Minimal polynomial of a field extension implies the powers of indeterminate are linearly independent

Question

Suppose $L:K$ is a field extension and $a \in L$ is algebraic over $K$ with minimal polynomial $m(a) \in K[x]$?

I have written in my notes that:

If $d=deg(m(a))$ then $\{1,x,x^2, \ldots , x^d\}$ is a linearly independent set.

Where does this come from? Isn’t it always an independent set as $x$ is just an indeterminate?

Answer 1

As noted in the comments, I think your result is misquoted. It is most likely supposed to be

If $\deg(m(a)) = d$, then $\{1,a,a^2, \ldots , a^{d-1}\}$ is a linearly independent set over $K$

If $\{1,a,a^2, \ldots , a^{d-1}\}$ were linearly dependent over $K$, then that would mean that there is a non-trivial linear combination of these elements that equals $0$: $$ k_0 + k_1a + k_2a^2 + \cdots + k_{d-1}a^{d-1} = 0, \qquad k_i\in K, \text{not all }0 $$ This runs directly counter to the fact that the minimal polynomial of $a$ over $K$ has degree $d$, as $$ k_0 + k_1x + k_2x^2 + \cdots + k_{d-1}x^{d-1} $$ would be a non-zero polynomial of lesser degree with $a$ as root.