For a polynomial $p(x)$ of degree $n$, irreducible in some field $F$, and if $a$ is some root of $p$ in an extension field, then $\{1, a, a^2, a^3 ... a^{n-1} \}$ are linearly independent.
I came across this post which assumes a linear combination of basis, then I would be able to produce you another polynomial $w$ of degree $n$ such that $w(a)$ is a root. This implies that $\gcd(p, w)=(x-a)$, and the author claims it contradicts the irreducibility of $p$.
I don't see how it contradicts the irreducibility of $p$, though. Because doesn't irreducibility just mean that you aren't able to factor the polynomial in the field? However, the factor that that $p $ and $w$ have in common is $(x-a)$, and the factoring is done in the extension field, isn't it, since $a$ is an element of the extended field?
We haven't done a lot on abstract algebra, so apologies if this is a basic misconception.