Originally Posted by
ThePerfectHacker There are some problems here. First, Q has no subfields other than itself. Second, $\displaystyle \sqrt{\alpha}$ is not well-defined. Third, we do not know even if $\displaystyle \sqrt{\alpha}$ exists.
Let me phrase the question differently: Let $\displaystyle F$ be an extension field over $\displaystyle K$. Let $\displaystyle \alpha \in F$ be algebraic over $\displaystyle K$. Suppose that $\displaystyle \beta \in F$ has property that $\displaystyle \beta^2 = \alpha$. Show that $\displaystyle \beta$ is algebraic over $\displaystyle K$.