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

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Extension
AI Thread Summary
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.
Mathematics news on Phys.org
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).
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top