I have the following question: "Is $U(\mathbb{Z}_2[x]/\langle x^{4}\rangle )$ a cyclic group ."
Attempt: I managed to show that it isn't a cyclic group by writing down all of its elements and computing their order.
Is there a quicker way to determine such questions? Let's say for an arbitrary $p(x)$ instead of $x^{4}$.
On
If $k$ is a field, $k[X]/(X^n)$ is local with maximal ideal $(X)$, and the units are the complement of this ideal, namely, those powerseries with nonzero constant term, which in your case is forcefully $1$.
This has cardinality $8$, since there are $2$ choices for the coefficient of $X$, $2$ for that of $X^2$, and $2$ for that of $X^3$. I claim every element has order $4$, which shows that the group cannot be cyclic. Indeed, the ring is of characteristic $2$, so if $X\mid p$, $(1+p)^4 = 1+p^4 = 1$.
I think it is not a cyclic group, since arbitrary element “y” is a class of atmost cubic polynomial, so if raised the 4th power of y, you get the element in ring Z_{2}.