Let $A$ be a ring with $0 \neq 1 $, which has $2^n-1$ invertible elements and less non-invertible elements. Prove that $A$ is a field.
[Math] A ring with few invertible elements
abstract-algebrafinite-ringsnoncommutative-algebraring-theory
abstract-algebrafinite-ringsnoncommutative-algebraring-theory
Let $A$ be a ring with $0 \neq 1 $, which has $2^n-1$ invertible elements and less non-invertible elements. Prove that $A$ is a field.
Best Answer
Step 1: The characteristic of $A$ is $2$
(Credit for this observation goes to Jyrki Lahtonen)
The mapping $x\mapsto -x$ is an involution on $A^\times$. Since $\lvert A^\times\rvert = 2^n - 1$ is odd, it has a fixed point. So $a = -a$ for an $a\in A^\times$. Multiplication with $a^{-1}$ yields $1 = -1$.
Step 2: $\lvert A\rvert = 2^n$
From the preconditions on $A$ we know $$2^n - 1 < \lvert A \rvert < 2(2^n - 1) = 2^{n+1} - 2.$$ By step 1, the additive group of $A$ is a $2$-group, so $\lvert A\rvert$ is a power of $2$. The only remaining possibility is $\lvert A\rvert = 2^n$.
Step 3: $A$ is a field
From step 2 and $\lvert A^\times\rvert = 2^n - 1$, we know that all non-zero elements of $A$ are invertible. Hence $A$ is a finite skew-field. Now by Wedderburn's little theorem, $A$ is a field.