Show there exists a unique monic polynomial $f(x)$ in $F[x]$ of minimum degree such that $f(a) = 0$

Question

Let $F$ be a field, $E$ an extension field of $F$, and a in $E$ an algebraic element. Show there exists a unique monic polynomial $f(x)$ in $F[x]$ of minimum degree such that $f(a) = 0$

All I have so far is that since a is algebraic over $E$, there exists some $p(x)$ in E such that $p(a) = 0$.

I'm really struggling with these ideas in general (fields, extension fields, and algebraic extensions).

Answer 1

Hint: Suppose that $p(x)$ and $q(x)$ are two monic polynomials of minimal degree such that $p(a) = q(a) = 0$. What can we say about the polynomial $p(x) - q(x)$?

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To summarize the comments below: $a$ is meant to be an element which is algebraic over $F$, rather than over $E$.

Answer 2

so $a \in E$ is algebraic over $F$ means there exists a monic minimal polynomial say $f(x) \in F[x]$ such that $f(a)=0$ but it is given $g(a)=0$ implies $f(x)$ divides $g(x)$ using division algorithm and minimality of $f$ having $a$ as a zero.Similarly g(x) divides f(x) , hence we get g(x)=h_1(x)*f(x)=h_1(x)*h_2(x)*g(x) implies h_1(x)*h_2(x)=1 but F[x] has only units same as F so h_1(x) and h_2(x) must be in F but f and g are minimal so h_1 and h_2 must be 1.