isomorphism between field extension Assume F is a field, E/F and H/F are two field extension. If E is isomorphic to H, then whether exist a isomorphism $\varphi:E \rightarrow H$ such that $\varphi|_F = Id_F$? I think it's worry but I can find some counterexamples. If F = Q, then it is true obviously, but other field? I think it maybe true if E/F is finite extension. Maybe someone give me some references or hints, thanks.
 A: Take $\beta=\sqrt[4]2\in\mathbb R$, so that $\beta^2=\sqrt2>0$. Let $F=\mathbb Q(\sqrt2)$, $E=\mathbb Q(\beta)$ and $H=\mathbb Q(i\beta)$. Then $E$ and $H$ both contain $F$. They are also isomorphic, as they are both isomorphic to $\mathbb Q[X]/(X^4-2)$. In fact, there are precisely two isomorphisms $E\to H$, given by $\beta\mapsto\pm i\beta$ (the two possible roots of $X^4-2$ in $H$), so both send $\sqrt2=\beta^2$ to $(\pm i\beta)^2=-\sqrt2$, and so neither is the identity on their common subfield $F$.
A: A rather general way of constructing an example is to take a Galois extension $E/K$ which admits an intermediate field $F$ such that $F/K$ is not Galois. In other words, there is $\varphi \in \operatorname{Gal}(E/K)$ such that $\varphi(F) \ne F$. Now just take $H = E$.
For instance you may take $K = \mathbb{Q}$, $E = H = K(\sqrt[3]{2}, \omega)$, where $\omega$ is a primitive third root of unity. $E/K$ is Galois, as $E$ is the splitting field over $K$ of the polynomial $x^{3} - 2$. Let $F = K(\sqrt[3]{2})$. There is an element $\varphi$ of $\operatorname{Gal}(E/K)$ such that $\varphi(\sqrt[3]{2}) = \omega \sqrt[3]{2}$, so that $\varphi(F) = K(\omega \sqrt[3]{2}) \ne F$.
A: Let $F=\mathbb{Q}(X)$, $E=F, H=\mathbb{Q}(Y)$. Now we consider $F\subset H$ by $X\mapsto Y^2$.
Then $E$ and $H$ is isomorphic as fields by $\mathbb{Q}(X)\to \mathbb{Q}(Y)$ defined by $X\mapsto Y$. (restriction on $F$  of this map is not id$_F$)
And there is no isomorphism $\phi:E\to H$ s.t. $\phi|_F=$id$_F$.
In fact if $\phi|_F=$id$_F$, then $\phi(X)=Y^2$ and $Y$ is not in Im$\phi$. Hence $\phi$ is not isomorphic.
