An element $\alpha\in \mathbb F$ is algebraic w.r.t a field extension $\mathbb E\subset \mathbb F$ if it's the root of some polynomial over $\mathbb E$.
One can prove the sum, product, and inverse of algebraic elements are also algebraic. Hence TFAE:
- The extension consists entirely of algebraic elements.
- The extension is generated by algebraic elements, i.e by roots of polynomials over the base field. Thus every element upstairs is a rational function in a root of a polynomial downstairs.
An element $\alpha\in \mathbb F$ is separable w.r.t a field extension $\mathbb E\subset \mathbb F$ if its minimal polynomial is separable.
Question 1. Is an element separable iff it's a simple root of a polynomial defined over the base field? (A separable element satisfies this property, but what about the converse?)
One can also prove the sum, product, and inverse of separable elements are separable.
Question 2. In light of the above, are the following conditions equivalent for?
- The extension consists entirely of simple roots of polynomials over the base fied.
- The extension is generated by simple roots of polynomials over the base field. Thus every element upstairs is a rational function in a simple root of a polynomial downstairs.
Best Answer
Question 1: The answer is yes, because the roots of an irreducible polynomial $f\in\mathbb{E}[X]$ possess equal multiplicities. This can be shown using formal derivatives.
If now $\alpha$ is a simple root of the polynomial $g\in\mathbb{E}[X]$, consider the minimal polynomial $f$ of $\alpha$. It divides $g$, hence has $\alpha$ as a simple root. Consequently all roots of $f$ are simple.
Question 2: The answer is yes.