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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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marked as duplicate by Dylan Moreland, Jack Schmidt, lhf, Zev Chonoles Oct 30 '11 at 19:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This comes up quite often. Searching for "domain vector space field" yields this question. – Dylan Moreland Oct 30 '11 at 18:56
Two upvotes?? Curious... – The Chaz 2.0 Oct 30 '11 at 19:31
@TheChaz It is that time of year when people are getting into courses which deal with this kind of problem, and are coming at it fresh, and don't realise that it was (inevitably) asked last year, and will inevitably be asked again in some equivalent form. – Mark Bennet Oct 30 '11 at 21:03
@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! :) – The Chaz 2.0 Oct 30 '11 at 23:04

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.

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