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Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field

Let $R$ be an integral domain containing a field $K$ as a subring. Suppose that $R$ is a finite dimensional vector space over K under the ring multiplication. Show that $R$ is a field.

I really have no idea how I suppose to attack this problem. Some sketch to the solution is needed. Thanks

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    @Mark: In the preview of your reply, I saw "It is that time of year..." and thought you were going to say something about Xmas... Alas! :)2011-10-30

2 Answers 2

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Since $R$ is an integral domain, you only need to show every nonzero element is invertible. Let $x\in R$ be nonzero. Then $x$ induces a map $\phi_x\colon R\to R : r\mapsto xr$, which is injective since $R$ is an integral domain. Thus $\ker\phi_x=\{0\}$, so $\dim\ker\phi_x=0$. Note that $\phi_x$ is a homomorphism, since for any $c\in K$, $r,s\in R$, $ \phi_x(cr+s)=x(cr+s)=x(cr)+xs=c(xr)+xs=c\phi_x(r)+\phi_x(s). $ By Theorem 5.3 of Lang, $\dim\operatorname{Im}(\phi_x)=\dim R$. By Corollary 5.4, $\operatorname{Im}(\phi_x)=R$, so $\phi_x$ is surjective. So there exists some $y\in R$ such that $\phi_x(y)=xy=1$, and $x$ is invertible, and $R$ is a field.

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Let $x \in R$. Then $x, x^2, .., x^n,..$ cannot be linearly independent over $K$, thus

$a_{i_1}x^{i_1}+a_{i_2}x^{i_2}+...+a_{i_n}x^{i_n} =0 \,.$

with all $a_{i_k} \in K^*$, and the powers in strictly increasing order.

This means that

$a_{i_2}x^{i_2-i_2}+a_{i_3}x^{i_3-i_1}+...+a_{i_n}x^{i_n-i_1} =-a_{i-1} \,.$

You can now multiply by the inverse of $-a_{i-1}$ and make an $x$ common factor.