Proving u - a is a Unit in a Ring with Unity

Let R be a ring with unity, where a is a nilpotent element of R and u is a unit element such that au ua. We aim to prove that u - a is a unit in R.

Step-by-Step Proof

1. Since a is nilpotent, it means there exists a smallest positive integer m1 such that am1 0. We will begin our proof by expressing u - a in a form that highlights its unit nature.

2. Consider the expression:

(u - a) (sum_{k0}^{m1-1} u^k a^{m1-1-k} )

which simplifies to:

u^{m1} - a^{m1} u^{m1}

Since u is a unit, um1 is also a unit. Therefore, we have found an element

(sum_{k0}^{m1-1} u^k a^{m1-1-k}) in R

that satisfies the property of a multiplicative inverse, thus making u - a a unit.

Advanced Approach - Factorization Argument

Another method is to use the factorization in commutative algebra. Suppose aN 0, then uN - uN - aN uN - aN can be expressed as a polynomial in u and a. Define

p uN-1uN-2dots aN-1

Thus, u - a uN - aN uN - ap. Given that u is a unit, uN is also a unit. Therefore, u - a has an inverse in u-NuN-1uN-2dots aN-1, making u - a a unit.

Unit Inverse in Commutative Algebra

By the nature of the unit and nilpotent properties, we know there exists a unit overline{u} such that uoverline{u} overline{u}u 1R. Since a is nilpotent and satisfies an 0R, we use the series expansion:

frac{1}{1-x} 1 - x x^2 - dots

To find a series that works for our case. Given that au ua, we have:

1 - overline{u}a (1 - overline{u}a (1 - overline{u}a (dotsm (1 - overline{u}a)}^n)) 1 - (overline{u}a)^n 1 - 0 1

And

(1 - overline{u}a (1 - overline{u}a (1 - overline{u}a (dotsm (1 - overline{u}a)}^n)))(1 - overline{u}a) 1

Hence, we identify the inverse as 1 - overline{u}a.

Conclusion

We have shown that u - a is a unit in the ring R by constructing a specific inverse element. This proof leverages the properties of nilpotent and unit elements and their interactions in a ring with unity. The methods used here highlight the algebraic structure and highlight the significance of unit and nilpotent elements.