sniffnoy | Tom solved the a^4 problem!

Remember that old thing, that annoyed everyone? Well, Tom solved it!

Problem: Given a ring R (not necessarily with identity) st a⁴=a ∀a∈R, prove R is commutative.

First, note that if x²=0, then x=x⁴=0.

Now, given an element y∈R st y²=y, we find that, ∀x∈R,

(xy-yxy)²=xyxy-xyxy+yxyxy-yxyxy=0, and similarly (yx-yxy)²=0. Thus xy=yxy=yx if y²=y.

Now, any element y of the form x²+x has the property that y²=y. For this we need that R is characteristic 2. ∀a∈R 2a=(2a)⁴=16a⁴=16a, so 14a=0. Also, 3a=(3a)⁴=81a⁴=81 a, so 78a=0. As 14a=0 and 78a=0, we get 8a=0, 6a=0, 2a=0. So, (x²+x)²=x⁴+2x³+x²=x+x².

So, ((x+y)²+x+y)y=y((x+y)²+x+y). Expanding, we get
x²y+xy²+yxy+y³+xy+y²=yx²+yxy+y²x+y³+yx+y².
Cancelling, we get
x²y+xy²+xy=yx²+y²x+yx.

We also know that (x²+x)y=y(x²+x), or x²y+xy=yx²+yx, so, subtracting, we get xy²=y²x ∀x,y∈R. As every element of R can be written as a square (being equal to its own fourth power), we find that R is commutative. QED.