Archive for the ‘Examples & Counter-Examples’ Category

Let R be a ring. Recall that an additive map \delta : R \longrightarrow R is called a derivation if

\delta(r_1r_2)=\delta(r_1)r_2 + r_1 \delta(r_2)

for all r_1,r_2 \in R. Thus if \delta is a derivation of R, then \delta(r^2)=\delta(r)r + r \delta(r) for all r \in R.

Definition. Let R be a ring. An additive map \delta : R \longrightarrow R is called a Jordan derivation if


for all r \in R.

So every derivation is a Jordan derivation . But the converse is not true and that is not surprising.

Example. Let S=\mathbb{C}[x] with the relation x^2=0. Let I=\mathbb{C}x, which is an ideal of S because x^2=0. Let

R = \begin{pmatrix} S & S \\ I & S \end{pmatrix}.

See that R, with matrix addition and multiplication, is a ring because I is an ideal of S. For any r = \begin{pmatrix}a & b \\ c & d \end{pmatrix} \in R define \delta(r)=\begin{pmatrix}0 & c \\ 0 & 0 \end{pmatrix}. Then \delta : R \longrightarrow R is a Jordan derivation but not a derivation.

Proof. It is obvious that \delta is additive. Let r = \begin{pmatrix}a & b \\ c & d \end{pmatrix} \in R. Then, since c^2=0, we have

\delta(r^2)=\delta(r)r+r \delta(r) = \begin{pmatrix}0 & (a+d)c \\ 0 & 0 \end{pmatrix}

and so \delta is a Jordan derivation. To see why \delta is not a derivation, let r_1=\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} and r_2=\begin{pmatrix}0 & 0 \\ x & 0 \end{pmatrix}. Then \delta(r_1r_2)=0_R but \delta(r_1)r_2 + r_1 \delta(r_2)=\begin{pmatrix}0 & x \\ 0 & 0 \end{pmatrix}. \ \Box


Is it true that in any ring R with 1, if ab=1 for some a,b \in R, then ba=1 ?
No, that is not true and here’s an example. Let V be a (countably) infinite dimensional vector space over some field F and let \{v_1,v_2, \cdots \} be a basis for V. Now consider R:=\text{End}_F(V), the ring of F-linear transformations of V. Define a,b \in R by a(v_1)=0, \ a(v_j)=v_{j-1}, \ j \ge 2, and b(v_j)=v_{j+1}, \ j \ge 1. See that ab=1 but ba \ne 1 because ba(v_1)=0.

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


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

Theorem. Let R be a ring and a \in R. Let Z(R) be the center of R. If Z(R) is reduced and ar - ra \in Z(R) for all r \in R, then a \in Z(R).

Proof. Let r \in R. Then


So (ar-ra)^2=0. Hence ar = ra because Z(R) is reduced.  Thus a \in Z(R). \ \Box

One class of rings with reduced centers is the class of semiprime rings. If Z(R) is not reduced, the result in the theorem need not hold. There is a nice example in Lam’s book, “A First Course in Noncommutative Rings”. Here it is:

Example. let k be a ring with 1 and let

R=\left \{ \begin{pmatrix} x & y & z \\ 0 & x & t \\ 0 & 0 & x \end{pmatrix}: \ x,y,z,t \in k \right \}.

Let a = e_{12}. Then ar-ra=te_{13} \in Z(R) for every r =\begin{pmatrix} x & y & z \\ 0 & x & t \\ 0 & 0 & x \end{pmatrix} \in R but obviously a \notin Z(R) because, for example, a does not commute with e_{23}.