## 9.12 Separable extensions

In characteristic $p$ something funny happens with irreducible polynomials over fields. We explain this in the following lemma.

Lemma 9.12.1. Let $F$ be a field. Let $P \in F[x]$ be an irreducible polynomial over $F$. Let $P' = \text{d}P/\text{d}x$ be the derivative of $P$ with respect to $x$. Then one of the following two cases happens

1. $P$ and $P'$ are relatively prime, or

2. $P'$ is the zero polynomial.

The second case can only happen if $F$ has characteristic $p > 0$. In this case $P(x) = Q(x^ q)$ where $q = p^ f$ is a power of $p$ and $Q \in F[x]$ is an irreducible polynomial such that $Q$ and $Q'$ are relatively prime.

Proof. Note that $P'$ has degree $< \deg (P)$. Hence if $P$ and $P'$ are not relatively prime, then $(P, P') = (R)$ where $R$ is a polynomial of degree $< \deg (P)$ contradicting the irreducibility of $P$. This proves we have the dichotomy between (1) and (2).

Assume we are in case (2) and $P = a_ d x^ d + \ldots + a_0$. Then $P' = da_ d x^{d - 1} + \ldots + a_1$. In characteristic $0$ we see that this forces $a_ d, \ldots , a_1 = 0$ which would mean $P$ is constant a contradiction. Thus we conclude that the characteristic $p$ is positive. In this case the condition $P' = 0$ forces $a_ i = 0$ whenever $p$ does not divide $i$. In other words, $P(x) = P_1(x^ p)$ for some nonconstant polynomial $P_1$. Clearly, $P_1$ is irreducible as well. By induction on the degree we see that $P_1(x) = Q(x^ q)$ as in the statement of the lemma, hence $P(x) = Q(x^{pq})$ and the lemma is proved. $\square$

Definition 9.12.2. Let $F$ be a field. Let $K/F$ be an extension of fields.

1. We say an irreducible polynomial $P$ over $F$ is separable if it is relatively prime to its derivative.

2. Given $\alpha \in K$ algebraic over $F$ we say $\alpha$ is separable over $F$ if its minimal polynomial is separable over $F$.

3. If $K$ is an algebraic extension of $F$, we say $K$ is separable1 over $F$ if every element of $K$ is separable over $F$.

By Lemma 9.12.1 in characteristic $0$ every irreducible polynomial is separable, every algebraic element in an extension is separable, and every algebraic extension is separable.

Lemma 9.12.3. Let $K/E/F$ be a tower of algebraic field extensions.

1. If $\alpha \in K$ is separable over $F$, then $\alpha$ is separable over $E$.

2. if $K$ is separable over $F$, then $K$ is separable over $E$.

Proof. We will use Lemma 9.12.1 without further mention. Let $P$ be the minimal polynomial of $\alpha$ over $F$. Let $Q$ be the minimal polynomial of $\alpha$ over $E$. Then $Q$ divides $P$ in the polynomial ring $E[x]$, say $P = QR$. Then $P' = Q'R + QR'$. Thus if $Q' = 0$, then $Q$ divides $P$ and $P'$ hence $P' = 0$ by the lemma. This proves (1). Part (2) follows immediately from (1) and the definitions. $\square$

Lemma 9.12.4. Let $F$ be a field. An irreducible polynomial $P$ over $F$ is separable if and only if $P$ has pairwise distinct roots in an algebraic closure of $F$.

Proof. Suppose that $\alpha \in \overline{F}$ is a root of both $P$ and $P'$. Then $P = (x - \alpha )Q$ for some polynomial $Q$. Taking derivatives we obtain $P' = Q + (x - \alpha )Q'$. Thus $\alpha$ is a root of $Q$. Hence we see that if $P$ and $P'$ have a common root, then $P$ does not have pairwise distinct roots. Conversely, if $P$ has a repeated root, i.e., $(x - \alpha )^2$ divides $P$, then $\alpha$ is a root of both $P$ and $P'$. Combined with Lemma 9.11.2 this proves the lemma. $\square$

Lemma 9.12.5. Let $F$ be a field and let $\overline{F}$ be an algebraic closure of $F$. Let $p > 0$ be the characteristic of $F$. Let $P$ be a polynomial over $F$. Then the set of roots of $P$ and $P(x^ p)$ in $\overline{F}$ have the same cardinality (not counting multiplicity).

Proof. Clearly, $\alpha$ is a root of $P(x^ p)$ if and only if $\alpha ^ p$ is a root of $P$. In other words, the roots of $P(x^ p)$ are the roots of $x^ p - \beta$, where $\beta$ is a root of $P$. Thus it suffices to show that the map $\overline{F} \to \overline{F}$, $\alpha \mapsto \alpha ^ p$ is bijective. It is surjective, as $\overline{F}$ is algebraically closed which means that every element has a $p$th root. It is injective because $\alpha ^ p = \beta ^ p$ implies $(\alpha - \beta )^ p = 0$ because the characteristic is $p$. And of course in a field $x^ p = 0$ implies $x = 0$. $\square$

Let $F$ be a field and let $P$ be an irreducible polynomial over $F$. Then we know that $P = Q(x^ q)$ for some separable irreducible polynomial $Q$ (Lemma 9.12.1) where $q$ is a power of the characteristic $p$ (and if the characteristic is zero, then $q = 1$2 and $Q = P$). By Lemma 9.12.5 the number of roots of $P$ and $Q$ in any algebraic closure of $F$ is the same. By Lemma 9.12.4 this number is equal to the degree of $Q$.

Definition 9.12.6. Let $F$ be a field. Let $P$ be an irreducible polynomial over $F$. The separable degree of $P$ is the cardinality of the set of roots of $P$ in any algebraic closure of $F$ (see discussion above). Notation $\deg _ s(P)$.

The separable degree of $P$ always divides the degree and the quotient is a power of the characteristic. If the characteristic is zero, then $\deg _ s(P) = \deg (P)$.

Situation 9.12.7. Here $F$ be a field and $K/F$ is a finite extension generated by elements $\alpha _1, \ldots , \alpha _ n \in K$. We set $K_0 = F$ and

$K_ i = F(\alpha _1, \ldots , \alpha _ i)$

to obtain a tower of finite extensions $K = K_ n / K_{n - 1} / \ldots / K_0 = F$. Denote $P_ i$ the minimal polynomial of $\alpha _ i$ over $K_{i - 1}$. Finally, we fix an algebraic closure $\overline{F}$ of $F$.

Let $F$, $K$, $\alpha _ i$, and $\overline{F}$ be as in Situation 9.12.7. Suppose that $\varphi : K \to \overline{F}$ is a morphism of extensions of $F$. Then we obtain maps $\varphi _ i : K_ i \to \overline{F}$. In particular, we can take the image of $P_ i \in K_{i - 1}[x]$ by $\varphi _{i - 1}$ to get a polynomial $P_ i^\varphi \in \overline{F}[x]$.

Lemma 9.12.8. In Situation 9.12.7 the correspondence

$\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F}) \longrightarrow \{ (\beta _1, \ldots , \beta _ n)\text{ as below}\} , \quad \varphi \longmapsto (\varphi (\alpha _1), \ldots , \varphi (\alpha _ n))$

is a bijection. Here the right hand side is the set of $n$-tuples $(\beta _1, \ldots , \beta _ n)$ of elements of $\overline{F}$ such that $\beta _ i$ is a root of $P_ i^\varphi$.

Proof. Let $(\beta _1, \ldots , \beta _ n)$ be an element of the right hand side. We construct a map of fields corresponding to it by induction. Namely, we set $\varphi _0 : K_0 \to \overline{F}$ equal to the given map $K_0 = F \subset \overline{F}$. Having constructed $\varphi _{i - 1} : K_{i - 1} \to \overline{F}$ we observe that $K_ i = K_{i - 1}[x]/(P_ i)$. Hence we can set $\varphi _ i$ equal to the unique map $K_ i \to \overline{F}$ inducing $\varphi _{i - 1}$ on $K_{i - 1}$ and mapping $x$ to $\beta _ i$. This works precisely as $\beta _ i$ is a root of $P_ i^\varphi$. Uniqueness implies that the two constructions are mutually inverse. $\square$

Lemma 9.12.9. In Situation 9.12.7 we have $|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| = \prod _{i = 1}^ n \deg _ s(P_ i)$.

Proof. This follows immediately from Lemma 9.12.8. Observe that a key ingredient we are tacitly using here is the well-definedness of the separable degree of an irreducible polynomial which was observed just prior to Definition 9.12.6. $\square$

We now use the result above to characterize separable field extensions.

Lemma 9.12.10. Assumptions and notation as in Situation 9.12.7. If each $P_ i$ is separable, i.e., each $\alpha _ i$ is separable over $K_{i - 1}$, then

$|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| = [K : F]$

and the field extension $K/F$ is separable. If one of the $\alpha _ i$ is not separable over $K_{i - 1}$, then $|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| < [K : F]$.

Proof. If $\alpha _ i$ is separable over $K_{i - 1}$ then $\deg _ s(P_ i) = \deg (P_ i) = [K_ i : K_{i - 1}]$ (last equality by Lemma 9.9.2). By multiplicativity (Lemma 9.7.7) we have

$[K : F] = \prod [K_ i : K_{i - 1}] = \prod \deg (P_ i) = \prod \deg _ s(P_ i) = |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})|$

where the last equality is Lemma 9.12.9. By the exact same argument we get the strict inequality $|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| < [K : F]$ if one of the $\alpha _ i$ is not separable over $K_{i - 1}$.

Finally, assume again that each $\alpha _ i$ is separable over $K_{i - 1}$. We will show $K/F$ is separable. Let $\gamma = \gamma _1 \in K$ be arbitrary. Then we can find additional elements $\gamma _2, \ldots , \gamma _ m$ such that $K = F(\gamma _1, \ldots , \gamma _ m)$ (for example we could take $\gamma _2 = \alpha _1, \ldots , \gamma _{n + 1} = \alpha _ n$). Then we see by the last part of the lemma (already proven above) that if $\gamma$ is not separable over $F$ we would have the strict inequality $|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| < [K : F]$ contradicting the very first part of the lemma (already prove above as well). $\square$

Lemma 9.12.11. Let $K/F$ be a finite extension of fields. Let $\overline{F}$ be an algebraic closure of $F$. Then we have

$|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| \leq [K : F]$

with equality if and only if $K$ is separable over $F$.

Proof. This is a corollary of Lemma 9.12.10. Namely, since $K/F$ is finite we can find finitely many elements $\alpha _1, \ldots , \alpha _ n \in K$ generating $K$ over $F$ (for example we can choose the $\alpha _ i$ to be a basis of $K$ over $F$). If $K/F$ is separable, then each $\alpha _ i$ is separable over $F(\alpha _1, \ldots , \alpha _{i - 1})$ by Lemma 9.12.3 and we get equality by Lemma 9.12.10. On the other hand, if we have equality, then no matter how we choose $\alpha _1, \ldots , \alpha _ n$ we get that $\alpha _1$ is separable over $F$ by Lemma 9.12.10. Since we can start the sequence with an arbitrary element of $K$ it follows that $K$ is separable over $F$. $\square$

Lemma 9.12.12. Let $E/k$ and $F/E$ be separable algebraic extensions of fields. Then $F/k$ is a separable extension of fields.

Proof. Choose $\alpha \in F$. Then $\alpha$ is separable algebraic over $E$. Let $P = x^ d + \sum _{i < d} a_ i x^ i$ be the minimal polynomial of $\alpha$ over $E$. Each $a_ i$ is separable algebraic over $k$. Consider the tower of fields

$k \subset k(a_0) \subset k(a_0, a_1) \subset \ldots \subset k(a_0, \ldots , a_{d - 1}) \subset k(a_0, \ldots , a_{d - 1}, \alpha )$

Because $a_ i$ is separable algebraic over $k$ it is separable algebraic over $k(a_0, \ldots , a_{i - 1})$ by Lemma 9.12.3. Finally, $\alpha$ is separable algebraic over $k(a_0, \ldots , a_{d - 1})$ because it is a root of $P$ which is irreducible (as it is irreducible over the possibly bigger field $E$) and separable (as it is separable over $E$). Thus $k(a_0, \ldots , a_{d - 1}, \alpha )$ is separable over $k$ by Lemma 9.12.10 and we conclude that $\alpha$ is separable over $k$ as desired. $\square$

Lemma 9.12.13. Let $E/k$ be a field extension. Then the elements of $E$ separable over $k$ form a subextension of $E/k$.

Proof. Let $\alpha , \beta \in E$ be separable over $k$. Then $\beta$ is separable over $k(\alpha )$ by Lemma 9.12.3. Thus we can apply Lemma 9.12.12 to $k(\alpha , \beta )$ to see that $k(\alpha , \beta )$ is separable over $k$. $\square$

[1] For nonalgebraic extensions this definition does not make sense and is not the correct one. We refer the reader to Algebra, Sections 10.42 and 10.44.
[2] A good convention for this chapter is to set $0^0 = 1$.

## Comments (4)

Comment #4535 by Ashutosh on

Second paragraph of Lemma 09H9 may be needs a line saying that now we prove that $K/F$ is separable.

Comment #7036 by ranjit vohra on

Does every field have a separable extension field?

Comment #7040 by on

If $k$ is a field, then the trivial extension $k/k$ is separable.

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