If $F\subset D \subset E$ then $D$ must be a field Let $F$ a field and $E$ a finite extension of $F$. I need to show that, if $D$ is a integral domain such that $F\subset D\subset E$ then, D is a field.
My attempt: As $E$ is a finite extension, it is a algebraic extension. Let $\alpha\in D$, it is sufficient to show that each element of $D$ has a inverse. So, take $\alpha\in D$ so $\alpha$ is algebraic over $F$. Suppose that $\deg(\alpha,F)=n$, then every element of $F(\alpha)$ can be written as $a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}$ where $a_i\in F$. Therefore, $F(\alpha)\subset D$ and as $F(\alpha)$ is a field and it contains the inverse of $\alpha$.
I have troubles with my proof because a don't see where the fact that $D$ is integral domain is important. 
 A: Take
$0 \ne \alpha \in D, \tag 1$
and let
$m(x) \in F[x] \tag 2$
be the minimal polynomial of $\alpha$ over $F$; writing
$m(x) = \displaystyle \sum_0^t m_i x^i, \; m_i \in F, 0 \le i \le t, \tag 3$
I claim that 
$m_0 \ne 0; \tag 4$
otherwise, we could write
$m(x) = \displaystyle \sum_1^t m_i x^i = x \sum_1^t m_i x^{i - 1}; \tag 5$
since
$m(\alpha) = 0, \tag 6$
we have
$\alpha \displaystyle \sum_1^t m_i \alpha^{i - 1} = 0; \tag 7$
since $\alpha \ne 0$, and $D$ is a integral domain, we conclude that
$\displaystyle \sum_1^t m_i \alpha^{i - 1} = 0; \tag 8$
since the degree of the polynomial $\sum_1^t m_i x^{i - 1}$ is $t - 1$, (8) contradicts the minimality of $m(x)$; thus we see that
$m_0 \ne 0, \tag 9$
which allows us to write
$\alpha \displaystyle \sum_1^t m_i \alpha^{i - 1} = -m_0 \ne 0; \tag{10}$
this shows that
$\alpha^{-1} = -m_0^{-1}\displaystyle \sum_1^t m_i \alpha^{i - 1} \in D, \tag{11}$
hence $D$ is a field.
A: $D$ is a finite-dimensional vector space over $F$. If $a\in D$, $a\ne0$
consider the map $\mu_a:D\to D$ defined by $\mu_a(x)=ax$. Then $\mu_a$
is an $F$-linear map. Its kernel is zero, so its image is all of $D$, by the
rank nullity formula. There is $b\in D$ with $\mu_a(b)=1$, that is $ab=1$.
A: Well every subring of a field is an integral domain anyway.
