## Twisted Rings, Zero Divisors, Nilpotents, Units

### Zero Divisors

If R is commutative, and there is no twist, units and nilpotents are well characterized.  This page extends some of those theorems to the noncommutative world.

If σ has a nontrivial kernel, then for c in that kernel, xc = 0.  In other words, x is a zero divisor.  This is usually not what we want, so for the rest of this page, assume σ is injective.

If R is a domain, consider the lowest term in the product of two polynomials or two series.  This looks like c*σ(d), which is always onnzero.  Therefore the twisted extension of a domain is a domain.

### Nilpotent

Let σ be an automorphism with a fixed period, such that σk is the identity map.  Let R be commutative, or at least, let the nilpotent elements of R commute past each other.  If cn = 0 in R, expand (cx)kn.  The result includes cn, and drops to 0.  Thus cx is nilpotent.  A similar result holds for c times any power of x.

Now consider a polynomial p with nilpotent coefficients.  If p has l terms, expand pn to find ln products, as each of l terms is placed in each of n positions.  The trick is to choose n high enough.  Let m be the highest exponent on any of the l coefficients of p.  Then set n = mlk.  At least mk instances of one of the terms of p will appear in each product.  Pull all these coefficients to the start, and at least m will be run through the same power of σ.  When σj(c) is raised to the m, the result is 0, just like cm.  Each product drops to 0, and p is nilpotent.

Adjoin any number of indeterminants, and select a σ for each one, as it commutes with R, such that each σ has a finite period.  The above proof remains valid, though the exponent must be much higher, to accommodate all the different ways a coefficient might run through the various automorphisms that are introduced by the (finitely many) indeterminants in a given polynomial.  I'll leave the details to you.  With these stipulations in mind, nilpotent coefficients produce nilpotent polynomials.

The converse is true in commutative rings, but not here.  Let R = Z[u,v], with uv = 0.  Let σ swap u and v.  Now ux is a nilpotent polynomial, even though u is not nilpotent in R.

Ad vx, another nilpotent polynomial, and get (u+v)x.  Rais this to the nth and get (un+vn)xn.  Thus the sum of nilpotent polynomials need not be nilpotent.

If p is a nilpotent polynomial, of any form, apply σ to all the coefficients and find another nilpotent polynomial with the same exponent.  This is because σ respects all the ring operations that take place when p is multiplied by itself again and again.

In general, σ extends to an endomorphism, and in this case a monomorphism, on the entire twisted ring.  If σ is surjective on R then it is surjective on the twisted ring.  Thus an automorphism extends to an automorphism.

### Units

The units in R are still units in the twisted extension.  This because R is a subring of its extension.

If a polynomial or series is a unit, its constant coefficient must be a unit.

If R is a domain, the lead terms in two polynomials combine to create a nonzero term in the product.  Therefore the units of the extension are precisely the units of R.

Like any other ring, 1-w is a unit when w is nilpotent.  Divide 1 by 1-w using synthetic division, and the process stops.

In the commutative world, a unit polynomial is always some unit in R plus a nilpotent polynomial - the converse of the above.  This is not the case here.  We don't even need a twist to demonstrate this fact; simply choose an appropriate noncommutative base ring R.  Start with Z and adjoin the indeterminant u, giving the polynomials in u.  Then adjoin 1/u, so that u is a unit in R.  Adjoin the indeterminant v, such that u and v do not commute.  Then let vn = 0.  Now 1-vx is a unit, since vx is nilpotent.  Multiply by u to get another unit, u-uvx.  But uvx is not nilpotent, since u and v do not commute.

Let σ be an R endomorphism, building the twisted series over x.  (Here σ need not be injective.)  As mentioned earlier, a series has to start with a unit to be a unit.  Premultiply by the inverse of this unit, so that our series f(x) starts with 1.  Divide this into 1 by synthetic division.  The quotient, g(x), begins with 1.  Subtract 1*f(x) from 1, and the linear term of the "remainder" is -a1x.  Multiply this by f(x) on the right and subtract.  The remainder begins with a squared term, and the process continues forever.  This builds g(x), such that g*f = 1.  By symmetry, f is also right invertible, hence f is a unit.  A series is a unit iff its constant term is a unit.