# Show that any subring $R\subseteq L$ is a field if $K\subseteq L$ is module-finite field extension and $K\subseteq R$.. [duplicate]

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.

• I presume $K$ is a field. In that case, let $r\in R$ and consider the elements $1,r,r^2,\ldots$. – Shivering Soldier Jan 16 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 at 19:19

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 $$y^{-1} \in R$$.