This follows immediately from Definition 1.

Suppose first that K is perfect. Since a field is a reduced ring, it follows from Lemma 5 ($V$, p. 35) that every extension of finite degree of $K$ is an etale algebra over $K$; therefore every algebraic extension of $K$ is separable.

Suppose now that $K$ is an imperfect field of characteristic $p \neq 0$. Let $\Omega$ be an algebraic closure of $K$. Since $K$ is imperfect, there exists $b \in K$ not belonging to $K^p$; put $a = b^{1/p}$. Then the extension $K(a)$ of $K$ is p-radical of finite degree. By $V$, p. 26, Prop. 3, there exists a single K-homomorphism of $K(a)$ into $\Omega$, and since $[K(a):K] > 1$, the algebra $K(a)$ is not etale over $K$ ($V$, p. 32, Prop. 4). In other words, the extension $K(a)$ of finite degree of $K$ is not separable.

a) ⇒ d): Under the hypothesis a) there exist two polynomials g and h in K[X] such that fg + f'h = 1 (IV, p. 12). Let a be a root of fin Ω; we have

whence $f'(a) \neq 0$; therefore $a$ is a simple root of fin Ω (IV, p. 17, Prop. 7).

d) ⇒ c): If d) holds, f splits in Ω[X] into a product of distinct factors of degree ≤ 1.

c) ⇒ b): Under the hypothesis c) there exist an element $A \neq 0$ in L and distinct elements $a_1, ..., a_n$ of L such that $f(X) = \lambda (X - a_1) ... (X - a_n)$. If $\deg(f) > 0$, we have (IV, p. 83, Prop. 11),

b) ⇒ a): Let c be the leading coefficient and D the discriminant of f; the resultant of $f'$ and f is equal to ± cD (IV, p. 84, Formula (54)), hence is non-zero; therefore (IV, p. 78, Cor. 2) the polynomials f and $f'$ are relatively prime in K[X].

a) ⇒ e): Let A be the K-algebra K[X]/(f) and x the image of X in A; by III, p. 573, Prop. 22, the A-module $\Omega_K(A)$ is generated by the elements $dx$, subject to the single relation $f'(x)\ dx = 0$. By V, p. 33, Th. 3, the K-algebra A is thus etale if and only if $f'(x)$ is an invertible element in A, which means that f and $f'$ are relatively prime in K[X].

a) $f$ is separable. b) There exists an extension $L$ of $K$ in which $f$ has a simple root. c) The derivative $f'$ off is not zero. d) The field $K$ is of characteristic $0$, or it is of characteristic $p \neq 0$ and $f \not\in K[X^p]$.

We note first that an irreducible polynomial in $K[X]$ is not constant. It is clear that a) implies b) (take an algebraic closure of $K$ for $L$). If $x$ is a simple root off in an extension $L$ of $K$, we have $f'(x) \neq 0$ (IV, p. 17, Prop. 7), so b) implies c), and the equivalence of c) and d) follows from V, p. 9, Cor.

Suppose finally that $f \neq 0$; let $x$ be a root of $f$ in an algebraically closed extension $\Omega$ of $K$. Since $f$ is the minimal polynomial of $x$ over $K$ and $\deg f' < \deg f$, we have $f'(x) \neq 0$, and so $x$ is a simple root of $f$ (IV, p. 17, Prop. 7). Therefore $f$ is separable and we have shown that c) implies a).

If the field $K$ is of characteristic $0$, $K$ is perfect and every irreducible polynomial of $K[X]$ is separable, by d) above. Suppose then that $K$ has characteristic $p \neq 0$.

Suppose first that $K$ is perfect. We have $K[X^p] = K[X]^p$, hence there exists no irreducible polynomial of $K[X]$ belonging to $K[X^p]$. By Prop. 4, every irreducible polynomial of $K[X]$ is then separable.

Suppose next that $K$ is imperfect, whence $K \neq K^p$. Let $a$ be an element of $K$ not belonging to $K^p$; the polynomial $X^p - a$ is irreducible in $K[X]$ (V, p. 24, Lemma 1), and it belongs to $K[X^p]$, and so is not separable.

Let $\Omega$ be an algebraic closure of $K$; iff is separable, $f$ has no repeated factors in $\Omega[X]$ (Prop. 3, d)). Conversely, if $L$ is a perfect extension of $K$ such that $f$ has no repeated factor in $L[X]$, then $f$ is separable in $L[X]$ (Cor. 1 and Remark 2), hence in $K[X]$ (Remark 1).

Let F be a subextension of E of finite degree over K. By V, p. 11, Cor., there exist elements $x_1, \ldots, x_m$ of A such that $F \subset K(x_1, \ldots, x_r)$ and we have

By the hypothesis on A, the algebras $K[x_1], \ldots, K[x_m]$ are etale over K; hence the same is true of $K[x_1] \otimes \cdots \otimes K[x_m]$ (V, p. 32, Cor. 1). Now F is isomorphic to a subalgebra of a quotient algebra of $K[x_1] \otimes \cdots \otimes K[x_m]$, and so is etale (V, p. 30, Prop. 3).

Next we observe that Condition b) implies that the extension E is of finite degree: by Remark 2 of V, p. 18 it suffices to prove that it is algebraic; now if z is an element of $E$ transcendental over $K$ then the subextensions $K(z^n), n \in \mathbf{N}$ are pairwise distinct.

To show that b) $\Rightarrow$ a) we now distinguish two cases: A) If the field $K$ is finite, the field $E$ is a vector space of finite dimension over K and hence is a finite set. Therefore $^1$ (V, p. 78, Lemma 1) there exists an element $x$ of $E$ generating the multiplicative group of $E$, and we have $E = K[x]$.

B) Suppose now that the field $K$ is infinite. If b) holds, the extension $E$ is of finite degree, so b) can also be expressed by saying that $E$ possesses only a finite number of subalgebras. This being so, the implication $b) \Rightarrow a)$ is a consequence of the following more general proposition (for which the hypothesis that the field $K$ be infinite is indispensable, cf. V, p. 153, Ex. 5 of § 7):

Let $A_1, \ldots, A_n$ be the subalgebras of $A$ distinct from $A$. If $x \notin A_1 \cup \ldots \cup A_n$, then the sub-algebra $K[x]$ cannot equal any of the $A_i$ and so must coincide with $A$. Further, since $V$ generates $A$, it is not contained in any of the subspaces $A_i$. Prop. 7 is therefore a consequence of the following lemma:

Lemma 1. — Let $A$ be a vector $K$-space, $V, A_1, \ldots, A_n$ subspaces of $A$. If $\mathrm{Card}(K) \geq n$ and if $V$ is not contained in any of the $A_i$, then $V$ is not contained in $A_1 \cup \ldots \cup A_n$.

Arguing by induction on $n$, we need only prove that if $V \subset A$, and $V \subset A_1 \cup \ldots \cup A_n$, then $V \subset A_1 \cup \ldots \cup A_{n-1}$. Let $x \in V, x \notin A_n$, and let $y$ be arbitrary in $V$. If $y \in Kx$, we have $y \in A_1 \cup \ldots \cup A_{n-1}$; if not, then the elements $x$ and $y + \lambda x, \lambda \in K$ are strictly greater than $n$ in number and belong to $A_1 \cup \ldots \cup A_n$, so two of them belong to the same $A_i$. Thus there exists $i, 1 \leq i \leq n$ such that either $x \in A_i$ and $y + \lambda x \in A_i$ for some $\lambda \in K$, or $y + \mu x \in A_i$ and $y + \mu x \in A_i$ for two distinct scalars $\lambda, \mu \in K$. In both cases we conclude that $x \in A_i$ and $y \in A_i$; but this implies that $i \neq n$, hence $y \in A_1 \cup \ldots \cup A_{n-1}$, as we wished to show.

This completes the proof of the equivalence of a) and b) in Th. 1. Finally if the extension $E$ is separable and of finite degree, condition b) holds, by V, p. 30, Prop. 3.

This follows at once from Prop. 6 (V, p. 39).

$^1$ The reader may convince himself that Th. 1 is used nowhere before the proof of Lemma 1 of V, p. 78.

Suppose first that F is separable over K; then E is separable over K by Prop. 1 (V, p. 36). Moreover, every element of F is separable over K (V, p. 39, Prop. 6) hence over E (V, p. 39, Cor. 2) and so F is separable over E (V, p. 39, Prop. 6).

Conversely, suppose that F is separable over E and E separable over K. Denote by x an element of F and by $f \in E[X]$ the minimal polynomial of x over E. Since E is algebraic over K, Th. 2 (V, p. 18) shows that there exists a subextension E' of E of finite degree over K such that $f \in E'[X]$; then f is at the same time the minimal polynomial of x over E and over E', and since x is separable over E (V, p. 39, Prop. 6) it is so also over E' (V, p. 39, Prop. 5). Write $F' = E'(x)$; then F is separable and of finite degree over E', and since E is separable over K, E' is separable and of finite degree over K (V, p. 36, Prop. 1). Hence F is separable and of finite degree over K, by V, p. 32, Cor. 2. Therefore (V, p. 39, Prop. 6) x is separable over K. We have now shown that every element of F is separable over K, hence F is separable over K (V, p. 39, Prop. 6).

a) If E is separable over K, then E' is separable over K'.

b) Conversely if E' is separable over K' and E and K' are linearly disjoint over K, then E is separable over K.

Assertion a) follows directly from Prop. 6 (V, p. 39).

Under the hypothesis b), let F be a subextension of E of finite degree over K. Then F and K' are linearly disjoint over K, hence the K'-algebra $F_{(K')} = K' \otimes_K F$ is isomorphic to $K'(F)$. Since $K'(F)$ is a subextension of E' of finite degree over K' and E' is algebraic and separable over K', the K'-algebra $K'(F)$ is etale. In other words, the K'-algebra $F_{(K')}$ is etale, and now Cor. 2 of Prop. 4 (V, p. 32) shows that F is etale over K. Thus we have shown E to be separable over K.

The case $p = 1$ is trivial by the Cor. of V, p. 37. Suppose from now on that $p \neq 1$.

By hypothesis E is algebraic over K and we have $E = K(S)$, hence

If E is of finite degree over K, it is a separable extension of K if and only if it is an etale algebra over K; the Cor. of V, p. 35 shows that this happens if and only if $E = K[E^p]$.

Suppose now that $E$ is separable and of infinite degree over $K$. Then $K[E^p]$ is the union of the subrings $K[{E'}^p]$ where $E'$ ranges over the set of subextensions of $E$ of finite degree over $K$; but such an extension $E'$ is separable over $K$ ($V$, p. 36, Prop. 1), whence $E' = K[{E'}^p] \subset K[E^p]$ by what has been said; finally we have $E = K[E^p]$. By induction on $n \geq 0$, the relation $E = K[E^p]$ implies that $E = K[E^{p^n}]$.

Let $K$ be a perfect field of characteristic exponent $p$, and let $E$ be an algebraic extension of $K$. Then $E$ is separable over $K$ ($V$, p. 36, Prop. 2) whence $E = K(E^p)$ by Prop. 11; but we have $K = K^p \subset E^p$, hence $E = K(E^p) = E^p$, and so $E$ is perfect.

We can immediately reduce to the case where $[E : K]$ is finite. Let $(x_i)_i$, be a basis of $E$ over $K$. For $E$ to be linearly disjoint from $K^{p^{-\infty}}$ it is necessary and sufficient that it should be so from $K^{p^n}$ for all $n \geq 0$, and this just means that the relation $\sum_{i \in I} x_i a_i^{p^{-n}} = 0$ implies $a_i = 0$ for all $i \in I$, for any family $(a_i)_i$, of elements of $K$. This in turn means that the family $(x_i^{p^n})_{i \in I}$ is free over $K$, or also that it is a basis of the vector space $E$ over $K$. Put differently, $E$ is linearly disjoint from $K^{p^{-n}}$ if and only if $E = K(E^{p^n})$. Now it suffices to apply Prop. 11.

By Prop. 6, a) ($V$, p. 39) every subextension of $E$ which is algebraic and separable over $K$ is contained in $E_s$. By Prop. 6, b) (loc. cit.) the extension $K(E_s)$ of $K$ is algebraic and separable, whence $K(E_s) \subset E$, and so $K(E_s) = E_s$.

With the notation of the preceding proposition, E, is called the relative separable (algebraic) closure of K in E. When K is perfect, E, is the relative algebraic closure of K in E (V, p. 36, Prop. 2).

Assume now the hypothesis b) and let $x \in E,$. Since x is separable over K, it is so also over F (V, p. 39, Cor. 2), but since E is p-radical over F, x is also p-radical over F, hence $x \in F$ (V, p. 39, Cor. 3).

Finally c) follows from a) and b) and Prop. 12.

For $K'(E_s)$ is a separable algebraic extension of K' by Prop. 10 (V, p. 42); since E is p-radical over E_s, the extension $K'(E)$ of $K'(E_s)$ is p-radical (V, p. 25, Cor.). Now it suffices to apply Prop. 13.

For each integer $n \geq 0$ denote by F, the subextension $K(E^{p^n})$ of E. The sequence $(F_n)_{n \geq 0}$ of vector subspaces of E is descending and E is of finite dimension over K. Hence there exists an integer $m \geq 0$ such that $F_m = F_n$ for all $n \geq m$. We thus have $K(F_m^p) = F_{m+1} = F_m$, hence $F_m$ is a separable extension of K (V, p. 42, Prop. 11); it is clear that E is p-radical over F, and so Prop. 13 implies $E_s = F_m = \bigcap_{n \geq 0} F_n$.

a) The relative separable algebraic closure Ωs of K in Ω is a separable closure of K.*

b) If E and E' are two separable closures of K, there exists a K-isomorphism of E onto E'.

Let F be a separable algebraic extension of Ω_s ; since Ω is algebraically closed, there exists an 0,-homomorphism u of F into Ω (V, p. 20, Th. 1). By Prop. 9 (V, p. 42) $u(F)$ is separable over K, hence $u(F) = \Omega_s$. Therefore F is a trivial extension of $\Omega_s$ and so $\Omega_s$ is separably closed, whence a).

Let E be a separable closure of K. Since E is an algebraic extension of K, there exists a K-homomorphism v of E into Ω (V, p. 20, Th. 1). Hence $v(E)$ is separable algebraic over K, whence $v(E) \subset \Omega_s$. By V, p. 42, Prop. 9, $\Omega_s$ is separable over $v(E)$ and since the field $v(E)$ is separably closed, we have $v(E) = \Omega_s$. It follows that v is a K-isomorphism of E onto $\Omega_s$. Now b) is an immediate consequence.

Let Ω be an algebraic closure of E ; we have $\Omega_s \subset E$ and it suffices to treat the case where $E = \Omega_s$. Since F is an algebraic extension of K, there exists a K-homomorphism u of F into Ω (V, p. 20, Th. 1). Since the field $u(F)$ is separable over K, we have $u(F) \subset \Omega_s$ and u defines a K-homomorphism of F into $\Omega_s = E$.

The field $\Omega$ is perfect and E is p-radical over $E_s$, by V, p. 44, Prop. 13; therefore Prop. 3 (V, p. 26) shows that every K-homomorphism of E, into $\Omega$ extends in a unique fashion to a K-homomorphism of E into $\Omega$; we thus have $[E : K]_s = [E_s : K]$. Since E, is a separable extension of finite degree of K, it is an etale algebra over K; so we have $[E_s : K]_s = [E_s : K]$ by V, p. 32, Prop. 4, and the result follows.

With the preceding notation, the degree of E over $E_s$ is called the inseparable degree of E over K and is denoted by $[E : K]_i$. We thus have

(1)

by Prop. 15.

When K is of characteristic 0, then $E = E_s$, and so $[E : K]_s = [E : K]$ and $[E : K]_i = 1$. If K is of characteristic $p \neq 0$, the number $[E : K]_i$ is a power of p because E is p-radical over $E_s$ (V, p. 44, Prop. 13 and p. 26, Prop. 4). It should be noted that $[E : K]_i$ is not necessarily equal to the highest power of $p$ dividing $[E : K]_s$, nor equal to the degree $[E_s : K]$ of the relativep-radical closure of E in K (V, p. 152, Ex. 3 and 2).

a) If $E \subset F$, then $[F : K]_s = [F : E]_s \cdot [E : K]_s$, and $[F : K]_i = [F : E]_i \cdot [E : K]_i$.

b) Let $K'$ be a subextension of $\Omega$; then we have

and equality holds if $K'$ is linearly disjoint from E over K.

c) We have $[K(E \cup F) : K]_s \leq [E : K]_s \cdot [F : K]_s$, and $[K(E \cup F) : K]_i \leq [E : K]_i \cdot [F : K]_i$, and equality holds if E and F are linearly disjoint over K.

The assertion about the separable degrees in a) follows from (9) (V, p. 32). Since $[F : K] = [F : E] \cdot [E : K]$, the assertion about the inseparable degrees follows from this and (1).

By Cor. 1 of Prop. 13 (V, p. 44) and Prop. 15, we have

(2)

when K' is linearly disjoint from E over K, then E, is linearly disjoint from K' over K and E is linearly disjoint from K'(E_s) over E_s (V, p. 15, Prop. 8). Assertion b) now follows from Prop. 5 (V, p. 14).

By a) we have

by b) we have Hence we obtain the inequality The assertion of c) about the inseparable degrees is proved in a similar fashion.