Vtyjkyuk

Let $F$ a field and $E$ a finite extension of $F$. I need to show that, if $D$ is a integral domain such that $Fsubset Dsubset E$ then, D is a field.

My attempt: As $E$ is a finite extension, it is a algebraic extension. Let $alphain D$, it is sufficient to show that each element of $D$ has a inverse. So, take $alphain D$ so $alpha$ is algebraic over $F$. Suppose that $deg(alpha,F)=n$, then every element of $F(alpha)$ can be written as $a_0+a_1alpha+cdots+a_n-1alpha^n-1$ where $a_iin F$. 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.

$D$ is a finite-dimensional vector space over $F$. If $ain D$, $ane0$
consider the map $mu_a:Dto D$ 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 $bin D$ with $mu_a(b)=1$, that is $ab=1$.

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

$alpha displaystyle sum_1^t m_i alpha^i - 1 = -m_0 ne 0; tag10$

this shows that

$alpha^-1 = -m_0^-1displaystyle sum_1^t m_i alpha^i - 1 in D, tag11$

hence $D$ is a field.

Well every subring of a field is an integral domain anyway.