If $F\subseteq K$ are fields, $\alpha \in K$ Prove $\alpha$ is algebraic over $F$ If $F\subseteq K$ are fields, $\alpha \in K$, and $K$ is an extension field of $F$.
Prove the following are equivalent:


*

*$\alpha$ is algebraic over $F$

*$F(\alpha)=F[\alpha]$ 

*$|F(\alpha):F|$ is finite


I'm trying to prove the properties from here: https://en.wikipedia.org/wiki/Algebraic_element

Edited the question because people voted to close for not understanding what I mean in the original question:
Intuition:
I did prove (2.) and (3.) independently, but I'm not sure how to show (1.)
 A: Hint:$(3)\implies (1)$ : Suppose degree of $\alpha$ over $F$ is $n$, that is, $[F(\alpha):F]=n$. Then consider these $n+1$ elements: $1,\alpha,\alpha^2,\ldots \alpha^n$, which are linearly dependent over $F$.
$(1)\implies (3)$ : If $\alpha$ is algebraic over $F$, then $F(\alpha)\cong\dfrac{F[x]}{(p(x))}$, where $p(x)$ is the minimal polynomial of $\alpha$ over $F$. Dimension of $\dfrac{F[x]}{(p(x))}$ as a vector space over $F$ is same as the degree of $p(x)$.
A: $(1)\implies(2)$
Let $f(x)$ be the minimal polynomial of $\alpha$, which is irreducible. Take a nonzero element $g(\alpha)\in F[\alpha]$, where $g(x)\in F[x]$. Since $g(\alpha)\ne0$, we have that $f(x)$ doesn't divide $g(x)$; hence a greatest common divisor of $f(x)$ and $g(x)$ is $1$, so by Bézout's identity there exist $p(x)$ and $q(x)$ such that $f(x)p(x)+g(x)q(x)=1$. Can you conclude $g(\alpha)^{-1}\in F[\alpha]$?
$(2)\implies(3)$
It should be easy to show that $F[\alpha]$ is a finite dimensional vector space over $F$.
$(3)\implies(1)$
If $|F(\alpha):F|=n$, then $1,\alpha,\dots,\alpha^n$ are …
