American Mathematical Monthly Problem 11451 by Greg Oman, Westerville, OH.
Let k and n be positive integers satisfying k >1. Let R be a ring (not necessarily having an identity, and not necessarily commutative). Assume that:
(i) There exists an element of R which is not nilpotent.
(ii) If x1, x2, ..., xk are anyk nonzero elements of R, then
k
P
j=1
xnj = 0.
Prove that R is a division ring, that is, the nonzero elements of R form a group under multiplication.
Solution by Darij Grinberg.
First, we notice that:
(a) For any two nonzero elements xand y of R, we have xn =yn. In fact, applying(ii)to (x1, ..., xk) =
x, x, ..., x
| {z }
n−1 times
, y
, we obtainxn+xn+...+xn
| {z }
n−1 times
+yn=
0,but applying (ii)to (x1, ..., xk) =
x, x, ..., x
| {z }
ntimes
, x
, we obtain xn+xn+...+xn
| {z }
ntimes
= 0, so that
xn+xn+...+xn
| {z }
ntimes
=xn+xn+...+xn
| {z }
n−1 times
+xn= 0 =xn+xn+...+xn
| {z }
n−1 times
+yn
and thus xn=yn, proving (a).
As a consequence of (a), the problem becomes trivial if n = 1 (because then all nonzero elements of R are equal, so that R is the field with two elements), so let us now assume that n ≥2.
We have:
(b) If some x∈R satisfies xn= 0, then x= 0.
In fact, there exists some y∈R such thatyn 6= 0 (by (i)), so thaty 6= 0, and thus, if x6= 0, then (a) yieldsxn=yn contradicting to xn = 0 and yn6= 0.This proves (b).
As a consequence:
(c) If somex∈R and y∈R satisfy xy= 0,then x= 0 or y= 0.
In fact, ifxandyare nonzero, thenxn=yn(by(a)), and thus (yn)n = yn
|{z}
=xn
(yn)n−1 = xn(yn)n−1 = 0 (since the termxy appears in this product, andxy= 0), so thatyn = 0 (by (b)), and thus y = 0 (by (b) again), contradicting to our assumption that y be nonzero. Hence, (c) is proven.
Now we will show that:
(d) For any nonzero x∈R and y ∈R, we have xny=y and yxn=y.
In fact, (a) yields xn = yn and also (y2)n = yn (since y2 is nonzero, because otherwise yn would be zero and therefore y = 0 according to (b), and contradicting y 6= 0), so that (xny−y)yn−1 = xn
|{z}
=yn
yn −yn = y2n−yn = y2n
| {z }
=yn
−yn = 0. Thus,
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(c) yields xny−y = 0 or yn−1 = 0. But yn−1 = 0 is impossible (since this would yield yn = 0, thus y = 0 according to (b), what is ruled out by y 6= 0). Therefore, xny−y= 0, so that xny=y. Similarly,yn−1(yxn−y) =yn xn
|{z}=yn
−yn =y2n−yn = 0 yieldsyn−1 = 0 or yxn−y= 0 (by (c)), out of which only yxn−y= 0 is possible, and we obtainyxn=y. Thus, (d) is proven.
Now, fix some nonzero x∈R. Then, I claim that the nonzero elements of Rform a group under multiplication, and thatxnis the unity of this group. In fact, associativity is clear, and xn is a unity because of (d). Besides, every nonzero element y ∈ R has an inverse, namely yn−1, since yyn−1 =yn =xn (by (a)) and yn−1y =yn =xn. This completes the solution.
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