Corollary 2

— Let $E$ be an extension of $K$, $A$ an algebra over $E$ and $x$ an element of $A$ algebraic over $K$. Then $x$ is algebraic over $E$, the minimal polynomial of $x$ over $E$ divides the minimal polynomial of $x$ over $K$ and the degree of $x$ over $E$ is at most equal to the degree of $x$ over $K$.

For let $f$ be the minimal polynomial of $x$ over $K$; we have $f(x) = 0$ and $f \in E[X]$, hence $x$ is algebraic over $E$ and $f$ is a multiple of the minimal polynomial of $x$ over $E$ (Th. 1, a).