A ring with few invertible elements

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.

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The $2^n-1$ thing must be critical for some combinatorical argument. For example, $\Bbb Z/(9)$ has units $\{1,2,4,5,7,8\}$ has more units than nonunits, but of course fails the count requirement. –  rschwieb Sep 11 '13 at 14:47
In my opinion, this question is quite interesting. I would like to see it reopened. –  azimut Sep 12 '13 at 11:27
The question has good mathematical content, but it is not written in a way to match the standards of this site. Questions should not merely state a problem; they should include a description of the context where the problem was encountered, and a description of what that asker has tried. Moreover, questions phrased as demands ("Prove that...") are considered to be mildly impolite by a significant number of users here. If the question is reopened, it should be edited to address these issues. –  Carl Mummert Sep 12 '13 at 12:10
The order of the unit group is odd. This implies that $u=-u$ for some unit $u$. Therefore $2=0$ in this ring. –  Jyrki Lahtonen Sep 12 '13 at 14:20
Why the four downvotes? It's a very cool question. –  Karl Kronenfeld Sep 12 '13 at 20:05

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.

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One can say a little bit more: whenever the group of units of a ring is both finite and odd order, the subring they generate is a direct product of (finite) fields of characteristic $2$. So in this question, the subring generated by the units is a finite field of order $2^n$. –  user641 Sep 15 '13 at 22:48
@drhab: No, you can't conclude that $n=1$. The finite fields $\mathbb F_{2^n}$ provide an example for every $n$. –  azimut Dec 6 '13 at 7:13