- Thread starter
- #1

- Jan 29, 2012

- 661

Let E be an extension field of F and let a, b be elements of E. Prove that F(a,b)=F(a)(b)=F(b)(a)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter Fernando Revilla
- Start date

- Thread starter
- #1

- Jan 29, 2012

- 661

Let E be an extension field of F and let a, b be elements of E. Prove that F(a,b)=F(a)(b)=F(b)(a)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter
- #2

- Jan 29, 2012

- 661

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).