Proof of Finiteness of a simple extension field

Question

I need to prove that

$a \in K $ is algebraic over the field F if and only if $[F(a):F] < \infty$

where K is an extension of the field F.

Answer 1

Proof: (=>) Let a $\in$ K be algebraic over F where K is the extension field of F. Since $a$ is algebraic over F, $a$ is a root of some nonzero polynomial f(X) $\in$ F[X] such that f(a) = 0. According to the corollary of "Fundamental Theorem of Kronecker" which states:

"If f(X) $\in$ F[X], then there exists a finite extension K of F where f(X) has a root. Moreover; [K:F] $\le $ deg(f(X))"

We can say that the degree of simple extension of F, i.e. F(a), over F is limited by [F(a):F]$\le$ deg(f(X))=n. So as n goes to infinity [F(a):F]$\lt \infty$

(<=) Let [F(a):F] = m $\le \infty$. Then $1, a, a^2, ..., a^m$ are vectors.

Claim: $1, a, a^2, ..., a^m$ are linearly dependent over F.

Proof: Let $\alpha_0 + \alpha_1 a + ..... + \alpha_m a^m = 0 $ . But there exists $\alpha_i \in$ F, not all zero since [F(a):F] = m. Therefore there exists a $p(X) = \alpha_0 + \alpha_1 x + ..... + \alpha_n x^n \in $F[X] such that p(a) = 0. Since there exists a $p(X)$ such that $p(a) = 0$ in F(a), $a$ is algebraic over F.

Answer 2

Hint 1

If $a$ is algebraic over $F$, and it is the root of a polynomial of degree $n$, then $[F(a):F] \le n$.

Hint 2

If $[F(a):F] = m < \infty$, then $1, a, a^2, \dots, a^m$ are $m+1$ vectors, and thus they are linearly dependent over $F$.