5.1. Primitive element theorem.

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Recall that a finite extension of fields E/F is simple if E = F[α] for some element α of E.

Such an α is called a primitive element of E. We shall show that (at least) all separable

extensions have primitive elements.

Consider for example Q[

√

2,

√

3]/Q. We know (see Exercise 13) that its Galois group over

Q is a 4-group < σ,τ >, where

_

σ

√

2 = −

√

2

σ

√

3 =

√

3

,

_

τ

√

2 =

√

2

τ

√

3 = −

√

3.

Note that

σ(

√

2 +

√

3) = −

√

2 +

√

3,

τ (

√

2 +

√

3) =

√

2 −

√

3,

(στ)(

√

2 +

√

3) = −

√

2 −

√

3.

These all differ from

√

2 +

√

3, and so only the identity element of Gal(Q[

√

2,

√

3]/Q) fixes

the elements of Q[

√

2 +

√

3]. According to the Fundamental √ Theorem, this implies that

2 +

√

3 is a primitive element:

Q[

√

2,

√

3] = Q[

√

2 +

√

3].

It is clear that this argument should work much more generally.

We say that an element α algebraic over a field F is separable over F if its minimum

polynomial over F has no multiple roots. Thus a finite extension E of F is separable if and

only if all its elements are separable over F.

Theorem 5.1. Let E = F[α1, ..., αr] be a finite extension of F, and assume that α2, ..., αr

are separable over F (but not necessarily α1). Then there is an element γ ∈ E such that

E = F[γ].

Proof. For finite fields, we proved this in (4.16). Hence we may assume F to be infinite.

It suffices to prove the statement for r = 2. Thus let E = F[α, β] with β separable over

F[α]. Let f and g be the minimum polynomials of α and β over F. Let α1 = α, . . . , αs be

the roots of f in some field containing E, and let β1 = β, β2, . . . , βt be the roots of g. For

j _= 1, βj _= β1, and so the the equation

αi + Xβj = α1 + Xβ1, j_= 1,

has exactly one solution, namely, X = αi−α1

β1−βj

. If we choose a c different from any of these

solutions (using that F is infinite), then

αi + cβj _= α + cβ unless i = 1 = j.

I claim that γ = α + cβ generates E over F.

The polynomials g(X) and f(γ − cX) have coefficients in F[γ][X], and have β as a root:

g(β) = 0, f(γ − cβ) = f(α) = 0.

In fact, β is their only common root, because the roots of g are β1, ..., βt, and we chose c so

that γ − cβj _= αi unless i = 1 = j. Therefore gcd(g(X), f(γ − cX)) computed in some field

FIELDS AND GALOIS THEORY 37

splitting fg is X − β, but we have seen (Proposition 3.1) that the gcd of two polynomials

has coefficients in the same field as the coefficients of the polynomials. Hence β ∈ F[γ], and

then α = γ − cβ also lies in F[γ].

Remark 5.2. Assume F to be infinite. The proof shows that γ can be chosen to be of

the form

γ = α1 + c2α2 + ・ ・ ・ + crαr, ci ∈ F.

In fact, all but a finite number of elements of this form will serve. If E = F[α1, . . . , αr] is

Galois over F, then an element of this form will be a primitive element provided it is moved

by every element of Gal(E/F) except 1. These remarks make it very easy to write down

primitive elements.

Our hypotheses are minimal: if two of the α’s are not separable, then the extension need

not be simple. Before proving this, we need another result.

Proposition 5.3. Let E = F[γ] be a simple algebraic extension of F. Then there are

only finitely many intermediate fields M,

F ⊂ M ⊂ E.

Proof. Let M be such a field, and let g(X) be the minimum polynomial of γ over M.

Let M_ be the subfield of E generated over F by the coefficients of g(X). Clearly M_ ⊂ M,

but (equally clearly) g(X) is the minimum polynomial of γ over M_. Hence

[E : M

_

] = degg = [E : M],

and so M = M_: M is generated by the coefficients of g(X).

Let f(X) be the minimum polynomial of γ over F. Then g(X) divides f(X) in M[X], and

hence also in E[X]. Therefore, there are only finitely many possible g’s, and consequently

only finitely many possible M’s.

Remark 5.4. (a) Note that the proposition in fact gives a description of all the intermediate

fields: each is generated over F by the coefficients of a factor g(X) of f(X) in E[X].

The coefficients of such a g(X) are partially symmetric polynomials in the roots of f(X)

(i.e., fixed by some, but not necessarily all, of the permutations of the roots).

(b) The proposition has a converse: if E is a finite extension of F and there are only finitely

many intermediate fields M, F ⊂ M ⊂ E, then E is a simple extension of F (see Dummit,

p508). This gives another proof of the theorem when E is separable over F, because Galois

theory shows that there are only finitely many intermediate fields in this case (embed E in

a Galois extension of F).

(c) The simplest nonsimple extension is k(X, Y ) ⊃ k(Xp, Y p) = F, where k is an algebraically

closed field of characteristic p. For any c ∈ k, we have

k(X, Y ) = F[X, Y ] ⊃ F[X + cY ] ⊃ F

with the degree of each extension equal to p. If F[X + cY] = F[X + c_Y ], c _= c_, then

F[X + cY ] would contain both X and Y , which is impossible because [k(X, Y ) : F] = p2.

Hence there are infinitely many distinct intermediate fields.3

3Zariski showed that there is even an intermediate field M that is not isomorphic to F(X, Y ), and Piotr

Blass showed in his UM thesis, 1977, using the methods of algebraic geometry, that there is an infinite

sequence of intermediate fields, no two of which are isomorphic.

38 J.S. MILNE

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