- Let $L/K$ be a finite extension.
- It is separable iff a given embedding $\sigma: K \to \overline K$ can be extended in $[L:K]$ ways (This number can be at most $[L:K]$.)
- We call the numbe of ways to embed $L$ in $\overline K$ via extending $\sigma$ to be the
*separability degree*of $L/K$.

- We will show for simple extensions $K(\alpha)/K$ that there are at most $[K(\alpha): K]$ ways to extend $\sigma: K \to \overline K$ into $\sigma': K(\alpha) \to \overline K$.
- We use two facts: first, $\sigma'$ is entirely determined by where it sends $\alpha$. Second, $\alpha$ can only go to another root of its minimal polynomial $p \in K[x]$. Thus, there are only finitely many choices, and the minimal polynomial has at most $degree(p)$ unique roots, and $[K(\alpha):K] = degree(p)$. Thus, there are at most $degree(p_\alpha) = [L:K]$ choices of where $\alpha$ can go to, which entirely determines $\sigma'$. Thus there are at most $degree(p) = [K(\alpha):K]$choices for $\sigma'$.
- Given a larger extension, write a sequence of extensions $L = K(\alpha_1)(\alpha_2)\dots(\alpha_n)$. Then, since $[L:K] = [K(\alpha):K][K(\alpha_1, \alpha_2):K(\alpha_1)]$and so on, can repeatedly apply the same argument to bound the number of choices of $\sigma'$.
- In detail, for the case $K(\alpha)/K$, consider the minimal polynomial of $\alpha$, $p(x) \in K[x]$. Then $p(\alpha) = 0$.
- Since $\sigma$ fixes $K$, and $p$ has coefficients from $K$, we have that $\sigma(p(x)) = p(\sigma(x))$.
- Thus, in particular, $\sigma(0) = \sigma(p(\alpha)) = p(\sigma(\alpha))$.
- This implies that $p(\sigma(\alpha)) = 0$, or $\sigma(\alpha)$ is a root of $p$.
- Since $\sigma': L \to \overline K$, $\sigma'$ can only map $\alpha$ to one of the other roots of $p$.
- $p$ has at most $deg(p)$ unique roots [can have repeated roots, or some such, so could have fewer that that ].
- Further, $\sigma'$ is entirely determined by where it maps $\alpha$. Thus, there are at most $[K(\alpha):K]$ ways to extend $\sigma$ to $\sigma'$.

- Given a tower $K \subseteq L \subseteq M \subseteq \overline K$, we fix an embedding $\kappa: K \to \overline K$. If both $L/K$ and $M/L$ are finite and separable, then $\kappa$ extends into $\lambda: L \to \overline K$ through $L/K$ in $[L:K]$ ways, and then again as $\mu: L \to \overline K$ in $[M:L]$ ways.
- This together means that we have $[L:K] \cdot [M:L] = [M:K]$ ways to extend $\kappa$ into $\mu$, which is the maximum possible.
- Thus, $M/K$ is separable.

- Let every $\alpha \in L$ have minimal polynomial that is separable (ie, has distinct roots).
- Then we must show that $L/K$ allows us to extend any embedding $\sigma: K \to \overline K$ in $[L:K]$ ways into $\sigma': L \to K$
- Write $L$ as a tower of extensions. Let $K_0 \equiv K$, and $K_{i+1} \equiv K_i(\alpha_i)$ with $K_n = L$.
- At each step, since the polynomial is separable, we have the maximal number of choices of where we send $\sigma'$. Since degree is multiplicative, we have that $[L:K] = [K_1:K_0][K_2:K_1]\dots[K_{n-1}:K_n$.
- We build $\sigma'$ inductively as $\sigma'_i: K \to K_i$ with $\sigma'_0 \equiv \sigma$.
- Then at step $i$, $\sigma'_{i+1}: K \to K(i+1)$ which is $\sigma'_{i+1}: K \to K_i(\alpha_{i+1})$ has $[K_{i+1}:K_i]$ choices, since $\alpha_{i+1}$ is separable over $K_i$ since its minimal polynomial is separable.
- This means that in toto, we have the correct $[L:K]$ number of choices for $\sigma_n: K \to K_n = L$, which is what it means to be separable by embeddings.

- Let $L/K$ be separable in terms of embeddings. Consider some element $\alpha \in L$, let its minimal polynomial be $p(x)$.
- Write $L = K(\alpha)(\beta_1, \dots, \beta_n)$. Since degree is multiplicative, we have $[L:K] = [K(\alpha):K][K(\alpha, \beta_i):K(\alpha)]$.
- So given an embedding $\sigma: K \to \overline K$,we must be able to extend it in $[L:K]$ ways.
- Since $\sigma$ must send $\alpha$ to a root of $\alpha$, and we need the total to be $[L:K]$, we must have that $p(x)$ has no repeated roots.
- If $p(x)$ had repeated roots, then we will have fewer choices of $\sigma(\alpha)$ thatn $[K(\alpha):K]$, which means the total count of choices for $\sigma'$ will be less than $[L:K]$, thereby contradicting separability.

- Let $L = K(\alpha_1, \dots, \alpha_n)$ be separable, so there are $[L: K]$ ways to extend a map $\kappa: K \to \overline K$ into $\lambda: L \to \overline L$.
- Since we have shown that separable by polyomial implies separable by embedding, we write $L = K(\alpha_1)(\alpha_2)\dots(\alpha_n)$. Each step is separable by the arguments given above in terms of counting automorphisms by where they send $\alpha_i$. Thus, the full $L$ is separable.

- https://math.stackexchange.com/questions/2227777/compositum-of-separable-extension
- https://math.stackexchange.com/questions/1248781/primitive-element-theorem-without-galois-group