$L/K$ is finite implies $L/K$ is algebraic

Question

I'm trying to prove the well known fact that a field extension $L/K$ is finite implies that $L/K$ is an algebraic field.

I'm aware of how to prove the statement if $L/K$ has the basis $1, \alpha, \alpha^2,...,\alpha^d$, then $L/K$ is algebraic. But how would I prove the statement if $L/K$ had a general basis $\alpha_1, \alpha_2,...,\alpha_d$, then $L/K$ is an algebraic field.

Answer 1

Suppose $\dim_K L=n$ and let $a\in L$; consider the elements $1,a,a^2,\dots,a^n$. If two of them are equal, say $a^h=a^k$ with $0\le h<k$, then $a$ is a root of $X^k-X^h\in K[X]$, so it is algebraic over $K$.

If those elements are pairwise distinct, then the set $\{1,a,a^2,\dots,a^n\}$ has more than $n$ elements, so it's linearly dependent: $n+1$ vectors in an $n$-dimensional vector space always form a linearly dependent set, because a linearly independent set can be extended to a basis.

Answer 2

If $L$ is a $n$-dimensional $K$-linear space, then for each $\alpha\in L$, $1,\alpha,\alpha^2,...,\alpha^n$ is $K$-linearly dependent.

Answer 3

The answer to this question is nothing more than linear algebra.

If you have a basis $\alpha_1,\dots,\alpha_d$, it means that $L/K$ is a $d$-dimensional vector space. Now let $\alpha\in L$. It is a fact about vector spaces that a linearly independent set cannot be bigger than any basis (indeed, any spanning set) for a space. Therefore, since the set $\{1,\alpha,\dots,\alpha^d\}$ has size greater than $d$, it must be linearly dependent, and a linear dependence between the elements is precisely a non-zero polynomial equation satisfied by $\alpha$.

Note that we are not claiming that $1,\alpha,\dots,\alpha^d$ is a basis (it can't be, since it is too big to be linearly independent). It is not even necessarily true that $1,\alpha,\dots,\alpha^{d-1}$ is a basis for $L/K$, though it will be if $\alpha$ is a primitive element for the extension.