Is it true that in any ring with if for some then

No, that is not true and here’s an example. Let be a (countably) infinite dimensional vector space over some field and let be a basis for Now consider the ring of -linear transformations of Define by and See that but because

**Definition 1**. A ring is called **Dedekind-finite** if

**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 is called **reversible** if

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

*Proof***. **Suppose that for some Then and thus So and hence It follows that So is Dedekind-finite. Finally, note that every reduced ring is reversible because if for some then and thus

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

*Proof.* We will assume that is left Noetherian. Suppose that for some Define the map by Clearly is an -module homomorphism and is onto because for all Now we have an ascending chain of left ideals of

Since is left Noetherian, this chain stabilizes at some point, i.e. there exists some such that Clearly is onto because is onto. Thus for some Then

Hence and therefore

**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 be a field and let be a finite dimensional -algebra. Then is Dedekind-finite.

*Proof*. Every left ideal of is clearly a -vector subspace of and thus, since any ascending chain of left ideals of will stop at some point. So is left Noetherian and thus, by Example 2, is Dedekind-finite.

**Remark 2**. Two important cases of Example 5 are the ring of matrices over a field, and, in general, semisimple rings. As a trivial result, is Dedekind-finite for any commutative domain because is a subring of , where is the quotient field of .

So the ring of matrices, where 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 Then 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 over a field is Dedekind-finite.

*Proof*. Suppose that for some Since is algebraic over there exist integers and some with such that We will assume that is as small as possible. Suppose that Then, since we have

which contradicts the minimality of So Let and see that But then and therefore

**Remark 3**. Regarding Examples 5 and 6, note that although any finite dimensional -algebra is algebraic over but being algebraic over does not necessarily imply that is finite dimensional over For example, if is the algebraic closure of in then it is easily seen that Thus the matrix ring is an algebraic -algebra which is not finite dimensional over So, as a -algebra, is Dedekind-finite by Example 6 not Example 5.

**Example 7**. Every PI-algebra is Dedekind-finite.

*Proof. *Let be the Jacobson radical of If then is a subdirect product of primitive algebras where are the primitive ideals of Since is PI, each is PI too and thus, by Kaplansky’s theorem, is a matrix ring over some division algebra and thus Dedekind-finite by Example 2. Thus is Dedekind-finite and so which is a subalgebra of is also Dedekind-finite. For the general case, let Now, is PI, because is PI, and Therefore, by what we just proved, is Dedekind-finite. Suppose that for some and let be the image of in respectively. Clearly and so Thus and so is invertible. Hence there exists such that But then and hence