Dedekind-finite ring

In mathematics, a ring is said to be a Dedekind-finite ring (also called directly finite rings^[1][2][3] and Von Neumann finite rings^[4][2][3]) if ab = 1 implies ba = 1 for any two ring elements a and b. In other words, all one-sided inverses in the ring are two-sided. Numerous examples of Dedekind-finite rings include Commutative rings, finite rings, and Noetherian rings.

A ring ${\displaystyle R}$ is Dedekind-finite if any of the following equivalent conditions hold:^{[3][better source needed]}

• All one sided inverses are two sided: ${\displaystyle xy=1}$ implies ${\displaystyle yx=1}$.
• Each element that has a right inverse has a left inverse: For ${\displaystyle x\in R}$, if there is a ${\displaystyle y\in R}$ where ${\displaystyle xy=1}$, then there is a ${\displaystyle z\in R}$ such that ${\displaystyle zx=1}$.
• Capacity condition: ${\displaystyle xy=1}$, ${\displaystyle xz=0}$ implies ${\displaystyle z=0}$.
• Each element has at most one right inverse.
• Each element that has a left inverse has a right inverse.
• Dual of the capacity condition: ${\displaystyle yx=1}$, ${\displaystyle zx=0}$ implies ${\displaystyle z=0}$.
• Each element has at most one left inverse.
• Each element that has a right inverse also has a two sided inverse.

• Any Commutative rings is Dedekind-finite^[3]
• Any finite ring is Dedekind-finite.^[4]
• Any Matrix rings ${\displaystyle M_{n}(F)}$ are Dedekind-finite.^[3]
• Any domain is Dedekind-finite.^[4]
• Any left/right Noetherian ring is Dedekind-finite.^[4][3]
• Given a group ${\displaystyle G}$, the group algebra ${\displaystyle kG}$ is Dedekind-finite.^[3]
• A Dedekind-finite ring with an idempotent ${\displaystyle e^{2}=e}$ implies that the corner ring ${\displaystyle eRe}$ is also Dedekind-finite.^[2]
• A ring with finitely many nilpotents is Dedekind-finite.^[3]
• Reversible rings, rings where ${\displaystyle xy=0}$ must imply ${\displaystyle yx=0}$, are Dedekind-finite.^[3]
• A unit-regular ring is Dedekind-finite.^[4]
• A ring without left/right zero divisors is Dedekind-finite.^[2]

A counter-example can be constructed by considering the polynomial ring ${\displaystyle R[x,y]}$, where the ring ${\displaystyle R}$ has no zero divisors and the indeterminates do not commute (that is, ${\displaystyle xy\neq yx}$), being divided by the ideal ${\displaystyle I=(xy-1)}$, then ${\displaystyle x+I\in R[x,y]/I}$ has a right inverse but is not invertible. This illustrates that Dedekind-finite rings need not be closed under homomorpic images^[2]

Dedekind-finite rings are closed under subrings^{[1][2][better source needed]}, direct products,^[3][2] and finite direct sums.^[2] This makes the class of Dedekind-finite rings a Quasivariety, which can also be seen from the fact that its axioms are equations and the Horn sentence ${\displaystyle ab=1\implies ba=1}$.^[2]

A ring is Dedekind-finite if and only if so is its opposite ring.^[2] If either a ring ${\displaystyle R}$, its polynomial ring ${\displaystyle R[X]}$ with indeterminates ${\displaystyle X}$, the free word algebra ${\displaystyle R[{\tilde {X}}]}$ over ${\displaystyle X}$ with coefficients in ${\displaystyle R}$, or the power series ring ${\displaystyle R[[X]]}$ are Dedekind-finite, then they all are Dedekind-finite.^[2] Letting ${\displaystyle {\text{Rad}}(R)}$ denote the Jacobson radical of the ring ${\displaystyle R}$, the quotient ring ${\displaystyle R/{\text{Rad}}(R)}$ is Dedekind-finite if and only if so is ${\displaystyle R}$, and this implies that local rings and semilocal rings are also Dedekind-finite.^[2] This extends to the fact that, given a ring ${\displaystyle R}$ and a nilpotent ideal ${\displaystyle I}$, the ring ${\displaystyle R}$ is Dedekind-finite if and only if so is the quotient ring ${\displaystyle R/I}$,^[2] and as a consequence, a ring is also Dedekind-finite if and only if the upper triangular matrices with coeffecients in the ring also form a Dedekind-finite ring.^[2]

• Dedekind-infinite set
• Von Neumann regular ring