Field Extension of $\mathbb{R}\left(x+\frac{1}{x}\right)$

Question

Can someone help me to prove $[\mathbb{R}(x):\mathbb{R}\left(x+\frac{1}{x}\right)]=2$? My intuition the basis is $\{x,x+\frac{1}{x}\}$ but I cannot prove. Thank you

Answer 1

The minimal polynomial of $x$ is $t^2-(x+1/x)t+1$.

To see that $x$ doesn’t lie in $\Bbb R(x+1/x)$, assume first that it does. $x =P(x+1/x)$ where $P$ is a polynomial.

$x= x^{-n}+a_{-n+1}x^{-n+1}+...+a_{n-1}x^{n-1}+x^n$, so $x^{-n}+a_{-n+1}x^{-n+1}+...+(a_1 - 1)x+...+a_{n-1}x^{n-1}+x^n = 0$. Therefore $a_1=1$ and all other $a_i =0$. But we must have $a_1=a_{-1}$ because the polynomial is symmetric in $x$ and $1/x$, which gives a contradiction.