$u^{-1}$ is integral over $R\subset S$ if and only if $u^{-1}\in R[u]$

Question

I need to prove the following:

Let $R\subset S$ be a commutative ring and $u$ be any invertible element in $S$. Then $u^{-1}$ is integral over $R$ is and only if $u^{-1}\in R[u].$

Proof: Let $u^{-1}$ is integral over $R \implies \exists\ $ a monic polynomial $f(x)\in R[x]$ such that $f(u^{-1})=0$. If \begin{align*} f(x) &=x^n+r_{n-1}x^{n-1}+\ldots+r_1x+r_0\\ \implies f(u^{-1}) &=u^{-n}+r_{n-1}u^{-n+1}+\ldots+r_1u^{-1}+r_0=0\\ \implies &u^{-1}+r_{n-1}+r_{n-2}u+\ldots+r_1u^{n-2}+r_0u^{n-1}=0 \end{align*} Therefore, $u^{-1}\in R[u]$.

Now how to do the converse.

Answer 1

Suppose $u^{-1} \in R[u]$. Then there exists some polynomial $f(x) \in R[x]$ such that $u^{-1}=f(u)$, or equivalently $$uf(u)-1=0$$ Write $f(x)= \sum_{i=0}^n a_ix^i$. Then $$0=\sum_{i=0}^n a_iu^{i+1}-1 = u^{n+1} \cdot (u^{-(n+1)}+a_0u^{-n} + a_1 u^{-(n-1)}+ \dots + a_n)$$ Since $u^{n+1}$ is a unit, you have $$u^{-(n+1)}+a_0u^{-n} + a_1 u^{-(n-1)}+ \dots + a_n=0$$ which is an integral relation of $u^{-1}$ over $R$.

Answer 2

It is almost the same thing: if $u^{-1}\in R[u]$ then there exists $a_0,\cdots,a_n\in R$ such that $$u^{-1} = a_0 + a_1u + \cdots + a_n u^n.$$

Multiplying by $(u^{-1})^n$ we get $$(u^{-1})^{n+1}= a_0(u^{-1})^{n} + \cdots a_{n-1}u^{-1}+a_n.$$

Thus $f(u^{-1}) = 0$, where $f(x) = x^{n+1} - a_0 x^n - \cdots - a_{n-1} x + a_n$, which means $u^{-1}$ is integral over $R$.