## Finite Extensions

My book gave a problem, If $\displaystyle F,E$ are fields and $\displaystyle E$ can be regarded as a finite extension over $\displaystyle F$. And $\displaystyle F\subseteq D\subseteq E$ where $\displaystyle D$ is an integral domain. Then prove that $\displaystyle D$ is a field.

This is how I proved it, first $\displaystyle D$ is a commutive ring with unity, thus if it can be shown that every element is a unit in $\displaystyle D$ then it is a field.
(The obvious way to prove this problem).

1)If $\displaystyle x\in D$ then $\displaystyle x\in E$.
2)$\displaystyle E$ is finite extension over $\displaystyle F$.
3)Thus $\displaystyle E$ is algebraic extension over $\displaystyle F$.
4)Thus, $\displaystyle F(x)$ is finite extension over $\displaystyle F$
5)Thus $\displaystyle x\in F(x)$ has inverse $\displaystyle x^{-1}$ because it is a field.
6)Thus, $\displaystyle x^{-1}=a_0+a_1x...+a_nx^n,a_i\in F$.
7)Thus, $\displaystyle x^{-1}\in D$ since $\displaystyle a_i,x\in D$ and $\displaystyle D$ are closed.
8)Thus, $\displaystyle D$ is a division ring.
Q.E.D.

But my problem is that the fact that $\displaystyle D$ has no zero divisors was never used in the proof. Thus, this is true for any commutative ring with unity.
Am I missing something?