Let $F/K$ be an algebraic field extension. Suppose $D$ is an integral domain with $K\subset D\subset F$. Show that $D$ is a field.

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    $\begingroup$ An algebraic field extension is not necessarily of finite degree; you cannot conclude that $\dim_KF=d$. It is also not necessary that $a^p\in D$ for any $p>0$. $\endgroup$ – Servaes Apr 8 '17 at 15:30
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    $\begingroup$ $D = K[a_1,\ldots,a_n]$ where the $a_i$ are algebraic, and $ K[a_1,\ldots,a_n]= K(a_1,\ldots,a_n)$ by the fundamental theorem $K[x]/(f(x)) \cong K[a]$ where $f$ is the minimal polynomial of $a$. $\endgroup$ – reuns Apr 8 '17 at 15:34
  • $\begingroup$ Sorry I mistake algebraic extension with simple extension... $\endgroup$ – Ivon Apr 8 '17 at 15:37

If $a\in D\setminus\{0\}$ then $D$ contains the $K$-algebra $K[a]$, and because $a$ is algebraic over $K$ we have $K[a]=K(a)$, therefore $a$ has an inverse in $D$.

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  • $\begingroup$ It reduces to show that any $a \in F$ is algebraic over $K$ $\endgroup$ – reuns Apr 8 '17 at 15:41
  • $\begingroup$ Exactly! The use of K[a] = K(a) is genius! $\endgroup$ – Ivon Apr 8 '17 at 15:42
  • $\begingroup$ @Ivan : How would you show $a$ is algebraic ? $\endgroup$ – reuns Apr 8 '17 at 15:43
  • $\begingroup$ @user1952009 a field extension L/K is called algebraic if every element of L is algebraic over K. $\endgroup$ – Ivon Apr 8 '17 at 15:44
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    $\begingroup$ @Ivan : I meant you need to show that the sum, product and quotient of algebraic numbers is algebraic, using the resultant or the adjugate matrix $\endgroup$ – reuns Apr 8 '17 at 15:58

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