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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.

azimut
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Claudiu Mindrila
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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
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    In my opinion, this question is quite interesting. I would like to see it reopened. – azimut Sep 12 '13 at 11:27
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    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
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    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
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    @JyrkiLahtonen: Yeah, that's it! So the characteristic is $2$, and hence the order of $A$ is a power of two. Furthermore, by the conditions on $A$, $2^n - 1 < \lvert A\rvert < 2\cdot (2^n - 1)$. The only possibility is $\lvert A\rvert = 2^n$ and therefore, $A$ is a field. – azimut Sep 12 '13 at 19:31
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    @azimut The question has been reopened. – Zhen Lin Sep 12 '13 at 20:00
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    Why the four downvotes? It's a very cool question. – Karl Kroningfeld Sep 12 '13 at 20:05
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    @KarlKronenfeld And so what? The answer to the question in your comment is given in an older comment (that is, if the question in your comment is not purely rhetorical). – Did Sep 12 '13 at 21:10
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    @Did I only read the comment(s) of one/two possible downvoters. – Karl Kroningfeld Sep 12 '13 at 21:43
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    @KarlKronenfeld Then addressing the rationalizations these users gave might be more productive than asking why the downvotes. (I also wish to mention that the fact these users indeed downvoted or not is irrelevant to the intrinsic merits of their arguments.) – Did Sep 13 '13 at 07:49
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    @JyrkiLahtonen: Now that the question is open (for the moment?), I took the opportunity and posted the solution as an answer. I hope you are fine with it, since a good part of it was done by you. – azimut Sep 13 '13 at 08:29
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    Of course, it's fine, @azimut. You made a key contribution I missed. A quick +1, lest somebody manages to close this question again. – Jyrki Lahtonen Sep 13 '13 at 09:10
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    @Did, it is not astute to assume that one activity of mine is productive and another is not, comparatively. My only intention was to learn to what extent voting patterns are based on the cover and not the content. Addressing, say, Carl Mummert's comment would change my role from observer to participant; ultimately it would have been unproductive. On the other hand, the knowledge gained from these observations would apply directly to my work as a reviewer. Therefore, I ask, why the seven downvotes?! – Karl Kroningfeld Sep 14 '13 at 00:19
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    @KarlKronenfeld This seems to introduce ad hoc distinctions (de facto you became a participant as soon as you posted a comment, as you perfectly know) to hide the fact that you wish to discredit the opportunity of downvotes per se although a cogent rationalization for them in this case is already on the page. To ask flatly "Why the four downvotes?" after said rationalization was posted is only possible if one thinks it has zero value. I disagree, if only for reasons of politeness. – Did Sep 14 '13 at 07:11
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    A more straightforward reading is that in addition to registering some disagreement with the downvotes, the comment (by inviting additional downvoters to speak) allows for the possibility that other more interesting reasons exist for some of the votes, instead of presuming that none are possible. This does not support the negative implications that @Did is assigning to the comment et seq, whether they count as observation or participation, or both. – zyx Sep 15 '13 at 01:40
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    "More straightforward" must be a kind of joke. – Did Sep 15 '13 at 06:10

1 Answers1

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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.

azimut
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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 $2^n$. –  Sep 15 '13 at 22:48
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    @drhab: No, you can't conclude that $n=1$. The finite fields $\mathbb F_{2^n}$ provide an example for every $n$. – azimut Dec 06 '13 at 07:13