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).

Published by -

Graduate student

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: