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

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In the discussion on proving that F(a,b) = F(a)(b) = F(b)(a) in field extensions, participants confirm that assuming a and b are algebraic over F is valid. A basis for F(a) over F is chosen, along with a basis for F(b) over F. The set formed by the products of these bases, {u_jv_k}, is examined. A dimensional argument is suggested to establish the equality of the field extensions. The conclusion emphasizes the importance of these bases in demonstrating the relationships among the fields.
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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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