# A generalization of “all members of a simple field extension of F are algebraic over F”

The following is a powerful powerful theorem: Let $F(\alpha)$ be a simple field extension of field $F$, where $\alpha$ is algebraic over $F$. Then every $x\in F(\alpha)$ is also algebraic over $F$. Also, $\deg(x,f)\leq \deg(\alpha,F)$.

The proof I think is most brilliant. I am not going to provide you with a proof here. I am just going to try and slightly generalize this theorem.

Let $\beta\in F(\alpha)$. Consider the vector space generated by $\{1,\alpha,\alpha^2,\dots\}$. Also, let $\{c_1,c_2,\dots,c_n\}$ be the set of vectors that span the vector space generated by $\{1,\alpha,\alpha^2,\dots\}$. Then $\deg(\beta,F)\leq n$.

How is this a generalization? Firstly, the set of vectors $\{c_1,c_2,\dots,c_n\}$ need tot span the whole of $F(\alpha)$. It only needs to span the subspace generated by $\{1,\alpha,\alpha^2,\dots\}$. Hence, we can find a better upper bound for the degree of the irreducible polynomial satisfying $\beta$. Secondly, we need only consider the set of vectors that spans the space, and not necessarily the basis of the space. Although this is likely to give us a worse estimate for the upper bound of the irreducible polynomial, it is still a generalization (sometimes, it may be impractical to determine the basis from the set spanning the vector space).