A confusion regarding the nature of elements in a field extension. I read this statement in a well-known textbook: "Since $[W:F]\leq mn$, every element in $W$ satisfies a polynomial of degree at most $mn$ over $F$." 
Let $F[a]$ be a field extension of $F$. Can $f\in F$ be algebraic over $F$, and how? I was under the impression that elements belonging only to $F[a]/F$ can be algebraic over $F$?
 A: Your question is confusing. The letter $f$ is usually reserved for a polynomial, but you are using it to mean an element of a field (you wrote $f\in F$). Also, any element $a\in F$ is clearly algebraic over $F$, because it is a root of the polynomial $g(x)=x-a\in F[x]$.
Perhaps it is worth emphasizing the definition of "algebraic element": given any field $F$, and any extension $L$ of $F$, then an element $a\in L$ is said to be algebraic over $F$ when there is a polynomial $g\in F[x]$ such that $g(a)=0$. Note that the field $L$ does not have to be equal to $F(a)$; that is, it can be the case that
$$\large L\supsetneq F(a)\supseteq F.$$
The element $a$ is obviously an element of $F(a)$. I'm not sure where you were going with this, so I don't know if I've addressed your concerns.

Let $F$ be a field, and let $L$ be an extension of $F$. For any $a\in L$, we have that
$$[F(a):F]=\deg(f)$$
where $f\in F[x]$ is the minimal polynomial of $a$. This polynomial is the monic polynomial in $F[x]$ of smallest degree such that $f(a)=0$. Consequently, if $g\in F[x]$ is any other polynomial with $g(a)=0$, we must have that $f\mid g$, and hence $\deg(f)\leq \deg(g)$.
Also, recall that degrees are multiplicative: if $K\supseteq L\supseteq F$ are fields, then
$$[K:F]=[K:L][L:F].$$
Thus, given the information that $[W:F]\leq mn$, and noting that $W\supseteq F(a)\supseteq F$ for any $a\in W$, we have that
$$mn=[W:F]=[W:F(a)][F(a):F]\geq [F(a):F]=\deg(f)$$
where $f$ is the minimal polynomial of $a$ over $F$. Thus, any $a\in W$ satisfies a polynomial in $F[x]$ of degree $\leq mn$.
A: All elements $f\in F$ are algebraic over $F$, that is: there is
some polynomial $p\in F[x]$ s.t $p(f)=0$. Namely $$p(x)=x-f\in F[x]$$
Note 1: $F\subseteq F[a]$ so if this isn't any contradiction to statement
that all elements of $F[a]$ are algebraic over $F$.
Note 2: Not only the elements of $F[a]$ are algebraic over $F$.
For example consider $$F=\mathbb{Q},a=\sqrt{2},b=\sqrt{3}$$ then
$$F[a]=\mathbb{Q}(\sqrt{2})$$ and $$b\not\in F[a]$$ but is algebraic
over $F$.
