## 9.14 Purely inseparable extensions

Purely inseparable extensions are the opposite of the separable extensions defined in the previous section. These extensions only show up in positive characteristic.

Definition 9.14.1. Let $F$ be a field of characteristic $p > 0$. Let $K/F$ be an extension.

An element $\alpha \in K$ is *purely inseparable* over $F$ if there exists a power $q$ of $p$ such that $\alpha ^ q \in F$.

The extension $K/F$ is said to be *purely inseparable* if and only if every element of $K$ is purely inseparable over $F$.

Observe that a purely inseparable extension is necessarily algebraic. Let $F$ be a field of characteristic $p > 0$. An example of a purely inseparable extension is gotten by adjoining the $p$th root of an element $t \in F$ which does not yet have one. Namely, the lemma below shows that $P = x^ p - t$ is irreducible, and hence

\[ K = F[x]/(P) = F[t^{1/p}] \]

is a field. And $K$ is purely inseparable over $F$ because every element

\[ a_0 + a_1t^{1/p} + \ldots + a_{p - 1}t^{p - 1/p}, a_ i \in F \]

has $p$th power equal to

\[ (a_0 + a_1t^{1/p} + \ldots + a_{p - 1}t^{p - 1/p})^ p = a_0^ p + a_1^ p t + \ldots + a_{p - 1}^ pt^{p - 1} \in F \]

This situation occurs for the field $\mathbf{F}_ p(t)$ of rational functions over $\mathbf{F}_ p$.

Lemma 9.14.2. Let $p$ be a prime number. Let $F$ be a field of characteristic $p$. Let $t \in F$ be an element which does not have a $p$th root in $F$. Then the polynomial $x^ p - t$ is irreducible over $F$.

**Proof.**
To see this, suppose that we have a factorization $x^ p - t = f g$. Taking derivatives we get $f' g + f g' = 0$. Note that neither $f' = 0$ nor $g' = 0$ as the degrees of $f$ and $g$ are smaller than $p$. Moreover, $\deg (f') < \deg (f)$ and $\deg (g') < \deg (g)$. We conclude that $f$ and $g$ have a factor in common. Thus if $x^ p - t$ is reducible, then it is of the form $x^ p - t = c f^ n$ for some irreducible $f$, $c \in F^*$, and $n > 1$. Since $p$ is a prime number this implies $n = p$ and $f$ linear, which would imply $x^ p - t$ has a root in $F$. Contradiction.
$\square$

We will see that taking $p$th roots is a very important operation in characteristic $p$.

Lemma 9.14.3. Let $E/k$ and $F/E$ be purely inseparable extensions of fields. Then $F/k$ is a purely inseparable extension of fields.

**Proof.**
Say the characteristic of $k$ is $p$. Choose $\alpha \in F$. Then $\alpha ^ q \in E$ for some $p$-power $q$. Whereupon $(\alpha ^ q)^{q'} \in k$ for some $p$-power $q'$. Hence $\alpha ^{qq'} \in k$.
$\square$

Lemma 9.14.4. Let $E/k$ be a field extension. Then the elements of $E$ purely-inseparable over $k$ form a subextension of $E/k$.

**Proof.**
Let $p$ be the characteristic of $k$. Let $\alpha , \beta \in E$ be purely inseparable over $k$. Say $\alpha ^ q \in k$ and $\beta ^{q'} \in k$ for some $p$-powers $q, q'$. If $q''$ is a $p$-power, then $(\alpha + \beta )^{q''} = \alpha ^{q''} + \beta ^{q''}$. Hence if $q'' \geq q, q'$, then we conclude that $\alpha + \beta $ is purely inseparable over $k$. Similarly for the difference, product and quotient of $\alpha $ and $\beta $.
$\square$

Lemma 9.14.5. Let $E/F$ be a finite purely inseparable field extension of characteristic $p > 0$. Then there exists a sequence of elements $\alpha _1, \ldots , \alpha _ n \in E$ such that we obtain a tower of fields

\[ E = F(\alpha _1, \ldots , \alpha _ n) \supset F(\alpha _1, \ldots , \alpha _{n - 1}) \supset \ldots \supset F(\alpha _1) \supset F \]

such that each intermediate extension is of degree $p$ and comes from adjoining a $p$th root. Namely, $\alpha _ i^ p \in F(\alpha _1, \ldots , \alpha _{i - 1})$ is an element which does not have a $p$th root in $F(\alpha _1, \ldots , \alpha _{i - 1})$ for $i = 1, \ldots , n$.

**Proof.**
By induction on the degree of $E/F$. If the degree of the extension is $1$ then the result is clear (with $n = 0$). If not, then choose $\alpha \in E$, $\alpha \not\in F$. Say $\alpha ^{p^ r} \in F$ for some $r > 0$. Pick $r$ minimal and replace $\alpha $ by $\alpha ^{p^{r - 1}}$. Then $\alpha \not\in F$, but $\alpha ^ p \in F$. Then $t = \alpha ^ p$ is not a $p$th power in $F$ (because that would imply $\alpha \in F$, see Lemma 9.12.5 or its proof). Thus $F \subset F(\alpha )$ is a subextension of degree $p$ (Lemma 9.14.2). By induction we find $\alpha _1, \ldots , \alpha _ n \in E$ generating $E/F(\alpha )$ satisfying the conclusions of the lemma. The sequence $\alpha , \alpha _1, \ldots , \alpha _ n$ does the job for the extension $E/F$.
$\square$

slogan
Lemma 9.14.6. Let $E/F$ be an algebraic field extension. There exists a unique subextension $F \subset E_{sep} \subset E$ such that $E_{sep}/F$ is separable and $E/E_{sep}$ is purely inseparable.

**Proof.**
If the characteristic is zero we set $E_{sep} = E$. Assume the characteristic if $p > 0$. Let $E_{sep}$ be the set of elements of $E$ which are separable over $F$. This is a subextension by Lemma 9.12.13 and of course $E_{sep}$ is separable over $F$. Given an $\alpha $ in $E$ there exists a $p$-power $q$ such that $\alpha ^ q$ is separable over $F$. Namely, $q$ is that power of $p$ such that the minimal polynomial of $\alpha $ is of the form $P(x^ q)$ with $P$ separable algebraic, see Lemma 9.12.1. Hence $E/E_{sep}$ is purely inseparable. Uniqueness is clear.
$\square$

Definition 9.14.7. Let $E/F$ be an algebraic field extension. Let $E_{sep}$ be the subextension found in Lemma 9.14.6.

The integer $[E_{sep} : F]$ is called the *separable degree* of the extension. Notation $[E : F]_ s$.

The integer $[E : E_{sep}]$ is called the *inseparable degree*, or the *degree of inseparability* of the extension. Notation $[E : F]_ i$.

Of course in characteristic $0$ we have $[E : F] = [E : F]_ s$ and $[E : F]_ i = 1$. By multiplicativity (Lemma 9.7.7) we have

\[ [E : F] = [E : F]_ s [E : F]_ i \]

even in case some of these degrees are infinite. In fact, the separable degree and the inseparable degree are multiplicative too (see Lemma 9.14.9).

Lemma 9.14.8. Let $K/F$ be a finite extension. Let $\overline{F}$ be an algebraic closure of $F$. Then $[K : F]_ s = |\mathop{Mor}\nolimits _ F(K, \overline{F})|$.

**Proof.**
We first prove this when $K/F$ is purely inseparable. Namely, we claim that in this case there is a unique map $K \to \overline{F}$. This can be seen by choosing a sequence of elements $\alpha _1, \ldots , \alpha _ n \in K$ as in Lemma 9.14.5. The irreducible polynomial of $\alpha _ i$ over $F(\alpha _1, \ldots , \alpha _{i - 1})$ is $x^ p - \alpha _ i^ p$. Applying Lemma 9.12.9 we see that $|\mathop{Mor}\nolimits _ F(K, \overline{F})| = 1$. On the other hand, $[K : F]_ s = 1$ in this case hence the equality holds.

Let's return to a general finite extension $K/F$. In this case choose $F \subset K_ s \subset K$ as in Lemma 9.14.6. By Lemma 9.12.11 we have $|\mathop{Mor}\nolimits _ F(K_ s, \overline{F})| = [K_ s : F] = [K : F]_ s$. On the other hand, every field map $\sigma ' : K_ s \to \overline{F}$ extends to a unique field map $\sigma : K \to \overline{F}$ by the result of the previous paragraph. In other words $|\mathop{Mor}\nolimits _ F(K, \overline{F})| = |\mathop{Mor}\nolimits _ F(K_ s, \overline{F})|$ and the proof is done.
$\square$

Lemma 9.14.9 (Multiplicativity). Suppose given a tower of algebraic field extensions $K/E/F$. Then

\[ [K : F]_ s = [K : E]_ s [E : F]_ s \quad \text{and}\quad [K : F]_ i = [K : E]_ i [E : F]_ i \]

**Proof.**
We first prove this in case $K$ is finite over $F$. Since we have multiplicativity for the usual degree (by Lemma 9.7.7) it suffices to prove one of the two formulas. By Lemma 9.14.8 we have $[K : F]_ s = |\mathop{Mor}\nolimits _ F(K, \overline{F})|$. By the same lemma, given any $\sigma \in \mathop{Mor}\nolimits _ F(E, \overline{F})$ the number of extensions of $\sigma $ to a map $\tau : K \to \overline{F}$ is $[K : E]_ s$. Namely, via $E \cong \sigma (E) \subset \overline{F}$ we can view $\overline{F}$ as an algebraic closure of $E$. Combined with the fact that there are $[E : F]_ s = |\mathop{Mor}\nolimits _ F(E, \overline{F})|$ choices for $\sigma $ we obtain the result.

If the extensions are infinite one can write $K$ as the union of all finite subextension $F \subset K' \subset K$. For each $K'$ we set $E' = E \cap K'$. Then we have the formulas of the lemma for $K'/E'/F$ by the first paragraph. Since $[K : F]_ s = \sup \{ [K' : F]_ s\} $ and similarly for the other degrees (some details omitted) we obtain the result in general.
$\square$

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