LECTURE 20: SEPARABLE EXTENSIONS
Multiplicity of zeros. In the previous lectures the notion of the index {E : F } of
a field extension played a central rôle. Here we investigate under
√ conditions
√ what
{E : F } = [E : F ] - we’ve already seen this is the case for Q( 2, 3) : Q where
the index and degree are both equal to 4. A finite extension which satisfies this
property is called a separable extension, as discussed below.
First consider a simple algebraic extension F (α) : F . We know that for every
distinct root β of irr(α, F ) there exists a conjugation isomorphism ψα,β : F (α) →
F (β), which extends the identity ι : F → F . Since isomorphisms must map α to
some conjugate over F , these are all possible extensions of the identity. Thus, we
have the following lemma.
Lemma. Let F (α) : F be a simple algebraic extension. Then {F (α) : F } is
precisely the number of distinct roots of irr(α, F ).
This suggests we ought to study the multiplicities of roots.
Definition. Let F be a field and f ∈ F [x]. We say a zero α ∈ F̄ of f is of
multiplicity ν if ν is the greatest integer such that (x − α)ν divides f in F̄ [x].
The key point is that if every root of f ∈ F [x] has multiplicity 1, then the number
of distinct roots of f is then clearly deg f .
Corollary. If F (α) : F be a simple algebraic extension and every root of irr(α, F )
has multiplicity 1, then {F (α) : F } = [F (α) : F ].
In all the examples we’ll be interested in, every irreducible polynomial over F
will have zeros of multiplicity 1 over F̄ and so we’ll be able to deduce the identity
{F (α) : F } = [F (α) : F ]. Our first step towards making this assertion precise is
the following theorem.
Theorem. Let f ∈ F [x] be irreducible. Then all the zeros of f have the same
multiplicity ν ∈ N. In particular, f has a factorisation in F̄ [x] of the form
f (x) = a
m
Y
(x − αj )ν
j=1
where α1 , . . . , αm ∈ F̄ are the distinct zeros of f and a ∈ F .
In the cases we’re primarily interested in it will actually only be possible for the
zeros of an irreducible polynomial to have multiplicity 1. However, if the field F is
infinite and of characteristic p > 0, then higher multiplicities can be observed (one
may consult the textbook for an example).
Proof (of Theorem). Let α, β ∈ F̄ be zeros of f so that α and β are conjugate over
F . We therefore have a conjugate isomorphism ψα,β : F (α) → F (β) which can be
extended to an automorphism ψ̃α,β : F̄ → F̄ . This in turn induces an automorphism
τ : F̄ [x] → F̄ [x] which fixes the linear polynomial x; in particular,
τ(
d
X
j=0
aj xj ) :=
d
X
j=0
1
ψ̃α,β (aj )xj
2
LECTURE 20: SEPARABLE EXTENSIONS
whenever a0 , . . . , ad ∈ F̄ . Now, since f ∈ F [x] (and so the coefficients of f lie in F ),
it follows that τ fixes f . On the other hand, by the definition of the conjugation
isomorphism we have
τ ((x − α)ν ) = (x − β)ν .
If ν is the multiplicity of α, then we have that f (x) = (x−α)ν g(x) for some g ∈ F̄ [x]
and so
f (x) = τ (f (x)) = τ ((x − α)ν g(x)) = (x − β)ν τ (g(x)).
Hence the multiplicity of α is less than or equal to that of β, but by a symmetric
argument we see that the multiplicities must be in fact equal.
Theorem. If E : F is a finite extension, then {E : F } | [E : F ].
Proof. Since E : F is finite, it follows that E = F (α1 , . . . , αn ) for some α1 , . . . , αn ∈
E. Let ni denote the number of distinct zeros of irr(αi , F (α1 , . . . , αi−1 )) and νi their
common multiplicity. Thus, it follows that
[F (α1 , . . . , αi ) : F (α1 , . . . , αi−1 )] = deg irr(αi , F (α1 , . . . , αi−1 ))
= ni νi
= {F (α1 , . . . , αi ) : F (α1 , . . . , αi−1 )}νi .
Thus, by the tower law we have
[E : F ] =
=
n
Y
[F (α1 , . . . , αi ) : F (α1 , . . . , αi−1 )]
i=1
n
Y
{F (α1 , . . . , αi ) : F (α1 , . . . , αi−1 )}νi
i=1
= {E : F }
n
Y
νi ,
i=1
concluding the proof.
Separable extensions.
Definition. Let F be a field.
• A finite extension E : F is a separable extension of F if {E : F } = [E : F ].
• An element α ∈ F̄ is separable over F if F (α) : F is a separable extension.
• A polynomial f ∈ F [x] is separable over F if every zero of f is separable
over F .
√ √
√ √
√ √
Example.
• Q( 2, 3) : Q is separable since [Q( 2, 3) : Q] = {Q( 2, 3) :
Q}√
= 4.
√
√
4
• Q( 2, i) : Q is separable since [Q( 4 2, i) : Q] = {Q( 4 2, i) : Q} = 8.
• From our earlier observations, if α ∈ F̄ and all the zeros of irr(α, F )
have multiplicity 1 (or, equivalently, there are deg(α, F ) distinct zeros of
irr(α, F ), then α is separable over F .
• From the previous example we see that an irreducible polynomial f ∈ F [x]
is separable over F if and only if all its zeros have multiplicity 1.
Theorem. If K : E and E : F are finite extensions, then K : F is separable if and
only if K : E and E : F are separable.
Proof. If K : F is separable, then it follows from the tower laws for the index and
degree of a finite extension that
[K : E][E : F ] = {K : E}{E : F }.
LECTURE 20: SEPARABLE EXTENSIONS
3
In addition, we know that {K : E} | [K : E] and {E : F } | [E : F ], which implies
the desired equality.
The converse is an immediate consequence of the tower laws.
Corollary. Suppose E : F is a finite extension. The following are equivalent:
i) E : F is separable.
ii) Every α ∈ E is separable over F .
Proof. i) =⇒ ii). If α ∈ E, then we know F (α) : F is separable by the previous
theorem.
ii) =⇒ i) Since E is finite, there exist α1 , . . . , αn ∈ E such that
F < F (α1 ) < F (α1 , α2 ) < · · · < F (α1 , . . . , αn ) = E
which are, by hypothesis, all separable elements. We claim αi is separable over
F (α1 , . . . , αi−1 ) for i = 1, . . . , n. Indeed, clearly
irr(αi , F (α1 , . . . , αi−1 )) | irr(αi , F )
and since αi is a root of multiplicity 1 for irr(αi , F ), it must be a root of multiplicity 1 for irr(αi , F (α1 , . . . , αi−1 )). This verifies the claim and thus, by iterative
application of the previous theorem, we see that E : F is separable.
Jonathan Hickman, Department of mathematics, University of Chicago, 5734 S. University Avenue, Eckhart hall Room 414, Chicago, Illinois, 60637.
E-mail address: [email protected]