Here I have two propositions from p.521 on Abstract Algebra written by Dummit Foote.
Let $\alpha$ be algebraic over the field $F$ and let $F(\alpha)$ be the field generated by $\alpha$ over $F$. Then $[F(\alpha):F]=\deg m_\alpha(x)=\deg\alpha$.
and
The element $\alpha$ is algebraic over $F$ if and only if the simple extension $F(\alpha)/F$ is finite. More precisely, if $\alpha$ is an element of an extension of degree $n$ over $F$ then $\alpha$ satisfies a polynomial of degree at most $n$ over $F$ and if $\alpha$ satisfies a polynomial of degree $n$ over $F$ then the degree of $F(\alpha)$ over $F$ is at most $n$.
In the precise statement of second proposition, it says $\deg\alpha=n$ implies $\alpha$ satisfies a polynomial $p(x)$ of degree at most $n$.
My question is why does $\deg p(x)\leq n$ but not $=n$? From the first proposition, $\deg m_\alpha(x)=n$ and obviously, $m_\alpha(x)\vert p(x)$ by the definition of the minimal polynomial.
"At most $n$" statement is also true, of course. But I think "Equals to $n$" makes more sense.
Is there something I missed in my argument?
The author probably means "an element of an (extension field of degree $n$)" and not "an element of an (extension field) of degree $n$". Thus, $\alpha$ is not required to generate the extension field $L$ of $F$ and $[L:F]=n$. At least $F(\alpha)$ is contained in $L$, so that $\mathrm{deg}(\alpha)=[F(\alpha):F]$ divides $[L:F]$.