I have the following problem:
Let $E$ be separable over $k$. Consider the tower $$k\subset k(\alpha_1)\subset...\subset k(\alpha_1,...,\alpha_n)$$Then since each $\alpha_i$ is separable over $k$ each $\alpha_i$ is separable over $k(\alpha_1,...\alpha_{i-1})$ for $i\geq 2$.
And I somehow don't see why this is ture. So I know that since $\alpha_i$ is separable over $k$ we have $$[k(\alpha_i):k]=[k(\alpha_i):k]_S$$ where $[k(\alpha_i):k]_S$ is the separable degree. But why is then $$[k(\alpha_1,...,\alpha_i):k(\alpha_1,...\alpha_{i-1})]=[k(\alpha_1,...,\alpha_i):k(\alpha_1,...\alpha_{i-1})]_S~~~~~~~~(1)$$ Could someone explain this to me?
Thanks a lot
The statement " each $\alpha_i$ is separable over $k(\alpha_1, \dots, \alpha_{i-1})$" means that we have to look at the extension $$k(\alpha_1, \dots, \alpha_{i-1})(\alpha_i)/k(\alpha_1, \dots, \alpha_{i-1})$$
Then we can either use the multiplicative property of the separable degree to get to the result or (what I would prefer) just use the definition of separable elements. An element $\beta \in L$ is called separable over $k$ (where $L/k$ is a field extension) iff its minimal polynomial has distinct roots in an algebraic closure $k^a$ of $k$. Now the minimal polynomial of each of the $\alpha_i$ is separable over $k$ (by assumption that $\alpha_i$ are separable over $k$).
This will also be true over $k(\alpha_1, \dots, \alpha_{i-1})$ as this field is bigger and contains more possible scalars, so $$\text{minpol}_{\alpha_i,k(\alpha_1, \dots, \alpha_{i-1})} \mid \text{minpol}_{\alpha_i,k}$$ the minimal polynomial over $k(\alpha_1, \dots, \alpha_{i-1})$ will be a divisor of the (separable) minimal polynomial of $\alpha_i$ over $k$. That means that it also has to have distinct roots in $k^a$ and that $\alpha_i$ is also separable over $k(\alpha_1, \dots, \alpha_{i-1})$