I'am trying to solve the following exercise in Dummit & Foote(Chapter 13 Section 2 Question 13),
Suupose $F={Q}(\alpha_1,\ldots, \alpha_n)$ where $\alpha_i^2\in Q$ for each i. Prove that $2^{1/3}$ is not in F.
Q denotes the set of rational numbers.
According to a solution that I read, it says $[Q(\alpha_1,\ldots, \alpha_k):Q(\alpha_1,\ldots,\alpha_{k-1})]=1 \quad \text{or} \quad 2.$ I could not understand this part. Why do we have the conclusion =1 or 2? Is it because of the condition given in the question $\alpha_i^2 \in Q$ for each i? If $\alpha_{l} \in Q(\alpha_1,\ldots,\alpha_{l-1})$, then $[Q(\alpha_1,\ldots, \alpha_l):Q(\alpha_1,\ldots,\alpha_{l-1})]=1.$ Is this true? A Little bit confused here.
Any help is appreciated.
The explanation you give is correct. $$ \mathbb Q(\alpha_1,\ldots,\alpha_{k}):\mathbb Q(\alpha_1,\ldots,\alpha_{k-1}) $$ is equal to the degree of the minimal polynomial of $\alpha_k$. Since $\alpha_k$ the root of a quadratic polynomial $x^2-q$ for some $q\in\mathbb Q$, the degree of its minimal polynomial is either 1 or 2, so the extension degree is either 1 or 2.
Also, to write $\mathbb Q$ properly in LaTeX, you should use \mathbb Q