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This is Exercise 1.50b in Fulton's Algebraic Curves. It's a follow up to showing that algebraic elements form a subfield. I found similar questions MSE, but with stronger hypothesis.

By module-finite extension we mean that $L$ is finitely-generated as a $K$-module.

My attempt: take $r\in R$ and let $\rho\in L$ be its inverse in the field $L$. I tried to show that $\rho \in R$, since we know that will happen.

By $K\subset L$ a finitely generated extension we mean that there are $l_1,\dots l_n$ in $L$ such that $r\in R$ can written as $r = l_1k_1+\dots+l_n k_n$.

When $r = l_jk_j$ we have $l_j = rk_j^{-1}\in L$, so $\rho = k_j^{-1}l_j^{-1}\in R$. This strategy blowed off really hard when the above equation for $r$ had two or more terms.

Daniel Teixeira
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  • I presume $K$ is a field. In that case, let $r\in R$ and consider the elements $1,r,r^2,\ldots$. – cqfd Jan 16 '21 at 17:38
  • This is false unless $K\subset R$: consider $K=L=\Bbb Q$, $R=\Bbb Z$. If you assume $K\subset R$, then this is a duplicate of this. – KReiser Jan 16 '21 at 19:19

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Proof: Note that since $L$ is a module finite extension of $K$, we have that $L$ is a finite field extension of $K$ which implies that $L$ is algebraic over $K$. Now, let $y \in R$. In particular, $y \in L$ which gives us that $y$ is algebraic over $K$. Let $P(x)=x^n+c_{n-1}x^{n-1}+ \dots c_{1}x+c_{0}$ be the minimum polynomial for $y$ over $K$. We then have $y^{n} +c_{n-1}y^{n-1} + \dots c_{1}y+c_{0}=0$, which gives $y^{n}+c_{n-1}y^{n-1} + \dots + c_{1}y=-c_{0}$. This implies, $y^{n-1}+c_{n-1}y^{n-2}+ \dots + c_{1} = y^{-1}c_{0}$. Clearly, $y^{-1}c_{0} \in R$ and since $c_{0}^{-1} \in R$, we get $y^{-1} \in R$.

mountainman1
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