Transcendental Extension and Algebraic Extension commute.

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?

• 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.) – AlgebraicsAnonymous Feb 8 at 14:33
• 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 $\overline{K}$. – 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 $$K(R)(S) = \bigcap_{\widetilde{S} \in \widehat{S}}\widetilde{S} = \bigcap_{\widetilde{R} \in \widehat{R}}\widetilde{R} = K(S)(R).$$