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

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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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Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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