This question already has an answer here:

Let $R$ be a finite ring (that does not necessarily contain the identity) such that more than three fourth of its elements are idempotents. Then show that $R$ must be commutative.

If $R$ is a ring with 1 then this case I have proved, but without assuming the existence of it how can it be shown?


marked as duplicate by rschwieb, Jack Schmidt, Start wearing purple, user1337, Rick Decker Aug 7 '13 at 18:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ As Julien's argument shows, most of Qiaochu's argument works without assuming 1. The use of Lagrange rather than direct product decompositions seems simpler to me, but the relation of central idempotents to direct product decompositions is very important. $\endgroup$ – Jack Schmidt Aug 7 '13 at 17:56
  • $\begingroup$ sorry to say @rschwieb,with your referrence I have gone through it...he has proven the fact assuming the existence of identity..but in the proposition too the existence of identity has not been mentioned at all..so the proof of the proposition is partially complete $\endgroup$ – sayak Aug 7 '13 at 17:56

The basic point is the following:

If an element commutes with more than half of the other elements, it already commutes with all elements.

This is basically Lagrange's theorem: Let $C_R(x) = \{r \in R | rx = xr\}$ be the centralizer of $x$. We assume $|C_R(x)| > \frac{1}{2} |R|$. $C_R(x)$ is a subgroup of $R$ and thus its order divides $|R|$. However, by our assumption $\frac{|R|}{|C_R(x)|} < 2$, so $C_R(x)$ already has to be whole $R$. Intuitively $C_R(x)$ is big enough to cover $R$; take an element of $R$ and look at $C_R(x) + r$. This set has the same number of elements as $C_R(x)$ and since there are more than $\frac{1}{2} |R|$ such elements, these two sets cannot be disjoint. But then they must actually be equal!

Now if more than half of the elements commute with more than half of the elements, then more than half of the elements commute with every element. Put another way: Every element commutes with more than half of the elements of $R$ and thus -- by the above -- every element commutes with every element; $R$ is commutative.

Now why do more than half of the elements commute with more than half of the elements? Now the idempotents come into play: Let $T = \{r \in R | r^2=r\}$ be the idempotents. Then $|T| > \frac{3}{4} |R|$ by assumption. Now fix $x \in T$. We will show that $x$ commutes with a special subset of $R$ and then count that subset to see that it has more than $\frac{1}{2} |R|$ elements.

The idea to get to this set is the following: If $x$ and $y$ and $x + y$ are idempotent, we get $(x + y)^2 = x^2 + xy + yx + y^2 = x + y$, so $xy = -yx$. For $x=y$ this says $x=-x$ or $2x = 0$. Let's concentrate on this case for a bit: Since we have a lot of idempotents, the "probability" is high for $x + x$ or equivalently $-x$ to also be idempotent. (If $-x$ is idempotent, we have $-x = (-x)^2=x^2=x$, thus $2x = x + x$ which is idempotent; if $x + x$ is idempotent, we have $x = -x$ idempotent by the above.) How many elements does $T \cap (-T)$ have? Well, by inclusion-exclusion for example, more than half of $|R|$. With a similar argument to the above (Lagrange), and with the above we now see that $T \cap (-T)$ is a subset of the subgroup $\{r \in R | 2r = 0 \}$ and thus this subgroup exhausts $R$. $R$ has characteristic 2!

Now the only thing that is left to ensure is that for our $x \in T$ there are enough idempotents $y$ such that $x + y$ is also idempotent, because then -- as we saw -- $xy = -yx = yx$. So we count the set $T \cap (T - x)$ or equivalently $T\cap (T + x)$. This is fairly easy using inclusion-exclusion again: Since we know $|T + x| = |T| > \frac{3}{4} |R|$ and $|T \cup {x + T}| \le |R|$ we get $|T \cap (x + T)| > \frac{1}{2} |R|$ as we wanted!

I hope this didn't become to convoluted; this theorem is a corollary of Theorem 4 of this paper.

I tried to make its application to this specific scenario a bit more transparent.

  • $\begingroup$ Oh, thats true; my bad. But we don't need this. I edited my answer accordingly. Thanks! $\endgroup$ – Julian Kniephoff Aug 7 '13 at 17:42
  • $\begingroup$ but how can you get x+y-yx-y^2=x+yx....the right hand side should be x-yx $\endgroup$ – sayak Aug 7 '13 at 19:48
  • $\begingroup$ I hope everything is correct now. You are right @sayak, however, $R$ has characteristic 2. I tried to write the proof in a way one could derive the fact from scratch just by looking at the standard proof of "idempotent/boolean rings are commutative", which makes it a bit messy, but more or less intuitive, I think. Sorry and thanks. ;) $\endgroup$ – Julian Kniephoff Aug 7 '13 at 22:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.