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My definition of separability is as follows (given in my course).

A polynomial $f(X)\in K(X)$ is separable if and only if $K(X)/(f(X))$ is separable. And an extension $L$ over $K$ is separable iff $[L:K] = [L:K]_s (=\#Hom_K(L,K^a))$.

Is this definition equivalent to the notion of separability given by the roots of the polynomial not being repeated?

I can see that in the case of a polynomial with is the minimal polynomial of an element in L, this is true. Indeed $$ K(X)/(f_\alpha(X)) \cong K[\alpha]$$ and we know that for a simple extension, $$ deg(f) = [L:K]=[L:K]_s = \#Hom_K(L,K^a) = \#roots$$ So the two notions are equivalent for a minimal polynomial. But does this remain true in the general case?

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If a polynomial is not irreducible (namely a minimal polynomial for some element), then $K[x]/(f(x))$ is not a field so you cannot speak about separable field extension. If $f(x)$ is a polynomial with only simple roots, then you can write it as $f(x)=\prod f_i(x)$ where the $f_i$ are irreducible and pairwise coprime. Using the Chinese remainder theorem you get that $K[x]/(f(x))\cong \prod K[x]/f_i(x)$ is a product of fields which are separable extensions of $K$.

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    And the product of separable extensions over $K$ is a separable extension over $K$? – 2017-01-17
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    @tomak it depends on how you define separable extension for rings. If you use the standard definition, as in https://en.wikipedia.org/wiki/Separable_algebra, then the product of field extensions is separable iff each of the fields in the product is a separable extension. – 2017-01-17