The next proposition is in a Japanese algebra text written by a Japanese mathematician who is famous in algebraic geometry.
But the proof is obviously not correct.
Then, is the next proposition true or false?
Let $k \subset k_1 \subset k_2$ be fields. If an element $\alpha\in k_2$ is algebraic over $k$, it is algebraic over $k_1$. If $\alpha$ has degree $n$ over $k$, its degree over $k_1$ divides $n$.
On
The first part is true and the second part is false.
If an element $\alpha\in k_2$ is algebraic over $k$, it is algebraic over $k_1$.
This is true because a polynomial over $k$ is a polynomial over $k_1$.
If $\alpha$ has degree $n$ over $k$, its degree over $k_1$ divides $n$.
This is not true. Let $k = \mathbb{Q}$, $k_1 = \mathbb{Q}(\sqrt[3]{2})$,$k_2 = \mathbb{Q}(\sqrt[3]{2}, \omega)$, where $\omega = e^{2\pi i / 3}$ as usual, and let $\alpha = \omega \sqrt[3]{2}$. Then:
The minimal polynomial of $\alpha$ over $k = \mathbb{Q}$ is $x^3 - 2$, which has degree $3$.
The minimal polynomial of $\alpha$ over $k_1 = \mathbb{Q}(\sqrt[3]{2})$ is $(x/\sqrt[3]{2})^2 + (x/\sqrt[3]{2}) + 1$ (because $\omega^2 + \omega + 1 = 0$), which has degree $2$.
And you can see that $2$ does not divide $3$.
The proposition is true. For $\alpha$ to be algebraic over $k$, there must be a polynomial with coefficients in $k$ with $\alpha$ as a root. Since each coefficient is also in $k_1$, the first assertion follows. For $\alpha$ to have degree $n$ over $k$, then the dimension of $k_2$ over $k$, denoted [$k_2$ : $k$], is equal to $n$; then there is a theorem which asserts [$k_2$ : $k$] = [$k_2$ : $k_1$] [$k_1$ : $k$], and the second assertion follows.