Separable extension
Idea
The idea of roots being “separable” in the sense that there are no repeated roots is important because it tells us we are working with a “simple” polynomial that can generate all other polynomials with this set of roots (see Nullstellensatz).
A separable element $\alpha$ is one whose minimal polynomial is separable. A separable extension is one in which every element is separable.
Definition
Definition. The number $[K:k]s\coloneqq \left|\sum{K/k}^{\bar{k}/k}\right|$ is called the seperable degree of the algebraic extension $K/k$.
Definition. A finite extension $K/k$ is called finite seperable iff $[K:k]_s=[K:k]$.
Definition. An algebraic element $\alpha\in K$ is called seperable iff the extension $k(\alpha)/k$ is finite seperable. An algebraic extension $K/k$ is called seperable iff all $\alpha\in K$ are seperable over $k$.
By (Property 4) a seperable element is one whose minimal polynomial has no repeated roots, hence the roots of its minimal polynomial are “separated” from each other.
Properties
- $[L:K]_s[K:k]_s=[L:k]_s$ if all three are finite.
- If $K/k$ is a finite extension then $[K:k]_s\leq [K:k]$.
- $\alpha$ is separable over $k$ iff $P_\alpha(T)$ has no multiple roots in $\bar{k}$.
- If $K/k$ is finite, then the notions of finite separable and separable coincide.
- If $\text{char}(k)=0$ then any algebraic extension $K/k$ is separable. If $\text{char}(k)=p$ and $K/k$ is finite then $[K:k]=p^v[K:k]_s$ for some nonnegative integer $v$.
- Every finite separable extension $K/k$ is of the form $k(\alpha)$ for some $\alpha\in K$.
- If $k$ and $K$ are finite fields then the extension $K/k$ is always separable.
- Suppose $K/k$ is a finite separable extension, $M/k$ any extension. Then $$K\otimes_k M\simeq \oplus M_i$$ is an $M$-algebra isomorphism, where $M_i$ are finite extensions of $M$ of the type $M_{P_i}$, $P_i\in M[T]$, $\sum \text{deg }P_i=[K:k]$. The set ${M_i}$ is unique up to permutation.
Proofs
2
If $\alpha$ is separable over $k$, then $[k(\alpha):k]s=[k(\alpha):k]$. But if $P\alpha$ had multiple roots, then the former would be less than the latter, as the repeated root would only determine a single map in $\sum_{k(\alpha)/k}^{\bar{k}/k}$ while accounting for more than one degree in $[k(\alpha):k]$.
Conversely, if $P_\alpha$ has no repeated roots then the number of roots is in bijection with elements in $\sum_{k(\alpha)/k}^{\bar{k}/k}$ hence $[k(\alpha):k]_s=[k(\alpha):k]$.
6
We can suppose $k$ is infinite, for otherwise $K$ is a finite element and this is generated by the generator of its multiplicative group. We will prove the following claim:
Claim. If $K$ is separable over $k$ and is generated by two elements then it is generated over $k$ by one element.
This will imply the proposition because starting with any finite number of generators we can repeatedly apply the claim to obtain a single generator. There can only be finitely many generators because $K/k$ is finite separable (in particular the extension is finite).
We will now prove the claim. Suppose $K=k(\alpha,\beta)$. Let ${\sigma_i}=\sum_{K/k}^{\overline{k}/k}$. Define $$P(T)=\prod_{i\neq j}(\sigma_i(\alpha)+\sigma_i(\beta)T-\sigma_j(\alpha)-\sigma_j(\beta)T).$$ This is nonzero because $i\neq j$, so either $\sigma_i\alpha\neq\sigma_j\alpha$ or $\sigma_i\beta\neq\sigma_j\beta$ (\alpha and $\beta$ generate $K/k$.) Hence there exists $t_0\in k$ such that $P(t_0)\neq 0$. Thus for any two $i\neq j$, we have $\sigma_i(\alpha+\beta t_0)\neq \sigma_j(\alpha+\beta t_0)$. Thus $[k(\alpha+\beta t_0):k]_s\geq [k(\alpha,\beta):k]_s$. Since both are separable, $[k(\alpha+\beta t_0):k]\geq[k(\alpha,\beta):k]$. But certainly the former cannot exceed the latter, so the two extensions are eqal.