# Thread: Ring with identity

1. ## Ring with identity

Let x be an element in a ring with identity, and suppose $\displaystyle x^{n} = 0$ for some positive integer n.
Now i want to proof that $\displaystyle (1+x)$ is a unit.

I tried to use $\displaystyle (1+x)(1-x+x^2-x^3+...) =1$ but i can't solve it. Any hints?

2. suppose n is odd.
then (-1)^n+1=0 so x+1 divides 1+x^n=1+0=1 by factor theorem
so there is an element y such that (1+x)(y)=1

what if n i even?

so after you have done this you can use long division for each aswer to give you the unique y needed and you see that the elements are both of the form you wrote them in and is composed of the identity element and x, and so y is in the ring with identity.

3. Originally Posted by bram kierkels
Let x be an element in a ring with identity, and suppose $\displaystyle x^{n} = 0$ for some positive integer n.
Now i want to proof that $\displaystyle (1+x)$ is a unit.

I tried to use $\displaystyle (1+x)(1-x+x^2-x^3+...) =1$ but i can't solve it. Any hints?

You have the idea. Use the well-known identity from geometric sequences:

$\displaystyle \frac{1-x^n}{1-x}=1+x+x^2+\ldots+x^{n-1}\Longrightarrow 1-x^n=(1-x)(1+x+\ldots+x^{n-1})$ , and change conveniently the sign if you will (though you don't have

to, since $\displaystyle x^n=0\Longleftrightarrow (-x)^n=0$ )

Tonio