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The definition of a field extension $K'/K$ is that there exists a field homomorphism $\phi: K \rightarrow K'$. Field morphism implies injectivity. Is not required that $K$ is actually contained in $K'$. My question is, if $E/F$ is a field extension and $F/E$ is a field extension, then is $E \simeq F $? I know that the question is true if $E/F$ is algebraic.

Werner Germán Busch
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  • But to say $E/F$ is a field extension means in particular that $F\subset E$, right? – Ted Shifrin Dec 03 '15 at 22:42
  • Not necessarily because the definition only involves a field morphism from $F$ to $E$ you then identify $E$ with the image via that morphism. – Werner Germán Busch Dec 03 '15 at 22:48
  • Ah, sorry. So we can have a field monomorphism $F\to F$ that is not an isomorphism. What if $F=k(X_1,X_2,\dots)$? – Ted Shifrin Dec 03 '15 at 22:59
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    This is the same question as this one (so the answer is no): http://math.stackexchange.com/questions/1015850/can-a-field-be-isomorphic-to-its-subfield-but-not-to-a-subfield-in-between?lq=1 –  Dec 04 '15 at 03:37
  • @ChristianRemling: I had in mind the map $F\to F$ sending $X_i$ to $X_{i+1}$ for all $i$. Definitely a monomorphism and not an isomorphism. So the answer to the original question is indeed no. – Ted Shifrin Dec 04 '15 at 21:25

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Thanks to André Nicolas for providing the counter-example and answering questions, and Christian Remling for pointing to the answer, I will write the details:

First we contruct fields $K \subseteq E \subseteq F$ over $\mathbb{Q}$ such that $K$ is isomorphic to $F$ but $E$ is not.

An example is: take $t_0, t_1, ...$ countable and algebraically independient transcendental constants over $\mathbb{Q}$. Then $K := \overline{\mathbb{Q}(t_1, t_2, t_3,...)} \subseteq E := K(t_0) \subseteq F := \overline{\mathbb{Q}(t_0, t_1, t_2,...)} $ work:

$\mathbb{Q}(t_0, t_1, t_2,...)$ is isomorphic to $ \mathbb{Q}(t_1, t_2, t_3,...)$ via

\begin{align}\mathbb{Q}(t_0, t_1, t_2,...) &\overset{\phi}{\longrightarrow} \mathbb{Q}(t_1, t_2, t_3,...)\\ t_i &\longrightarrow t_{i+1} \ (\forall i \geq \ 0 )\\ & \phi |_\mathbb{Q} = Id_\mathbb{Q} \end{align}

so their algebraic closures $F$ and $K$ are isomorphic. But $K$ is not isomorphic to $E$. Suppose there was an isomorphism $\psi : K \rightarrow E$. Let $t :=\psi^{-1}(t_0)$. Then $P(x)= x^2-t$ has a root in $K$ (because $K$ is algebraically closed). So there exists $\omega \in K$ such that $ P(\omega) =0$. Let $\omega_0 = \psi(\omega) \in E$ and thus applying $ \psi$ on both sides

\begin{align} \psi(P(\omega))=0\\ \psi(\omega^2-t) = 0\\ \psi(\omega)^2-\psi(t) = 0\\ \omega_0^2-t_0 = 0 \end{align}

Since $\omega_o \in E, \omega_0 = \frac{Q_1(t_0)}{Q_2(t_0)}, Q_1(x), Q_2(x) \neq 0 \in K[X]$

So

\begin{align} \omega_0^2 = t_0\\ \frac{Q_1(t_0)^2}{Q_2(t_0)^2} = t_0\\ \ Q_1(t_0)^2 = t_0.Q_2(t_0)^2\\ 2.\deg_{t_0}(Q_1)= 2.\deg_{t_0}(Q_2) + 1 \end{align}

Absurd.

So $K$ and $E$ are not isomorphic.

Finally, $E/K$ is a field extension with the inclusion homomorphism and $K/E$ is a field extension with field homomorphism $ E \hookrightarrow F \overset{\cong}{\underset{\tilde{\phi}}{\longrightarrow}} K$ but $E \ncong F$

Werner Germán Busch
  • 803