(R)=\{0\},$ then $R$ is a subdirect product of primitive algebras $R/P_i,$ where $P_i$ are the primitive ideals of $R.$ Since $R$ is PI, each $R/P_i$ is PI too and thus, by Kaplansky’s theorem, $R/P_i$ is a matrix ring over some division algebra and thus Dedekind-finite by Example 2. Thus $\prod R/P_i$ is Dedekind-finite and so $R,$ which is a subalgebra of $\prod R/P_i,$ is also Dedekind-finite. For the general case, let $S=R/J(R).$ Now, $S$ is PI, because $R$ is PI, and $J(S)=\{0\}.$ Therefore, by what we just proved, $S$ is Dedekind-finite. Suppose that $ab = 1$ for some $a,b \in R$ and let $c,d$ be the image of $a,b$ in $S$ respectively. Clearly $cd=1_S$ and so $dc=1_S.$ Thus $1-ba \in J(R)$ and so $ba=1-(1-ba)$ is invertible. Hence there exists $e \in R$ such that $e(ba)=1.$ But then $eb=(eb)ab=e(ba)b=b$ and hence $ba=(eb)a=e(ba)=1. \Box$