# Cyclic multiplicative subgroups

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.

• This is false for $R=\mathbb {Q}$. Maybe you want $R$ to be finite? – darij grinberg Dec 2 '18 at 7:42
• 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 ... – reuns Dec 2 '18 at 7:56