Lecture. Separable extensions and field embeddings (cont.) [l8]

Apr 22, 2026

Example 1. $x^p - a$ [1705]

If $F$ is a field of characteristic $p$ and $a \in F$, then $x^{p} - a$ is inseparable. To prove that, we divide it into two cases:

If $a$ is a $p$-th power in $F$, let $\alpha \in \overline{F}$ such that $\alpha^{p} = a$, then $\alpha \in F$.Now $x^{p} - a = x^{p} - \alpha^{p} = (x - \alpha)^{p}$, $\alpha$ has multiplicity $p > 1$. So just by definition, $x^{p} - a$ is inseparable.
If $a$ is NOT a $p$-th power in $F$, then $\alpha \not\in F$. By the HW problem in section 1.4, $m_{\alpha, F}(x) = x^{p} - a$. Now taking the derivative of $x^{p} - a$ gives $px^{p-1} = 0$ and then by part 3 of Proposition 1704, $x^{p} - a$ is inseparable.

It follows that $F(\alpha) / F$ is inseparable in both cases.

Consider $F = \mathbb{F}_{p}(t) = \left\{ \frac{a(t)}{b(t)} \mid a(t), b(t) \in \mathbb{F}_{p}[t], b(t) \neq 0 \right\}$. $t \in F$ is not a $p$-th power in $F$ since if $t = \left( \frac{a(t)}{b(t)} \right)^{p}$ for some $a(t), b(t) \in \mathbb{F}_{p}[t]$ with $b(t) \neq 0$, then we would have $$ t \cdot \left( b(t) \right)^{p} = \left( a(t) \right)^{p}. $$ Suppose $\deg a(t) = m, \deg b(t) = n$. Then comparing degrees of both sides gives $np + 1 = mp \implies 1 = p(m - n)$. Since $p, m, n$ are integers, this is impossible. Therefore, $t$ cannot be a $p$-th power in $F$.

So again we have $x^{p} - t$ is irreducible in $F[x]$ and $F[x] / (x^{p} - t)$ is inseparable.

Definition 2. $F$-embedding [1706]

Let $K, L$ be field extensions of $F$. An $F$-embedding $\varphi : K \to L$ is a field embedding such that $\varphi |_{F} = \operatorname{id}_{F}$.

Proposition 3. the number of $F$-embeddings [1707]

Let $K / F$ be a finite field extension. Then $$ \# \{ F\text{-embeddings } \varphi : K \to \overline{F} \} \leq [K : F], $$ and the equality holds if and only if $K / F$ is separable.

Proof. sketch [1707A]

We prove by induction on $[K : F]$.

Trivial case: $[K : F] = 1$. In this case, $K = F$, the proposition is obviously true.

Assume the statement holds for all field extensions of degree $\leq n - 1$. Given a field extension $K / F$ with $[K : F] = n$, take $\alpha \in K - F$, by Proposition 17010, $$ \begin{align*} &\# \{ F\text{- embeddings } \varphi : K \to \overline{F} \} \\ = &\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \cdot \#\{ F(\alpha)\text{-embeddings }\varphi : K \to \overline{F(\alpha)} = \overline{F} \}. \end{align*} $$ By Proposition 1709, $$ \#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \} \leq [F(\alpha) : F] $$ $[K : F(\alpha)] < n$, so by the induction hypothesis, $$ \# \{ F(\alpha)\text{-embeddings } \varphi : K \to \overline{F} \} \leq [K : F(\alpha)]. $$ Therefore, by Theorem 1207, $$ \#\{ F\text{-embeddings } \varphi : K \to \overline{F} \} \leq [F(\alpha) : F][K : F(\alpha)] = [K : F]. $$ The equality holds if and only if the equality holds for $$ \#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \} \leq [F(\alpha) : F]. $$ So only if $$ \begin{align*} &\# \{ \text{roots of }m_{\alpha, F}(x) \text{ in } \overline{F} \} = [F(\alpha) : F] = \deg m_{\alpha, F}(x) \\ \iff &\text{all roots of }m_{\alpha, F}(x)\text{ are simple} \\ \iff &\alpha\text{ is separable over }F. \end{align*} $$ Since $\alpha \in K - F$ is arbitrarily chosen, it implies that the equality holds if and only if all $\alpha \in K - F$ is separable over $F$, so if and only if $K / F$ is separable.

Remark 4. another definition of separable extension [1708]

In some literature, a finite extension $K / F$ is defined to be separable if $$ \# \{ F\text{-embeddings } \varphi : K \to \overline{F} \} = [K : F]. $$

To prove Proposition 1707, we need the following two propositions:

Proposition 5. [1709]

Let $\alpha \in \overline{F}$ with minimal polynomial $m_{\alpha, F}(x) \in F[x]$.

$\# \{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in } \overline{F} \}$
$\#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} \leq [F(\alpha) : F]$ and the equality holds if and only if $\alpha$ is separable over $F$.

Proof. sketch [1709A]

The map $$\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \to \overline{F}, \qquad \varphi \mapsto \varphi(\alpha)$$ is injective since the image of $\varphi$ is completely determined by $\varphi(\alpha)$. To associate it with the roots of $m_{\alpha, F}(x)$ in $\overline{F}$, suppose $\varphi(\alpha) = \beta \in \overline{F}$. Then since $\varphi |_{F} = \operatorname{id}_{F}$ and $m_{\alpha, F}(x) \in F[x]$, we have $\varphi(m_{\alpha, F}(\alpha)) = m_{\alpha, F}(\beta) = \varphi(0) = 0$, hence $\beta$ is a root of $m_{\alpha, F}(x)$ in $\overline{F}$. Thus $\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \leq \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \}$.

Conversely, given a root $\gamma$ of $m_{\alpha, F}(x)$ in $\overline{F}$, then the kernel of the field homomorphism $F[x] \to \overline{F}, a(x) \mapsto a(\gamma)$ is $\left( m_{\alpha, F}(x) \right)$. Thus we obtain a field embedding $$\tau : F[x] / \left( m_{\alpha, F}(x) \right) \to \overline{F}, \qquad \overline{a(x)} \mapsto a(\gamma).$$ Moreover, by Proposition 1407, $$\sigma : F[x] / \left( m_{\alpha, F}(x) \right) \to F(\alpha), \qquad \overline{a(x)} \mapsto a(\alpha)$$ is a field isomorphism. Then we can check that the field homomorphism $\tau \circ \sigma ^{-1} |_{F} = \operatorname{id}_{F}$ and $\tau \circ \sigma ^{-1} (\alpha) = \gamma$, so $\tau \circ \sigma ^{-1}$ is the desired field embedding. Thus $\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \geq \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \}$.
By $\# \{ \text{roots of }m_{\alpha, F}(x) \text{ in } \overline{F} \} \leq \deg m_{\alpha, F}(x) = [F(\alpha) : F]$, and the equality holds if and only if $\alpha$ is separable over $F$.

Proposition 6. [17010]

Let $K / F$ and $L / K$ be finite field extensions. Then $$ \begin{align*} &\# \{ F\text{-embeddings }\phi : L \to \overline{F} \} \\ =&\# \{ F\text{-embeddings }\varphi : K \to \overline{F} \} \cdot \# \{ K\text{-embeddings }\varphi : L \to \overline{K} \}. \end{align*} $$

Proof. sketch (TODO) [17010A]

Proposition 7. [17011]

Let $K / F$ and $L / K$ be algebraic field extensions. Then $L / F$ is separable if and only if $L / K$ and $K / F$ are both separable.

Proof. sketch (TODO) [17011A]

$\implies$: Suppose $L / F$ is separable. Given $\alpha \in K$, we have that $\alpha \in L$ is hence separable over $F$, so $K / F$ is separable. To show that $L / K$ is separable, consider $m_{\alpha, F}(x)$. Since $\alpha$ is separable over $F$, all roots of $m_{\alpha, F}(x)$ are simple. Moreover, $m_{\alpha, K}(x) \mid m(\alpha, F)(x)$ in $K[x]$, so all roots of $m_{\alpha, K}(x)$ are simple as well $\implies$ $\alpha$ is separable over $K$. Hence $L$ is separable over $K$.

$\impliedby$: Suppose that $K / F$ and $L / K$ are both separable.

Case 1: $[L : F] < \infty$. By Proposition 1707, $$ \#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} = [K : F], \, \#\{ K\text{-embeddings } \varphi : L \to \overline{K} \} = [L : K]. $$ Then Proposition 17010 implies $$ \#\{ F\text{-embeddings }\varphi : L \to \overline{F} \} = [K : F][L : K] = [L : F]. $$ By Proposition 1707, this implies that $L / F$ is separable.

Case 2: $[L : F] = \infty$.