Let \(R\) be a finite ring with identity, and let \(a\in R\) be an element which is not a zero divisor. Show that \(a\) is invertible.
Solution
1. The map\[ L_a:R\longrightarrow R, \qquad L_a(x)=ax \]is injective.
If \[ L_a(x)=L_a(y), \] then \[ a(x-y)=0. \] Since \(a\) is not a zero divisor, this forces \[ x-y=0, \] so \(x=y\).
2. There is an element \(b\in R\) such that\[ ab=1. \]
The set \(R\) is finite, so every injective self-map of \(R\) is surjective. By step 1, \(L_a\) is surjective. Hence the identity element lies in its image: \[ L_a(b)=ab=1 \] for some \(b\in R\).
3. The map\[ R_a:R\longrightarrow R, \qquad R_a(x)=xa \]is injective and therefore surjective.
If \[ xa=ya, \] then \[ (x-y)a=0. \] Since \(a\) is not a zero divisor, \(x=y\). Thus \(R_a\) is injective, and finiteness of \(R\) makes it surjective.
4. There is an element \(c\in R\) such that\[ ca=1. \]
By surjectivity of \(R_a\) from step 3, the identity element lies in its image.
5. The elements \(b\) and \(c\) from steps 2 and 4 are equal.
Using associativity, \[ c = c(ab) = (ca)b = b. \]
6. The element \(a\) is invertible.
By steps 2, 4, and 5, the single element \[ b=c \] satisfies \[ ab=ba=1. \] Thus it is the inverse of \(a\).
7. Q.E.D.
Step 6 proves the claim.