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My book (Fraleigh) states the following:

Assumption: All algebraic extensions of $F$ are assumed to be contained in a fixed algebraic closure $\overline{F}$ of $F$.

Isomorphism Extension Theorem- Let $E$ be an algebraic extension of field $F$. Let $\sigma$ be an isomorphism of $F$ onto a field $F'$. Let $\overline{F'}$ be an algebraic closure of $F'$. Then...

As per the assumption, there exists an extension $H$ of field $F$ in which an algebraic closure of $F$ is contained. Similarly, there must be a similar field extension for $F'$, inside which $\overline{F'}$ is contained. What field extension is this. Can we say it is isomorphic to $H'$? How?

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  • $\begingroup$ "There exists an extension $H$ of field $F$ in which the algebraic closure of $F$ is contained" No, the algebraic closure need not be contained in $H$, that's not what you're told. $\endgroup$ – Git Gud Apr 28 '14 at 10:13
  • $\begingroup$ @GitGud- Oh god sorry for the mess. I suppose $E$ is contained inside an algebraic closure of $F$, and and that closure is probably isomorphic to $\overline{F'}$. $\endgroup$ – algebraically_speaking Apr 28 '14 at 10:20

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