Let $m(x)=\prod(x-r_i)$ . Why for every $i$ we have an automorphism of $F'/F$ which sends $r$ into $r_i$? Let $D$ be a domain, $F$  its field of fractions, $E$ a finite dimensional extension of $F$ and $D'$ the subring of $E$ of $D$-integral elements and assume that $D$ is integrally closed. Let $r$ be in $E$ and let $m(x)$, $f(x)$ be the minimum polynomial and the field polynomial of $r$ over $F$. I have to show that if $r$ is integral over $D$, then $m(x)$ is in $D[x]$.
Let $F'$ be an algebraic closure of $F$ containing $E$. We have $m(x)=\prod(x-r_i)$ in $F'[x]$ where $r_1=r$. Since $r$ is $D$-integral, then there exists a monic polynomial $g(x)$ in $D[x]$ such that $g(r)=0$. 
I do not understand why for every $i$ we have an automorphism of $F'/F$ which sends $r$ into $r_i$?
Would you help me, please?
Thank you in advance.
 A: Let us first prove the following:

Lemma 1: If $K$ is a field, $K(\alpha)$ is a simple algebraic field extension and $f$ is the minimal polynomial of $\alpha$ over $K$, then for any field extension $L / K$ and $\beta \in L$ such that $f(\beta)=0$, there exists a field homomorphism $\sigma: K(\alpha) \to L$ such that $\sigma(\alpha)=\beta$ and $\sigma$ leaves $K$ pointwise fixed.

Proof: We have $K[x]/(f) \cong K(\alpha)$ via $\bar{x} \mapsto \alpha$ by the first isomorphism theorem. Now consider the evaluation map $K[x] \to L, g \mapsto g(\beta)$ clearly the kernel of this map is $(f)$, so we get an induced homomorphism $K[x]/(f) \to L$ that sends $\bar{x}$ to $\beta$. Composing with the isomorphism $K(\alpha) \cong K[x]/(f)$ gives us the desired $\sigma$.

Lemma 2: If $K' / K$ is an algebraic extension and $L$ is an algebraically closed field and $\sigma: K \to L$ is a field homomorphism, then there exists an extension of $\sigma$ to $K'$

Proof: Consider the set
$\Sigma := \{(F,\tau)|K \subset F \subset K', F$ is an intermediate field, $\tau$ is an extension of $\sigma\}$
We impose a partial order on $\Sigma$ given by $(F,\tau) \leq (F',\tau')$ iff $F \subset F'$ and $\tau'_{|F} = \tau$
It is obvious that this is a partial order and $\Sigma$ is not empty because $(K,\sigma) \in \Sigma$.
If we have a completely ordered chain $(F_1,\tau_1) \leq (F_2,\tau_2) \leq \dots$, then we can set $F = \bigcup_{i=1}^n F_i$ and define $\tau:F \to L$ in the obvious way. Then it is easy to see that $(F,\tau)$ is in $\Sigma$ and is an upper bound for the chain.
By Zorn's lemma, we have a maximal element $F_0,\tau_0$. I claim that $F_0=K'$. Indeed, if there is some $\alpha \in K'$ such that $\alpha \notin F_0$, then let $f$ be the minimal polynomial of $\alpha$ over $F_0$. As $\tau_0$ is injective, we may regard $F_0$ as a subfield of $L$. Now as $L$ is algebraically closed, $f$ has a root in $L$, thus
by lemma 1, we can extend $\tau_0$ to $F_0(\alpha)$ contradicting the maximality of $F_0$. Thus $F_0 = K'$ and $\tau_0$ is the desired extension.

Lemma 3: If $L /K$ is an algebraic extension, and $\sigma: L \to L$ is a field homomorphism that leaves $K$ point-wise fixed, then $\sigma$ is surjective, thus an automorphism.

Proof: Let $\alpha \in L$ and $f$ the minimal polynomail of $\alpha$ over $K$, then for any root $\beta$ of $f$, $\sigma(\beta)$ is again a root of $f$, as $\sigma$ fixes $K$. Let $S$ be the set of roots of $f$ over $L$. $S$ is a finite set, bounded by the degree of $f$. Now $\sigma$ is a field homomorphism, thus it is injective, so we if we restrict $\sigma$ to $S$, then $\sigma$ is still injective, but $\sigma(S) \subset S$, so $\sigma_{|S}:S \to S$ is an injective map from a finite set to itself, thus $\sigma_{|S}$ is bijective, thus there exists $\beta \in S \subset L$, such that $\sigma(\beta) = \alpha$.
Now we are done, we just have to put everything together: if $F'/F$ is an algebraic closure, $r_1, r_2$ are roots of the same irreducible polynomial over $F$, then by lemma 1, we have a homomorphism $F(r_1) \to F'$ that sends $r_1$ to $r_2$ and fixes $F$. By lemma 2, we can extend this homomorphism to a $F$-homomorphism $F' \to F'$ and lemma 3 shows that this extension is actually an automorphism.
