Prove that M is an extension of N of degree 2

Question

Let $L$ be a field of rational functions in one variable over $\mathbb{C}; i.e$

$$L = \mathbb{C}(y) = \{ \frac{f(y)}{g(y)} | f(y), g(y) \in \mathbb{C}(y)\}$$

Let $M = \mathbb{C}(\beta)$ with $\beta = y^5$ and let $N = \mathbb{C}(\alpha)$ with $\alpha = -\frac{(y^5-1)^2}{y^5} \in M.$ Notice that now we have

$$\mathbb{C} \subseteq N = \mathbb{C}(\alpha) \subseteq M = \mathbb{C}(\beta) \subseteq L = \mathbb{C}(y)$$ a)Prove that M is an extension of N of degree 2.

My attempts

$\alpha = -\frac{(y^5-2)^2}{y^5}$ with $\beta = y^5$

Here Polynomial f(x) = $x^2+(\alpha-2)x+1=0$ has a root $\beta,\frac{1}{\beta}$

and we know minimal polynomial $m_{\alpha}(x)$ | f(x). So, deg $m_{\alpha}(x) =$ 1 or 2.

but how can we eliminate the case with deg $m_{\alpha}(x) =1$.

I am struggling to prove that $\beta$ or $\frac{1}{\beta} \notin \mathbb{C}(\alpha)$. It would be great if someone can give hint in this direction. Thanks in advance

Answer 1

We have the identity $2-\alpha= \beta + \beta^{-1}$ and therefore $\mathbb{C}(\alpha)=\mathbb{C}(\beta + \beta^{-1})$. Now assume $\beta\in\mathbb{C}(\beta + \beta^{-1})$. The equation $\beta\cdot g(\beta + \beta^{-1}) = f(\beta + \beta^{-1});\ f,g\in \mathbb{C}[X]$ will lead to a contradiction when we compare coefficients.