Let $\xi\in GR(4^{m})$ and $\xi$ is of order $2^{m}-1$. Let $a=l-k$ for l and k are distinct in the range $[0,2^{m}-2]$, where $m\geq 2$. Why $2\xi^{a}=0$ is a contradiction? Could you help me to explain the reasons?
A problem in Galois rings
1
$\begingroup$
abstract-algebra
ring-theory
finite-rings
galois-rings
-
0$GR(4^{m})=\{a_{0}+a_{1}x+a_{2}x^{2}+...+a_{m-1}x^{m-1}|a_{i}\in \mathbb{Z}_{4},i=0,1,2,...,m-1\}$ – 2012-07-27
1 Answers
2
Since $\xi$ is invertible, so is $\xi^a$. Multiplication of the equation $2\xi^a = 0$ with the inverse of $\xi^a$ yields the contradiction $2 = 0$ (note that the characteristic of your base ring is $4$).