Originally Posted by
Primality Thanks very much for the help. I guess the answer to my question as stated is "no" although I'm not sure the replies above show this completely but am sure they can be easily adapted. (I don't think I've assumed that $\displaystyle A$ has a multiplicative unit...if you can prove there is a ring with $\displaystyle x^k=x$ doesn't have a multiplicative unit then this just shows that the answer is "no" and the slight problem with taking ring with a non-zero and non-unital idempotent is that I'd specifically said $\displaystyle k \geq 4$.)
I'll state the problem I'm really interested in:
Suppose $\displaystyle f(x) \in \mathbb{Z}[x]$ and assume we know that $\displaystyle f(x) = ((f(x))^k$ in the ring $\displaystyle \frac{\mathbb{Z}[x]}{(x^r-1,n)}$ where $\displaystyle r,n,k \in \mathbb{Z}$ and $\displaystyle k \geq 4$. Suppose we also have that $\displaystyle (k-1) \mid (b-c)$ for integers b,c. Then show that $\displaystyle (f(x))^b = (f(x))^c$ in $\displaystyle \frac{\mathbb{Z}[x]}{(x^r-1,n)}$.
I thought I must be missing something obvious (which led to the original question), but perhaps not...Can anyone help?