Algebraic Degree of a & b: F(a,b)=mn

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If a and b are algebraic over a field F with degrees m and n, respectively, and m and n are relatively prime, then the extension degree [F(a,b):F] equals mn. Clarification is sought regarding whether a and b are indeed algebraic over F and if the notation F(a,b)=mn is intended to express [F(a,b):F]=mn. The relationship between the degrees of the extensions is questioned, particularly whether [F(a,b):F] equals the product of the individual degrees [F(a):F] and [F(b):F], which is generally false. Understanding the implications of the degrees being relatively prime is crucial in this context. The discussion emphasizes the need for precise definitions and conditions in algebraic field extensions.
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Homework Statement



If a and b are algebraic over F of degree m and n, both relatively prime, then

F(a,b)=mn, (i.e. [F(a,b):F]=mn)

any comments are helpfull.
 
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Your post is very incomplete. Are a and b algebraic over a field F? And by "F(a,b)=mn" do you mean "[F(a,b):F]=mn"?

In which case, that m and n are relatively prime is important. Are you trying to say that [F(a,b):F]=[F(a):F][F(b):F]? This is false in general (why?).
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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