Possible Duplicate:
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
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.
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.
