Let $K$ be an extension field of $F$, and let $\alpha \in K$ be algebraic. Suppose that $f(x) \in F[x]$. Prove that $f(x)$ is the unique monic polynomial of least degree with f($\alpha$) = $0$.
My thinking is that if $f(x)$ isn't unique, then there exists another polynomial of $g(x)$ of the same degree. And, if that's the case, then these two polynomials would divide each other, which means $f(x)$ is not irreducible. Is this correct? Or am I missing pieces, or completely misunderstanding things? Thank you for your help.
Best Answer
Either $f(x)$ divides $g(x)$ or the vice versa .
Let WLOG;$g(x)|f(x)\implies f(x)=g(x)q(x)+r(x)$ where $\deg r(x)<\deg g(x)$...
$f(a)=0\implies r(a)=0$ which is false as $\deg r(x)<\deg g(x)\le\deg f(x)$
Hence $r=0\implies f(x)=cg(x);c\text{is a constant}$
But that is false as $f,g$ are monic.So $f=g$