MHB Esteban's question at Yahoo Answers (Field extension)

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The discussion centers on proving that for an extension field E of F, the equality F(a,b) = F(a)(b) = F(b)(a) holds for elements a and b in E. It establishes that if S1 and S2 are subsets of E, then F(S1 ∪ S2) equals F(S1)(S2). The proof utilizes the concept of Moore families and the closure of subfields in E. By applying this to the specific case of S1 being {a} and S2 being {b}, the equality is confirmed. This mathematical relationship is crucial in understanding the structure of field extensions.
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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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