$\alpha\in L$ algebraic over $K$ implies $\beta\in L$ algebraic over $K$ iff $\beta$ algebraic over $K(\alpha)$.

Question

The fact that $\beta\in L$ algebraic over $K$ implies $\beta$ is algebraic over $K(\alpha)$ is obvious. Since $\alpha\in L$ is algebraic over $K$, we have $[K(\alpha):K]=n<\infty$. Assume $\beta$ is algebraic over $K(\alpha)$. Then $[K(\beta):K(\alpha)]=d<\infty$. We have $$ [K(\beta):K]=[K(\beta):K(\alpha)][K(\alpha):K]=dn<\infty. $$

Hence $\beta$ is algebraic over $K$. Is this correct?

Answer 1

It may not be true that $K(\beta)$ is an extension of $K(\alpha)$. So using the term $[K(\beta):K(\alpha)]$ is not proper.

You can modify your reasoning as follows:

$$[K(\alpha, \beta):K] = [K(\beta, \alpha):K(\alpha)][K(\alpha):K]$$

which is finite by the assumption that $\beta$ is algebraic over $K(\alpha)$ and $\alpha$ is algerbaic over $K$. So $[K(\alpha, \beta):K]$ is finite and therefore so is $[K(\beta):K]$ and we are done.