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15. Symmetric polynomials
15.1 The theorem 15.2 First examples 15.3 A variant: discriminants
1. The theorem
Let Sn be the group of permutations of { 1 ,... , n}, also called the symmetric group on n things.
For indeterminates xi, let p ∈ Sn act on Z[x 1 ,... , xn] by
p(xi) = xp(i)
A polynomial f (x 1 ,... , xn) ∈ Z[x 1 ,... , xn] is invariant under Sn if for all p ∈ Sn
f (p(x 1 ),... , p(xn)) = f (x 1 ,... , xn)
The elementary symmetric polynomials in x 1 ,... , xn are
s 1 = s 1 (x 1 ,... , xn) =
i xi s 2 = s 2 (x 1 ,... , xn) =
i<j xixj s 3 = s 3 (x 1 ,... , xn) =
i<j<k xixj^ xk s 4 = s 4 (x 1 ,... , xn) =
i<j<k<xixj^ xkx
... st = st(x 1 ,... , xn) =
i 1 <i 2 <...<it xi^1 xi^2... xit
... sn = sn(x 1 ,... , xn) = x 1 x 2 x 3... xn
[1.0.1] Theorem: A polynomial f (x 1 ,... , xn) ∈ Z[x 1 ,... , xn] is invariant under Sn if and only if it is a
polynomial in the elementary symmetric functions s 1 ,... , sn.
[1.0.2] Remark: In fact, the proof shows an algorithm which determines the expression for a given
Sn-invariant polynomial in terms of the elementary ones.
213
214 Symmetric polynomials
Proof: Let f (x 1 ,... , xn) be Sn-invariant. Let
q : Z[x 1 ,... , xn− 1 , xn] −→ Z[x 1 ,... , xn− 1 ]
be the map which kills off xn, that is
q(xi) =
xi (1 ≤ i < n) 0 (i = n)
If f (x 1 ,... , xn) is Sn-invariant, then
q(f (x 1 ,... , xn− 1 , xn)) = f (x 1 ,... , xn− 1 , 0)
is Sn− 1 -invariant, where we take the copy of Sn− 1 inside Sn that fixes n. And note that
q(si(x 1 ,... , xn)) =
si(x 1 ,... , xn− 1 ) (1 ≤ i < n) 0 (i = n)
By induction on the number of variables, there is a polynomial P in n − 1 variables such that
q(f (x 1 ,... , xn)) = P (s 1 (x 1 ,... , xn− 1 ),... , sn− 1 (x 1 ,... , xn− 1 ))
Now use the same polynomial P but with the elementary symmetric functions augmented by insertion of xn, by g(x 1 ,... , xn) = P (s 1 (x 1 ,... , xn),... , sn− 1 (x 1 ,... , xn))
By the way P was chosen, q(f (x 1 ,... , xn) − g(x 1 ,... , xn)) = 0
That is, mapping xn −→ 0 sends the difference f − g to 0. Using the unique factorization in Z[x 1 ,... , xn], this implies that xn divides f − g. The Sn-invariance of f − g implies that every xi divides f − g. That is, by unique factorization, sn(x 1 ,... , xn) divides f − g.
The total degree of a monomial c xe 11... xe nn is the sum of the exponents
total degree (c xe 11... xe nn ) = e 1 +... + en
The total degree of a polynomial is the maximum of the total degrees of its monomial summands.
Consider the polynomial f − g sn
f (x 1 ,... , xn) − g(x 1 ,... , xn) sn(x 1 ,... , xn)
It is of lower total degree than the original f. By induction on total degree (f − g)/sn is expressible in terms of the elementary symmetric polynomials in x 1 ,... , xn. ///
[1.0.3] Remark: The proof also shows that if the total degree of an Sn-invariant polynomial
f (x 1 ,... , xn− 1 , xn) in n variables is less than or equal the number of variables, then the expression for f (x 1 ,... , xn− 1 , 0) in terms of si(x 1 ,... , xn− 1 ) gives the correct formula in terms of si(x 1 ,... , xn− 1 , xn).
- First examples
[2.0.1] Example: Consider
f (x 1 ,... , xn) = x^21 +... + x^2 n
216 Symmetric polynomials
= 4xyz(x + y + z) = 4s 3 (x, y, z) · s 1 (x, y, z)
Thus, with 3 variables, x^4 + y^4 + z^4
= s 1 (x, y, z)^4 − s 2 (x, y, z) · (4s 1 (x, y, z)^2 − 2 s 2 (x, y, z)) + 4s 3 (x, y, z) · s 1 (x, y, z)
Abbreviating si = si(x, y, z, w), we anticipate that
x^4 + y^4 + z^4 + w^4 −
s^41 − 4 s^21 s 2 + 2s^22 + 4s 1 s 3
= constant · xyzw
We can save a little time by evaluating the constant by taking x = y = z = w = 1. In that case
s 1 (1, 1 , 1 , 1) = 4 s 2 (1, 1 , 1 , 1) = 6 s 3 (1, 1 , 1 , 1) = 4
and 1 + 1 + 1 + 1 −
= constant
or constant = 4 − (256 − 384 + 72 + 64) = − 4
Thus, x^4 + y^4 + z^4 + w^4 = s^41 − 4 s^21 s 2 + 2s^22 + 4s 1 s 3 − 4 s 4
By the remark above, since the total degree is just 4, this shows that for arbitrary n
x^41 +... + x^4 n = s^41 − 4 s^21 s 2 + 2s^22 + 4s 1 s 3 − 4 s 4
Garrett: Abstract Algebra 217
- A variant: discriminants
Let x 1 ,... , xn be indeterminates. Their discriminant is
D = D(x 1 ,... , xn) =
i<j
(xi − xj )
Certainly the sign of D depends on the ordering of the indeterminates. But
D^2 =
i 6 =j
(xi − xj )^2
is symmetric, that is, is invariant under all permutations of the xi. Therefore, D^2 has an expression in terms of the elementary symmetric functions of the xi.
[3.0.1] Remark: By contrast to the previous low-degree examples, the discriminant (squared) has as
high a degree as possible.
[3.0.2] Example: With just 2 indeterminates x, y, we have the familiar
D^2 = (x − y)^2 = x^2 − 2 xy + y^2 = (x + y)^2 − 4 xy = s^21 − 4 s 2
Rather than compute the general version in higher-degree cases, let’s consider a more accessible variation on the question. Suppose that α 1 ,... , αn are roots of an equation
Xn^ + aX + b = 0
in a field k, with a, b ∈ k. For simplicity suppose a 6 = 0 and b 6 = 0, since otherwise we have even simpler methods to study this equation. Let f (X) = xn^ + aX + b. The discriminant
D(α 1 ,... , αn) =
i<j
(αi − αj )
vanishes if and only if any two of the αi coincide. On the other hand, f (X) has a repeated factor in k[X] if and only if gcd(f, f ′) 6 = 1. Because of the sparseness of this polynomial, we can in effect execute the Euclidean algorithm explicitly. Assume that the characteristic of k does not divide n(n − 1). Then
(Xn^ + aX + b) −
X
n
· (nXn−^1 + a) = a(1 −
n
)X + b
That is, any repeated factor of f (X) divides X + (^) (nbn−1)a , and the latter linear factor divides f ′(X).
Continuing, the remainder upon dividing nXn−^1 + a by the linear factor X + (^) (nbn−1)a is simply the value of
nXn−^1 + a obtained by evaluating at (^) (n−−bn1)a , namely
n
−bn (n − 1)a
)n− 1
nn(−1)n−^1 bn−^1 + (n − 1)n−^1 an
· ((n − 1)a)^1 −n
Thus, (constraining a to be non-zero)
nn(−1)n−^1 bn−^1 + (n − 1)n−^1 an^ = 0
if and only if some αi − αj = 0.
