 Solution

Choose a generator $u$ of the cyclic group $U(R)$, so $u^p=1$. Part (a)(iii) gives a unital ring homomorphism ${\mathbb{F}}_2\to R$. Its image lies in the centre of $R$, so even if $R$ is not a commutative ring, evaluating a polynomial at $u$ defines a ring homomorphismBecause $\varphi (X^p-1)=0$, it factors through the quotient ring to giveIts image is the subring $S={\mathbb{F}}_2[u]$. Every unit of $S$ is a unit of $R$, because an inverse in $S$ also lies in $R$. Conversely, every member of $U(R)$ is $u^j$ for some $j$, and both $u^j$ and its inverse $u^{p-j}$ lie in $S$. HenceThe inverse argument matters: merely belonging to a subring would not by itself make an ambient unit invertible there.