Esteban's question at Yahoo Answers (Field extension)

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The discussion addresses the proof that for an extension field E of F, the equality F(a, b) = F(a)(b) = F(b)(a) holds true for elements a and b in E. The proof utilizes the concept of Moore families and the closure properties of subfields within E. By demonstrating that the union of subsets leads to the smallest subfield containing those subsets, the conclusion is established definitively.

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Hello Esteban,

In general, if $S_1,S_2$ are subsets of $E$, let us prove that $F(S_1\cup S_2)=F(S_1)(S_2)$.

We know that the intersection of subfields of $E$ is a subfield of $E$ so, the colection of subfields of $E$ form a Moore family. The corresponding Moore closure $X\to \bar{X}$ associates to every subset of $E$ the smallest subfield $\bar{X}$ of $E$ containing $X$, so $F(S)=\overline{F\cup S}$. Then, $$\begin{aligned}F(S_1\cup S_2)&=\overline{F\cup S_1\cup S_2 }\\&=\overline{\overline{F\cup S_2}\cup S_2}\\&=F(S_1)(S_2)\end{aligned}$$ Now we can particularize $S_1=\{a\}$ and $S_2=\{b\}$ (or reciprocally).
 

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