Does $\alpha$ need to be transcendental over F? In the book there is this exercise: Let E be an extension fiel of F, with $\alpha, \beta \in E$. Suppose $\alpha$ is transcendental over F but algebraic over $F(\beta)$. Show that $\beta$ is algebraic over $F(\alpha)$. The suggested solution was: !enter image description here It doesn't seem that they use that $\alpha$ is transcendental in F, or do they? Or is the statement valid without this assumption?

Actually, they use that hypothesis in the part

> _a polynomial in $\alpha$ with coefficients that are polynomials in $\beta$ can be formally rewritten as a polynomial in $\beta$ with coefficients that are polynomials in $\alpha$_

to conclude that this new polynomial (seen in $(F[\alpha])[x]$) is not the zero polynomial. This is necessary to conclude that $\beta$ satisfy a **non-zero** polynomial over $F(\alpha)$ so $\beta$ is algebraic over $F(\alpha)$.