Summary: Is the statement true?
$K$, $E$, $k$ are subfields of a larger field $\Omega$. If every member of $K$ is algebraic over $E$ and every member of $E$ is algebraic over $k$, then every element of $K$ is algebraic over $k$.
We know the tower property of algebraic field extension:
$K/E$ and $E/k$ are algebraic extension. Then $K/k$ is also algebraic.
However, when I am reading a textbook, the author seemed to apply the following statement, whick looks like the tower property, but the inclusion is removed:
$K$, $E$, $k$ are subfields of a larger field $\Omega$. If every member of $K$ is algebraic over $E$ and every member of $E$ is algebraic over $k$, then every element of $K$ is algebraic over $k$.
Is the statement true? I think it is right. Proof: Let $u\in K$, and $f(x)=\sum_{i=0}^{n} a_ix^i$ be its irreducible polynomial over $E$. Then $u$ is algebraic over $k(a_0,a_1,\cdots, a_n)$, and $k(a_0, a_1, \cdots a_n)$ is algebraic over $k$ since $E$ is. Now apply the tower property to $k(a_0,a_1, \cdots, a_n, u)\supseteq k(a_0,a_1, \cdots, a_n)\supseteq k$.