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$[\mathbb{R}(x):\mathbb{R}(x+\frac{1}{x})]=2$ since $x$ satisfies a quadratic polynomial over $\mathbb{R}(x+\frac{1}{x})$. Similarly, $[\mathbb{R}(x):\mathbb{R}(x^2+\frac{1}{x^2})]=4$. But, as $(x+\frac{1}{x})^2=(x^2+\frac{1}{x^2})+2$, so $\mathbb{R}(x):\mathbb{R}(x^2+\frac{1}{x^2}):\mathbb{R}(x+\frac{1}{x})$ which clearly contradicts Tower law. What is wrong here?

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    $\begingroup$ ? What actually are you asserting after "so"? $\endgroup$ – Lord Shark the Unknown Mar 31 at 6:34
  • $\begingroup$ I got confused. I got it. Thanks@LordSharktheUnknown $\endgroup$ – Shanghaikid Mar 31 at 7:47
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I can't see what you are asserting, but here are some facts: $$|k(x):k(x+1/x)|=2,$$ $$|k(x):k(x^2+1/x^2)|=4,$$ $$|k(x+1/x):k(x^2+1/x^2)|=2.$$ Here $k$ is any field, and $x$ is transcendental over $k$.

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