My solution

American Mathematical Monthly Problem 11407 by Erwin Just, New York, NY.

Let p be a prime such that p > 3. Show that any ring R (not necessarily commu- tative and not necessarily having a multiplicative identity) satisfying

p

X

i=1

x²ⁱ⁻¹ = 0 for every x∈R (11407.1) must be the trivial ring (i. e., the ring consisting of one element only).

Solution by Darij Grinberg.

First, every x∈R satisfies

x=x^2p+1, (11407.2)

since

x−x^2p+1 =

p

X

i=1

x²ⁱ⁻¹

| {z }

=0 by (11407.1)

−x²

p

X

i=1

x²ⁱ⁻¹

| {z }

=0 by (11407.1)

= 0−0 = 0.

Now, lety∈R.Then, (11407.2) yieldsy=y^2p+1,so that (y^2p)² =y^4p =y^2p−1y^2p+1 = y^2p−1y=y^2p. Thus, y^2p is idempotent; hence,

y^2pk

=y^2p (11407.3)

for every integer k ≥1.Applying (11407.1) to x=y^2p yields

p

P

i=1

(y^2p)²ⁱ⁻¹ = 0; thus, 0 =

p

X

i=1

y^2p2i−1

| {z }

=y^2pby (3)

=

p

X

i=1

y^2p =py^2p. (11407.4)

On the other hand, applying (11407.2) tox= 2y^2p yields 2y^2p = (2y^2p)^2p+1; thus, 2y^2p = 2y^2p2p+1

= 2^2p+1 y^2p2p+1

| {z }

=y^2pby (3)

= 2^2p+1y^2p, so that

0 = 2^2p+1y^2p−2y^2p = 2^2p+1−2

y^2p. (11407.5)

But 2^2p+1−2 is coprime to p(in fact, p is a prime and greater than 3, so that 4^p−1 ≡ 1 modp by Fermat’s Little Theorem, and thus

2^2p+1−2 = 8· 4^p−1

|{z}

≡1 modp

−2≡8−2 = 6 modp,

so that 2^2p+1 −2 is coprime to p since 6 is coprime to p, which again follows from p being prime and greater than 3). Hence, Hence, by Bezout’s theorem, there exist integers a and b such that a(2^2p+1−2) +bp= 1. This leads to

y^2p = 1y^2p = a 2^2p+1−2 +bp

y^2p =a 2^2p+1−2 y^2p

| {z }

=0 by (5)

+b py^2p

|{z}

=0 by (4)

= 0 + 0 = 0.

Applying (2) to x=y yields y=y^2p+1,so that y=y^2py = 0y= 0.

Thus, we have proven that y= 0 for every y∈R.Hence, R= 0,proving our claim.

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