Let $A$ be a ring with $0\neq1$ having $2^n-1$ invertible elements and at most $2^n-1$ non-invertible elements. Then $A$ must be a field? How many elements does the ring need to have? Because $2^n-1$ is odd, the characteristic of the ring is $2$. Thank you!
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One question: why must |A| be a power of two? – user265311 Aug 29 '15 at 13:06
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@DanielFischer: There is a slight gap between this Question and the duplicate cited, namely here the number of non-invertible elements is at most $2^n - 1$ and there the number of non-invertible elements is strictly less than $2^n - 1$. However this only affects the second step of azimut's Answer there, bracketing $2^n - 1 \lt |A| \le 2(2^n - 1)$ where $|A|$ is the size of ring $A$. The conclusion $|A| = 2^n$ remains. – hardmath Aug 29 '15 at 13:13
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1@hardmath Well, yes, here we have the bounds $2^n - 1 < \lvert A\rvert \leqslant 2\cdot 2^n - 1$, I figured that doesn't make much of a difference. – Daniel Fischer Aug 29 '15 at 13:19
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@user265311: The order of the additive group of $A$ is $|A|$. If $|A|$ had an odd prime factor $p$, then by the structure theorem for finite abelian groups, there would be some element $u \neq -u$. – hardmath Aug 29 '15 at 13:19
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@user265311 Since the characteristic is $2$, it is a $\mathbb{F}_2$ vector space. – Daniel Fischer Aug 29 '15 at 13:19