Let $R[t]$ be the polynomial ring over the nonzero commutative ring $R$ and $f_R$ be the associated polynomial function.
If $|\{\alpha \in R : f_R(\alpha)=0\}| \leq \deg(f)$ for every $0 \neq f \in R[t]$, show that $R \times$ is isomorphic to a cyclic group.
Theorem $7.12$ (Fundamental theorem of algebra) Over any field $F$, a monic polynomial $f(x) \in F[x]$ of degree $m$ can have no more than $m$ roots in $F$. If it does have $m$ roots $\{ \beta_1, \ldots, \beta_m \}$, then the unique factorization of $f(x)$ is $f(x) = (x-\beta_1) \cdots (x-\beta_m)$. Since the polynomial $x^n -1$ can have at most $n$ roots in $F$, we have an important corollary:
Theorem $7.13$ (Cyclic multiplicative subgroups) In any field $F$, the multiplicative group $F^*$ of nonzero elements has at most one cyclic subgroup of any given order $n$. If such a subgroup exists, then its elements $\{1, \beta, \dots, \beta^{n-1}\}$ satisfy $x^n -1 = (x-1)(x-\beta)\cdots(x-\beta^{n-1})$.
It seems that I can get the result directly if I have $|\{\alpha \in R : f_R(\alpha) = 0\}| = \deg(f)$, which occurs only when $|\{\alpha \in R : f(\alpha) = 0\}| \geq \deg(f)$.
Is this direction correct? I have no idea how to prove the last inequality, though.