When asked to describe the field extension F(a), where F is a subfield of E and a is an element of E, many people instinctively respond with the following algorithm: they generate (in E) the sequence $a, a^2, a^3, ...$ , stopping at the first $a^n$ that results in a set of n elements that are closed under multiplication. Then they assert that the vector space ${c_0 + c_1a + c_2a^2 + ... + c_na^n}$ is not only a vector space but also a field and assert that this field is the smallest subfield of E that contains a.
I have a hunch this algorithm is correct but I am asking how to prove it is correct. All I know for sure is that F(a) is isomorphic to F[x]/<p> where p is the minimal polynomial of a in F[x]. Is there some way to prove this algorithm does always produce the same result, something isomorphic to F[x]/<p> where p is the minimal polynomial of a in F[x]? Or even that this algorithm does, indeed always produce a field at all and not just a vector space that is not a field? Thank you in advance.
It indeed follows from $F(a)\cong F[x]/(p)$. Let's say $p=a_nx^n+\ldots+a_0$ with $a_i\in F$ for all $i$. Then (in $F[x]/(p)$) you can express $x^n$ as a linear combination of $\{1,x,x^2\dots x^{n-1}\}$ by as follows (note that in $F[x]/(p)$, you identify $p$ with $0$): $$x^n = x^n - a_n^{-1}p = -a_n^{-1}a_{n-1}x^{n-1}-\ldots-a_n^{-1}a_0$$
More generally if $N\geq n$:
$$x^N=x^N-a_n^{-1}px^{N-n}$$
reduces the degree by at least $1$ and you can just keep doing that until everything is in linear combination of $\{1,x,x^2\dots x^{n-1}\}$ since the degree decreases every step. Now note that the isomorphism from $F[x]/(p)$ to $F(a)$ maps $x\to a$, meaning $\{1,a,a^2\dots a^{n-1}\}$ will be the basis for $F(a)$.