Let $\alpha$ be an element of an extension field $K$ of the field $F$ that is algebraic over $F$. Then, if $f$ is an irreducible element of $F[x]$, and $\alpha$ is a root of $f$, then $f$ is the monic polynomial of lowest degree in $F[x]$ that has $\alpha$ as a root.
How do I show that such irreducible polynomial is unique? Suppose that there are two irreducible polynomials $f,g$ satisfying $f(\alpha)=0$ and $g(\alpha)=0$. Then by division algorithm, by dividing $f$ by $g$, say, I find a polynomial $r$ with less degree than both $f$ with $\alpha$ as a root. Now I divide say the quotient by $r$, and continue. Then, by successive division remainders I get in each step, I get either constant reminder or 0 remainder. Constant remainder means that $\alpha$ not root of the last remainder, but that means that $\alpha$ not root of $f$, last remainder zero means $f$ reducible.
Am I correct above?