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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.

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    $\begingroup$ This is false for $R=\mathbb {Q}$. Maybe you want $R $ to be finite? $\endgroup$ – darij grinberg Dec 2 '18 at 7:42
  • $\begingroup$ The maximum number of roots depends on being an integral domain or not. If $R$ is an integral domain with fraction field $K = Frac(R)$ and $f \in R[t], f(\alpha) = 0$ then $\frac{f(t)}{t-\alpha} \in K[t]$ so $|\{\alpha \in R : f(\alpha)=0\}| \le |\{\alpha \in K : f(\alpha)=0\}| \le \deg(f)$. If $R$ is not an integral domain then ... $\endgroup$ – reuns Dec 2 '18 at 7:56

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