Let $S/R$ be a ring extension and $\beta \in S^*$. Show that for all $\alpha \in R[\beta]\cap R[\beta^{-1}]$ is integral over R.

Question

I don't really know where to start, I've written $\alpha$ as a polynomial evaluated in $\beta$ and in $\beta^{-1}$ but there doesn't seem to be much information there.

Answer 1

• If $\beta$ is transcendental over $R$ then the statement is trivial as $R[x] \cap R[x^{-1}] = R$.

• Thus let $f(x) = \sum_{n=0}^{d} b_n x^n \in R[x]$ be the minimal polynomial of $\beta$ so that $\beta = -\frac{1}{b_d} \sum_{n=0}^{d-1} b_n \beta^{-(d-1-n)}$ $$\alpha= \sum_{m\ge 0} a_m \beta^{m}= \sum_{m\ge 0} \frac{a_m}{b_d^m} (-\sum_{n=0}^{d-1} b_n \beta^{-(d-1-n)})^m$$ $\alpha \in R[\beta]$ means $a_m \in R$ and $\alpha \in R[\beta^{-1}] $ means $\frac{a_m}{b_d^m} \in R$. Thus $\alpha \in R[b_d \beta]$ where $b_d \beta$ is integral over $R$ since $\sum_{n=0}^{d} b_n b_d^{d-1-n} (b_d\beta)^{n}=0$.