Rings, Nilpotent and Idempotent Elements

Nilpotent and Idempotent Elements

The element x in R is nilpotent if some integer n satisfies xn = 0.  The integer represents successive multiplications and need not be in R.

A reduced ring has no nonzero nilpotents.

In a commutative ring, any linear combination of nilpotent elements is nilpotent.  Use the multinomial theorem and make n bigger than the sum of all the exponents that drive the individual nilpotents to 0.

If x is nilpotent, 1-x is a unit.  This is because synthetic division terminates.  That is, 1-xn is divisible by 1-x.

An element x is idempotent if x2 = x.  If R is free of zero divisors, write x2-x = 0, whence x = 0 or 1.

Note that an idempotent x has an orthogonal counterpart 1-x.  Each squared is itself, their sum is 1, and their product is 0.  This works even in characteristic 2.

Idempotents and Component Rings

Let R be the finite direct product of subrings R1 R2 … Rn.  Let ei be 1 in the ith component and 0 elsewhere.  Thus the sum of all ei is 1 in R, and the product ei*ej is 0 when i does not equal j.  Note also that each ei is idempotent, and these idempotents commute with R.

Conversely, let a finite set of orthogonal idempotents sum to 1, such that these idempotents commute with R.  Let Ri be the principal ideal generated by ei.  Show that each Ri is a subring, with ei acting as 1, and that R is isomorphic to the direct product of these subrings.  The map is accomplished via x*1, where 1 is replaced with the sum of idempotents.  Then, x*1 times y*1 expands into a large cross product, but all the mixed terms drop out, and we simply find e1*xy + e2*xy + … + en*xy, which is the projection of the product onto the component rings.  All the algebra works; I'll leave it to you.

The algebra doesn't work when the orthogonal idempotents don't commute with R.  Let a and b be idempotents in a ring with characteristic 2.  Include x and y, such that xa = x, yb = y, ay = y, bx = x, and the other nonidempotent products are all 0.  Verify this is a ring, with a+b = 1, that does not split into a finite product of subrings.  Again, I'll leave the algebra to you.