Posts Tagged ‘Dedekind-finite’

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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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