Let K be a field and f a member of K[x] a separable polynomial. Prove that the simple extension M= K[x]/(f) (where (f) is the ideal generated by f) is separable over K. Deduce that if a1,...an are separable elements over K, then the extension K(a1,...,an) is separable over K. Conclude that if f is any separable polynomial, then the splitting field of f over K is separable over K.

Any ideas? I think for the first part I should be using a theorem about the number of homomorphisms from M to a splitting field, but I don't quite see how it fits in...

Many thanks.

2. Originally Posted by KSM08
Let K be a field and f a member of K[x] a separable polynomial. Prove that the simple extension M= K[x]/(f) (where (f) is the ideal generated by f) is separable over K.
This is the sketch of the proof for f being an "irreducible" separable polynomial in K[x].

Let $\displaystyle \alpha$ be a root of f. Then M is a simple extension of K such that

$\displaystyle M \cong K(\alpha) \cong K[x]/(f)$.

Every element of $\displaystyle K(\alpha)$ including $\displaystyle \alpha$ is separable over K, since the irreducible polynomial of $\displaystyle \alpha$, i.e., f is separable by hypothesis.

Deduce that if a1,...an are separable elements over K, then the extension K(a1,...,an) is separable over K.
Let $\displaystyle S=\{a_1, a_2, \cdots, a_n\}$ be a set of separable elements over K; let $\displaystyle Y=\{f_k\}_{k=1}^{n}$ be the corresponding set of irreducible polynomials of $\displaystyle S$. Let $\displaystyle Z$ be the set of all zeroes of polynomials in Y. Then K(Z) is the splitting field of polynomials in Y implying that it is separable over K. It follows that K(S), which is a subfield of K(Z), is separable over K.

Conclude that if f is any separable polynomial, then the splitting field of f over K is separable over K.
Let $\displaystyle f \in K[x]$ be separable; let $\displaystyle Y=\{f_k\}_{k=1}^{n}$ be irreducible factors of f. Then, the splitting field of Y over K is K(Z), where Z is the set of all zeroes of polynomials in Y. The splitting field of f over K, i.e., $\displaystyle K(Z)$ is separable over K, since f splits in its splitting field K(Z) such that $\displaystyle f(x)=c\prod_{k=1}^{n}\prod_{j=1}^{k_m}(x-\alpha_{kj})$, where $\displaystyle c \in K$ and each $\displaystyle f_k$ has $\displaystyle k_m$ distinct roots.