I want to show the following:

Let $L|K$ be an algebraic field extension of $K$. Let $T$ be transcendental over $K$. Then $$ K(L)(T) = K(T)(L). $$

We defined the adjunction $K(A)$ of a subset $A \subseteq E$ to be the smallest field which contains $A$ and extends $K$, where $E|K$ is a field extension. In other words

$$ K(A) = \bigcap_{K \subseteq M \subseteq E} M, $$ where $M$ is a field and $A \subseteq M$.

My question is: Is there an elegant proof? And how does it look like?

  • $\begingroup$ The elegant thing to do would probably to state the theorem without the assumption that $L$ or $T$ are any particular supersets of $K$. Otherwise, it's just straightforward. So: Well done! (At least, as good as possible.) $\endgroup$ – AlgebraicsAnonymous Feb 8 at 14:33
  • $\begingroup$ Ok it took me some time to figure out what you mean, but now I see it. So actually I want $T \notin \overline{K}$ or at least I don't want to assume that it lies in $\overline{K}$. $\endgroup$ – lugggas Feb 9 at 11:42

I think a found it by my self.

Let $\mathcal{M} := \{ M|L : T \in M\}$ and $\mathcal{S} := \{S | K(T) : L \subseteq S\}$. Then one has $$ L(T) = \bigcap_{M \in \mathcal{M}} M \text{ and}$$ $$ K(T)(L) = \bigcap_{S \in \mathcal{S}} S. $$

$\underline{\subseteq}:$ Let $S \in \mathcal{S}$. From the definition of $\mathcal{S}$ one gets $S$ is a field and $L \subseteq S$, whence $S|L$. Also from the definition one gets $S|K(T)$, whence $T \in S$. So in summary $$ S \in \mathcal{M}. $$ Now let $x \in L(T)$. Then $$ x \in \bigcap_{M \in \mathcal{M}}M \Rightarrow \forall M \in \mathcal{M}: x \in M \Rightarrow \forall S \in \mathcal{S} \subseteq \mathcal{M}: x \in S \Rightarrow x \in \bigcap_{S\in\mathcal{S}}S=L(T)$$

$\underline{\supseteq}:$ First one easily confirms that $L(T) \in \mathcal{S}$. Hence if $x \notin L(T)$ there is an $S \in \mathcal{S}$ (namely $S = L(T)$) s.th. $x \notin S$ . With that one gets $x \notin K(T)(L)$.

This can be generalized:

Let $K$ be a field and $R,S$ be arbitrary sets. Then $$ K(R)(S) = K(S)(R). $$

$\textit{proof}.$ Define $$ \widehat{R} := \{ \widetilde{R}|K(S) : R \subseteq \widetilde{R} \} \text{ and}$$ $$ \widehat{S} := \{ \widetilde{S}|K(R) : S \subseteq \widetilde{S} \}.$$ Then $$ \widetilde{R} \in \widehat{R} \implies \widetilde{R} | K(S) \land R \subseteq \widetilde{R}\implies S \subseteq \widetilde{R} \land \widetilde{R}|K(R) \implies \widetilde{R} \in \widehat{S} \text{ and}$$ $$ \widetilde{S} \in \widehat{S} \implies \widetilde{S} | K(R) \land S \subseteq \widetilde{S}\implies R \subseteq \widetilde{S} \land \widetilde{S}|K(S) \implies \widetilde{S} \in \widehat{R}.$$

Hence we have $\widehat{R} = \widehat{S}$ which implies $$ K(R)(S) = \bigcap_{\widetilde{S} \in \widehat{S}}\widetilde{S} = \bigcap_{\widetilde{R} \in \widehat{R}}\widetilde{R} = K(S)(R). $$


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