5.8. The general polynomial of degree n.

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When we say that the roots of

aX2 + bX + c

are

−b Ѓ}

√

b2 − 4ac

2

we are thinking of a, b, c as variables: for any particular values of a, b, c, the formula gives

the roots of the particular equation. We shall prove in this section that there is no similar

formula for the roots of the “general polynomial” of degree ≥ 5.

We define the general polynomial of degree n to be

f(X) = Xn − t1Xn−1 + ・ ・ ・ + (−1)ntn ∈ F[t1, ..., tn][X]

where the ti are variables. We shall show that, when we regard f as a polynomial in X with

coefficients in the field F(t1, . . . , tn), its Galois group is Sn. Then Theorem 5.23 proves the

above remark (at least on characteristic zero).

Symmetric polynomials. Let R be a commutative ring (with 1). A polynomial

P(X1, ...,Xn) ∈ R[X1, . . . ,Xn] is said to be symmetric if it is unchanged when its variables

are permuted, i.e., if

P(Xσ(1), . . . ,Xσ(n)) = P(X1, . . . ,Xn), all σ ∈ Sn.

For example

p1 =

_

iXi = X1 + X2 + ・ ・ ・ + Xn,

p2 =

_

i<j XiXj = X1X2 + X1X3 + ・ ・ ・ + X1Xn + X2X3 + ・ ・ ・ + Xn−1Xn,

p3 =

_

i<j<k XiXjXk, = X1X2X3 + ・ ・ ・

・ ・ ・

pr =

_

i1<···<ir Xi1...Xir

・ ・ ・

pn = X1X2 ・ ・ ・Xn

are all symmetric, because pr is the sum of all monomials of degree r made up out of distinct

Xi’s. These particular polynomials are called the elementary symmetric polynomials.

Theorem 5.25 (Symmetric polynomials theorem). Every symmetric polynomial

P(X1, ...,Xn) in R[X1, ...,Xn] is equal to a polynomial in the elementary symmetric

polynomials with coefficients in R, i.e., P ∈ R[p1, ..., pn].

FIELDS AND GALOIS THEORY 47

Proof. We define an ordering on the monomials in the Xi by requiring that

Xi1

1 Xi2

2

・ ・ ・Xin

n >Xj1

1 Xj2

2

・ ・ ・Xjn

n

if either

i1 + i2 + ・ ・ ・ + in > j1 + j2 + ・ ・ ・ + jn

or equality holds and, for some s,

i1 = j1, . . . , is = js, but is+1 > js+1.

For example,

X1X3

2X3 > X1X2

2X3 > X1X2X2

3 .

Let Xk1

1

・ ・ ・Xkn

n be the highest monomial occurring in P with a coefficient c _= 0. Because

P is symmetric, it contains all monomials obtained from Xk1

1

・ ・ ・Xkn

n by permuting the X’s.

Hence k1 ≥ k2 ≥ ・ ・ ・ ≥ kn.

The highest monomial in pi is X1 ・ ・ ・Xi, and it follows that the highest monomial in

pd1

1

・ ・ ・ pdn

n is

Xd1+d2+···+dn

1 Xd2+···+dn

2

・ ・ ・Xdn

n .

Therefore

P(X1, . . . ,Xn) − cpk1−k2

1 pk2−k3

2

・ ・ ・ pkn

n < P(X1, . . . ,Xn).

We can repeat this argument with the polynomial on the left, and after a finite number of

steps, we will arrive at a representation of P as a polynomial in p1, . . . , pn.

Let f(X) = Xn + a1Xn−1 + ・ ・ ・ + an ∈ R[X], and let α1, . . . , αn be the roots of f(X) in

some ring S containing R, i.e., f(X) =

_

(X − αi) in S[X]. Then

a1 = −p1(α1, . . . , αn), a2 = p2(α1, . . . , αn), . . . , an = Ѓ}pn(α1, . . . , αn).

Thus the elementary symmetric polynomials in the roots of f(X) lie in R, and so the theorem

implies that every symmetric polynomial in the roots of f(X) lies in R. For example, the

discriminant

D(f) =

_

i<j

(αi − αj)2

of f lies in R.

The general polynomial.

Theorem 5.26 (Symmetric functions theorem). When Sn acts on E = F(X1, ...,Xn) by

permuting the Xi’s, the field of invariants is F(p1, ..., pn).

Proof. Suppose f = g

h , g, h ∈ F[X1, . . . ,Xn], is symmetric, i.e., fixed by all σ ∈ Sn.

Then H =

_

σ∈Sn σh is symmetric, and so therefore is Hf. Both Hf and H are polynomials,

and therefore lie in F[p1, . . . , pn]. Hence their quotient f = Hf

H lies in F(p1, . . . , pn).

Corollary 5.27. The field F(X1, ...,Xn) is Galois over F(p1, ..., pn) with Galois group

Sn (acting by permuting the Xi).

Proof. We have shown that F(p1, . . . , pn) = F(X1, . . . ,Xn)Sn, and so this follows from

(3.12).

Theorem 5.28. The Galois group of the general polynomial of degree n is Sn.

48 J.S. MILNE

Proof. Let f(X) be the general polynomial of degree n,

f(X) = Xn − t1Xn−1 + ・ ・ ・ + (−1)ntn ∈ F[t1, ..., tn][X].

Consider the homomorphism

F[t1, . . . , tn] → F[p1, . . . , pn], ti → pi.

We shall prove shortly that this is an isomorphism, and therefore induces an isomorphism

on the fields of fractions

F(t1, . . . , tn) → F(p1, . . . , pn), ti → pi.

Under this isomorphism, f(X) corresponds to

g(X) = Xn − p1Xn−1 + ・ ・ ・ + (−1)npn.

But g(X) =

_

(X −Xi) in F(X1, . . . ,Xn)[X], and so F(X1, . . . ,Xn) is the splitting field of

g(X) ∈ F(p1, . . . , pn)[X]. Therefore the last corollary shows that the Galois group of g is

Sn, which must also be the Galois group of f.

It remains to show that the homomorphism ti → pi is an isomorphism. Let E ⊃

F(t1, . . . , tn) be a splitting field of f, and let α1, ..., αn be the roots of f in E. Consider the

diagram

E ⊃ F[α1, . . . , αn] α←i←−Xi F[X1, . . . ,Xn]

∪ ∪

F[t1, . . . , tn]

ti−_→→pi F[p1, . . . , pn].

The top and bottom maps are well-defined because F[X1, ...,Xn] and F[t1, ..., tn] are polynomial

rings. The diagram commutes because ti = pi(α1, ..., αn). Hence the lower horizontal

map is injective, and, since it is obviously surjective, it is an isomorphism.

Remark 5.29. In the final section of this course, we’ll discuss algebraic independence.

Then it will be obvious that the map ti → pi : F[t1, . . . , tn] → F[p1, . . . , pn] is an isomorphism,

which simplifies the proof.

Remark 5.30. Since Sn occurs as a Galois group over Q, and every finite group occurs

as a subgroup of some Sn, it follows that every finite group occurs as a Galois group over

some finite extension of Q, but does every finite Galois group occur as a Galois group over

Q itself?

The Hilbert-Noether program for proving this was the following.

Hilbert proved that if G occurs as the Galois group of an extension E ⊃ Q(t1, ..., tn) (the ti

are variables), then it occurs infinitely often as a Galois group over Q. For the proof, realize

E as the splitting field of a polynomial f(X) ∈ k[t1, . . . , tn][X] and prove that for infinitely

many values of the ti, the polynomial you obtain in Q[X] has Galois group G. (This is quite

a difficult theorem—see Serre, Lectures on the Mordell-Weil Theorem, Chapter 9.)

Noether conjectured the following: Let G ⊂ Sn act on F(X1, ...,Xn) by permuting the Xi;

then F(X1, . . . ,Xn)G ≈ F(t1, ..., tn) (for variables ti).

Unfortunately, Swan proved in 1969 that the conjecture is false for C47. Hence this approach

can not lead to a proof that all finite groups occur as Galois groups over Q, but it

doesn’t exclude other approaches. [For more information on the problem, see Serre, ibid.,

Chapter 10, and Serre, Topics in Galois Theory, 1992.]

FIELDS AND GALOIS THEORY 49

Remark 5.31. Take F = C, and consider the subset of Cn+1 defined by the equation

Xn − T1Xn−1 + ・ ・ ・ + (−1)nTn = 0.

It is a beautiful complex manifold S of dimension n. Consider the projection

π : S → Cn, (x, t1, . . . , tn) → (t1, . . . , tn).

Its fibre over a point (a1, . . . , an) is the set of roots of the polynomial

Xn − a1Xn−1 + ・ ・ ・ + (−1)nan.

The discriminant of Xn − T1Xn−1 + ・ ・ ・ + (−1)nTn, regarded as a polynomial in X, is a

polynomial D(f) ∈ C[T1, . . . , Tn]. Let Δ be the zero set of D(f) in Cn. Then over each

point of Cn \ Δ, there are exactly n points of S, and S \ π−1(Δ) is a covering space over

Cn \ Δ with group of covering transformations Sn.

A brief history. As far back as 1500 BC, the Babylonians (at least) knew a general formula

for the roots of a quadratic polynomial. Cardan (about 1515 AD) found a general formula

for the roots of a cubic polynomial. Ferrari (about 1545 AD) found a general formula for the

roots of quartic polynomial (he introduced the resolvant cubic, and used Cardan’s result).

Over the next 275 years there were many fruitless attempts to obtain similar formulas for

higher degree polynomials, until, in about 1820, Ruffini and Abel proved that there are none.

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