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I would like verification of my proof of the above claim. That is, I am trying to prove that if $K\subseteq\overline{F}$ is a splitting field for $\{f_i\}_I$ over $F,$ where $\overline{F}$ is an algebraic closure, then given $\sigma:K\to\overline{F}$ an isomorphism leaving $F$ fixed, it follows that $\sigma$ is an automorphism. The proof I am trying to verify uses Zorn's lemma. A lemma which is quickly becoming a favorite of mine. The part where it says 'previous result' is quoting a result that was done by my professor, which said that if $\sigma:E\to E$ is an embedding leaving $F$ fixed, then it is actually an automorphism. This result holds for any extension $E$ of $F$

$\textbf{Proof:}$

Let $K$ be a splitting field for $\{f_i\}_I\subseteq F[x]$ over $F$. Let $\overline{F}$ be an algebraic closure for $F.$

let $$Z=\{K'\subseteq K:\text{ every isomorphism of }K'\text{ onto a subfield of }\overline{F}\text{ leaving}$$ $$F\text{ fixed is an automorphism}\}.$$ We index as usual by inclusion, and assume that $S\subseteq Z$ is a chain. Now we consider $\bigcup S.$ If $a,b,c\in\bigcup S,$ then there exists a field $E\in S$ with $a,b,c\in E$, so all our field properties hold for $a,b,c$ in $E,$ and hence in $\bigcup S,$ so $\bigcup S$ is a field. Now let $\tau$ be an embedding of $\bigcup S$ into $\overline{F}$ leaving $F$ fixed. Then $\tau|_E$ an an automorphism of $E$ as $E\in Z,$ therefore $\tau(a)\in E\subseteq\bigcup S,$ hence since $a$ was arbitrary $\tau$ is actually an embedding of $\bigcup S$ into $\bigcup S.$ Then by a previous result it follows that since $\tau $ leaves $F$ fixed that $\tau$ is actually an automoprhism. Thus every embedding of $\bigcup S$ into $\overline{F}$ leaving $F$ fixed is an automorphism, hence $\bigcup S\subseteq Z$ is an upper bound for $S.$ By Zorn's Lemma let $M\in Z$ be maximal. We wish to show that $M=K.$

Suppose $M\not= K.$ Since $K$ is a smallest subfield of $\overline{F}$ such that $\{f_i\}_I$ all split in $K$ it follows that some $f_i$ must not split in $M.$ Let $f_i=c(x-\beta_1)...(x-\beta_m).$ Now we consider $M'=M(\beta_1,...,\beta_m).$ Let $\rho$ be an embedding of $M'$ into $\overline{F}.$ Since $M\in Z$ we find that $$\rho|_M:M\to M.$$ As products of powers of $\beta_1,...,\beta_m$ form a basis for $M'$ over $M$ is we show $\rho(\beta_j)\in \{\beta_1,...,\beta_m\}$ for all $j,$ then we'll know $\rho$ is an into map. As $\rho(\beta_j)$ is a root of $\rho(f_i)=f_i$ (F is fixed). So $$\rho(\beta_i)\in\{\text{roots of }f_i\}=\{\beta_1,...,\beta_m\}.$$ it follows that $\rho$ is an embedding from $M'$ into $M'$ leaving $F$ fixed, hence is an automorphism. Well this contradicts the maximality of $M,$ therefore $M=K.$ This completes the proof. $\blacksquare$

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