Reversible code if reciprocal value of root is a root I know that a code C is called reversible if $(a_0, ..., a_{n-1}) \in C$ implies that $(a_{n-1}, ..., a_1, a_0) \in C$.
Now, how can I show that a cyclic code C = g is reversible iff with each root of g also the reciprocal value of that root is a root of g?  
 A: This is true only if you assume the multiplicities of the roots also match (i.e., modify the statement to: $a$ is a root of $g(x)$ with multiplicity $m$ if and only if $\dfrac 1 a$ is a root of $g(x)$ with multiplicity $m$).
For $c = (a_0, \ldots, a_{n - 1}) \in C$, the associated polynomial is $p_c(x) = a_0 + a_1 x + \cdots a_{n - 1} x^{n - 1}$.
For the reverse word $d = (a_{n - 1}, \ldots, a_0)$, the associated polynomial is $$p_d(x) = a_{n - 1} + a_{n - 2} x + \cdots + a_0 x^{n - 1} = x^{n - 1}\left[ a_{n-1}\left(\dfrac 1 x\right)^{n-1} + a_{n - 1} \left(\dfrac 1 x\right)^{n - 2}  + \cdots + a_0\right] = x^{n - 1}p_c\left( \dfrac 1 x \right)$$
from which it is clear that the roots of $p_d(x)$ are exactly the reciprocals of the roots of the (non-zero) roots of $p_c(x)$.
Now if $C = \langle g \rangle$ is reversible, it is obvious from the above argument that $g(r) = 0 \iff g\left(\dfrac 1 r\right) = 0$. Conversely if $g(r) = 0 \iff g\left(\dfrac 1 r\right) = 0$, and $r$ and $\dfrac 1 r$ have the same multiplicities, then again it is clear that $x^{n - 1}g\left(\dfrac 1 x\right) \in C$. Thus, $C$ is reversible.
