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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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).