Why is the following set linearly independent over the Field $L$?

Question

The author is trying to prove that if $[L:K]$ is finite, then there must exist finitely many elements in $L$ fuch that $L = K(a_1, ..., a_s)$ and $L$ must be algebraic over $K$.

The first part I understand, but I don't get why $L$ is algebraic over $K$. The author proves it this way:

Let $x$ be any element of $L$ and let $n=[L:K]$. The set ${1, x, ..., x^n}$ has $n+1$ elements, so it must be linearly independent over $K$. (...)

---

From the assumption ${1, x, ..., x^n}$ is linearly dependent over $K$ the theorem is easily proved. Though how can one prove that such set is linearly independent over $K$ for all of the elements in $L$?

I would really appreciate any help/thoughts!

Answer 1

There's a typo in your statement:

Let $x$ be any element of $L$ and let $n=[L:K]$. The set $1,x,\ldots,x^n$ has $n+1$ elements, so it must be linearly independent over $K$.

In fact, the highlighted word must be dependent. Please check whether the typo was in your book of whether you made it here. They are dependent simply because there are too many of them, viz. $n+1$ elements.

Now, their dependence means that there exists their linear combination $$c_0\cdot1+c_1\cdot x+\cdots+c_n\cdot x^n=0,$$ which is precisely what we need.