I am currently trying to solve the following problem: Let $j:K\rightarrow L$ be a field extension with [$L:K$] < $\infty$.
Let $x_1,x_2\in L$ be arbitrary. Show the following multiplicativity of the degree of separability:
\begin{equation} [K(x_1,x_2):K]_s = [K(x_1,x_2):K(x_1)]_s\cdot [K(x_1):K]_s. \end{equation}
My definitions are:
- $[L:K]_s$ = #{K-homomorphisms $L\rightarrow K^a$} where $K^a$ is the algebraic closure of $K$.
and
- $[K(x):K]_s$ = #{different roots of $\iota_a^*\:m_x$} where $m_x$ is the minimal polynomial of x with $\iota:K\rightarrow K(x).$
I have given a hint, that I should identify the K-homomorphism $f:K(x_1,x_2)\rightarrow K^a$ with pairs $(x_1^*,x_2^*)\in K^a$, where $x_1^*$ is a root in $K^a$ of the minimal polynomial of $x_1$ over $K$ and $x_2^*$ a root in $K^a$ of the minimal polynomial of $x_2$ over $K(x_1)$.
With my second definition, it makes some intuitive sense why this multiplicative identity holds. You can write $K(x_1,x_2) = (K(x_1))(x_2)$ so we can look at $[K(x_1,x_2):K(x_1)]_s$ as the number of different roots of the minimal polynomial of $x_2$ in $K^a$ over $K(x_1)$. I now don't understand how to bring all of those facts together to show this. I feel like that this is somewhat trivial, but I can't seem to make a bullet-proof argument. Any push in the right direction would be much apreciated!