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Let $A\subset B$ be an integral extension. If $F$ and $E$ are the fields of fractions of $A$ and $B$, respectively, I want to show that $E$ is an algebraic extension of $F$.

I know that since $A \subset E$, it follows that $F \subset E$. Now I consider an arbitrary element $b_1/b_2\in E$. I would like to show that $b_1/b_2$ is algebraic over $F$. Since $B$ is algebraic over $A$, there exist monic polynomial $f_1(x), f_2(x)\in A[x]$ such that $f_1(b_1) = f_2(b_2) = 0$. But how do I obtain a polynomial $f(x) \in F[x]$ such that $f(b_1/b_2) = 0$?

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It is useful to simplify your work here (and there are many ways to do so). Here is one possible thing to do:

  • Show that elements of the form $b/1$ and $1/c$ are algebraic over $F$.
  • Note then that product of two elements algebraic over $F$ is algebraic over $F$. And, this completes the proof (why?).
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  • $\begingroup$ $b\in B$. There is $a_1,\cdots,a_n\in A$ st $b^n+a_1b^{n-1}+\cdots+a_n=0$. Taking $a'_i=\dfrac{a_i}{1}\in S^{-1}A$, we've $\left(\dfrac{b}{1}\right)^n+a'_1\left(\dfrac{b}{1}\right)^{n-1}+\cdots+a'_n=\dfrac{b^n}{1}+\dfrac{a_1b^n-1}{1}+\cdots+ \dfrac{a_n}{1}=\dfrac{b^n+a_1b^{n-1}+\cdots+a_n}{1}=\dfrac{0}{1}$. So, $\dfrac{b}{1}$ is algebraic over $Frac(A)$. Now, take $c\in B-\{0\}$. There's $a''_1,\cdots,a''_n\in A$ st $c^n+a''_1c^{n-1}+\cdots+a''_n=0$, $a''_n\neq 0$. $\endgroup$ Dec 5, 2019 at 17:14
  • $\begingroup$ Then, $1+a''_1\dfrac{1}{c}+\cdots+a''_n\dfrac{1}{c^n}=\dfrac{0}{c^n}=\dfrac{0}{1}$, therefore $\dfrac{1}{a''_n}+\dfrac{a''_1}{a''_n}\dfrac{1}{c}+\cdots+\dfrac{1}{c^n}=\dfrac{0}{a''_n}=\dfrac{0}{1}.$ Each $\dfrac{a''_i}{a''_n}$ is in $Frac(A)$, then $\dfrac{1}{c}$ is algebraic over $Frac(A)$. $\endgroup$ Dec 5, 2019 at 17:14
  • $\begingroup$ please, is that way that you expect? Many thanks in advance! $\endgroup$ Dec 5, 2019 at 17:14

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