Unit in a ring (abstract algebra)

[cummings12332]

Homework Statement

Is (x^2-1) a unit in F[x]? where F is a field.


2. The attempt at a solution
I might say yes, cause we can find the taylor expansion of 1/(x^2-1), is my idea right?

 

[Dick]

Homework Statement

Is (x^2-1) a unit in F[x]? where F is a field.


2. The attempt at a solution
I might say yes, cause we can find the taylor expansion of 1/(x^2-1), is my idea right?

The taylor expansion is not a polynomial. It has an infinite number of terms.

 

[hapefish]

it has been a long time since I studied Ring Theory, but here is what I remember that might be relevant:

A unit is an element that has an inverse. So in order for $x^2-1$ to be a unit, there would have to exist an inverse of $x^2-1$ in your field. Your suggestion of $\frac{1}{x^2-1}$ is a reasonable candidate, but I do not believe it is an element of your field. This is because I always took $F[x]$ to represent the ring of finite degree polynomials over the field, F, and the Taylor expansion of $\frac{1}{x^2-1}$ is infinite.

In my opinion, the answer needs to be "no." I think proving this hinges on the fact that we're in a field (and hence an integral domain) so there is no terms that can multiply by $x^2$ to make the leading coefficient zero.

Good Luck!