Lecture. Separable extensions and field embeddings [l7]

Apr 20, 2026

I did’t go to both Lecture 6 and Lecture 7, so I don’t know where it stopped at the end of Lecture 6. But whatever, let’s just pretend that it splits here.

Separable extensions and field embeddings

Definition 1. root & multiplicity [1701]

Let $F$ be a field and $\overline{F}$ be an algebraic closure of $F$. Let $f(x) \in F[x]$. A root $\alpha \in \overline{F}$ of $f(x)$ is said to be of multiplicity $m$ if $$ (x - \alpha)^{m} \mid f(x), \text{ and } (x - \alpha)^{m+1} \nmid f(x) \text{ in } \overline{F}[x]. $$ A root of multiplicity $1$ is called a simple root. A root of multiplicity $\geq 2$ is called a multiple root.

Proposition 2. roots of an irreducible have the same multiplicity [1702]

Let $f \in F[x]$ be an irreducible polynomial. Then all roots of $f(x)$ in $\overline{F}$ have the same multiplicity.

Proof. sketch [1702A]

WLOG, assume $f(x)$ is monic in $F[x]$.

Let $\alpha, \beta \in \overline{F}$ be roots of $f(x)$ with multiplicities $a, b$ respectively. Suppose $a < b$. Since $f(x)$ is monic and irreducible in $F[x]$, $m_{\alpha, F}(x) = m_{\beta, F}(x) = f(x)$. Therefore, by Theorem 1408, there exists a field isomorphism $$ \varphi : F(\alpha) \to F(\beta) $$ such that $\varphi(\alpha) = \beta$ and $\varphi(r) = r$ for all $r \in F$.

Over $F(\alpha)$, $f(x) = m_{\alpha, F}(x) = (x - \alpha)^{a}g(x)$ for some $g(x) \in F(\alpha)[x]$ such that $(x - \alpha) \nmid g(x)$. Over $F(\beta)$, $f(x) = m_{\beta, F}(x) = (x - \beta)^{b}h(x)$ for some $h(x) \in F(\beta)[x]$ such that $(x - \beta) \nmid h(x)$. Then $$ \begin{align*} \varphi \left( f(x) \right) &= (x - \varphi(\alpha))^{a} \varphi \left( g(x) \right) \\ &= (x- \beta)^{a} \varphi \left( g(x) \right) \\ &= (x - \beta)^{b}h(x). \end{align*} $$ It follows that $\varphi \left( g(x) \right) = (x - \beta)^{b-a}h(x)$. Since $b > a$, $(x - \beta) \mid \varphi \left( g(x) \right)$. Apply $\varphi ^{-1}$ on $\varphi \left( g(x) \right)$, it turns out that $$ g(x) = \varphi ^{-1} \varphi \left( g(x) \right) = (x - \alpha)^{b-a} \varphi ^{-1} \left( h(x) \right). $$ Hence $(x - \alpha) \mid g(x)$ in $F(\alpha)$, contradicting the fact that $(x - \alpha) \nmid g(x)$. So we must have $a = b$, i.e. the multiplicities of $\alpha$ and $\beta$ are equal. Because $\alpha, \beta$ are arbitrary, all roots of $f(x)$ in $\overline{F}$ have the same multiplicity.

Definition 3. separable extensions [1703]

Let $F$ be a field and $\overline{F}$ be an algebraic closure of $F$.

A polynomial in $F[x]$ is called separable if all its roots in $\overline{F}$ are simple. A polynomial which is not separable is called inseparable.
Let $K / F$ be a field extension and $\alpha \in K$ be an element algebraic over $F$. We say $\alpha$ is separable over $F$ if its minimal polynomial over $F$ is separable.
An algebraic extension $K / F$ is called separable if every element in $K$ is separable over $F$. An algebraic extension $K / F$ is called inseparable if it is not separable.

Let $f(x) \in F[x]$. Its derivative $f'(x)$ is defined as $$ f'(x) = na_{n}x^{n-1} + \cdots + 2a_{2}x + a_{1}. $$ One can check that for $f(x), g(x) \in F[x]$, we have $$ \left( f(x) + g(x) \right)' = f'(x) + g'(x), \qquad \left( f(x)g(x) \right)' = f'(x)g(x) + f(x)g'(x). $$

Proposition 4. some propositions [1704]

$\alpha \in \overline{F}$ is a multiple root of $f(x) \in F[x]$ if and only if it is a root of both $f(x)$ and $f'(x)$.
$f(x) \in F[x]$ is separable if and only if $\gcd \left( f(x), f'(x) \right) = 1$.
An irreducible polynomial in $F[x]$ is separable if and only if $f'(x) \neq 0$.
If $F$ is a field of characteristic $0$, then every irreducible polynomial in $F[x]$ is separable and every field extension of a field of characteristic $0$ is separable.

Proof. sketch [1704A]

Omitted.
Let $d(x) = \gcd(f(x), f'(x))$.

Claim: $\alpha \in \overline{F}$ is a common root of $f(x)$ and $f'(x)$ if and only if $\alpha$ is a root of $d(x)$. Then, just follow part 1.

Let’s prove the claim.

$\impliedby$: It is clear that $d(\alpha) = 0 \implies f(\alpha) = f'(\alpha) = 0$.

$\implies$: Since $F[x]$ is a PID, there exists $a(x), b(x) \in F[x]$ such that $a(x)f(x) + b(x)f'(x) = d(x)$. Then $d(\alpha) = a(\alpha)f(\alpha) + b(\alpha)f'(\alpha) = 0$.
A direct proposition of part 2.
By part 3.
$\operatorname{ch}(F) = 0 \implies f'(x) \neq 0$, then use part 3. For every $K / F$ and $\alpha \in K$, by part 4, $m_{\alpha, F}(x)$ is separable, so $\alpha$ is separable over $F$, hence $K / F$ is separable.