Prove F(a,b)=F(a)(b)=F(b)(a) in Field Extensions

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Discussion Overview

The discussion centers around proving the equality of field extensions, specifically that F(a,b) = F(a)(b) = F(b)(a) for elements a and b in an extension E of a field F. The scope includes theoretical aspects of field extensions and algebraic properties.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes to prove that F(a,b) = F(a)(b) = F(b)(a) for elements a and b in an extension E of F.
  • Another participant questions whether it is valid to assume that a and b are algebraic over F.
  • A subsequent reply confirms the assumption that a and b can be considered algebraic over F.
  • Another participant suggests picking bases for F(a) and F(b) over F and using a dimensional argument involving the set formed by the products of these bases.

Areas of Agreement / Disagreement

There is no consensus on the proof method, and the discussion includes multiple viewpoints regarding the assumptions about a and b being algebraic over F.

Contextual Notes

The discussion does not resolve the implications of assuming a and b are algebraic, nor does it clarify the specifics of the dimensional argument proposed.

Scherie
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Let E be an extension of F and let a, b belong to E. Prove that F(a,b) =F(a)(b) = F(b)(a).
 
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May we assume that $a$ and $b$ are algebraic over $F$?
 
Deveno said:
May we assume that $a$ and $b$ are algebraic over $F$?

Yes
 
Pick a basis $\{u_j\}$ for $F(a)$ over $F$, and a basis $\{v_k\}$ for $F(b)$ over $F$.

Consider the set: $\{u_jv_k\}$, and use a dimensional argument.
 

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