9.6. Symmetric Functions
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9. FIELD EXTENSIONS – SECOND LOOK 9.6. Symmetric Functions Let K be any field, and let x1; : : : ; xn be variables. For a vector ˛ D .˛1; : : : ; ˛n/, with nonnegative integer entries, let x˛ D x˛1 : : : x˛n
1
n . The total degree of the monic monomial x˛ is j˛j D P ˛i . A polynomial is said to be homogeneous of total degree d if it is a linear combination of monomials x˛ of total degree d .
Write Kd Œx1; : : : ; xn for the set of polynomials in n variables that are homogeneous of total degree d or identically zero. Then Kd Œx1; : : : ; xn is a vector space over K and KŒx1; : : : ; xn is the direct sum over over d 0 of the subspaces Kd Œx1; : : : ; xn; see Exercise 9.6.1.
The symmetric group Sn acts on polynomials and rational functions in n variables over K by .f /.x1; : : : ; xn/ D f .x.1/; : : : ; x.n//. For 2 Sn, .x˛/ D x˛1 : : : x˛n . A polynomial or rational function is
.1/
.n/
called symmetric if it is fixed by the Sn action. The set of symmetric polynomials is denoted KS Œx1; : : : ; xn, and the set of symmetric rational functions is denoted KS .x1; : : : ; xn/.
Note that for each d , Kd Œx1; : : : ; xn is invariant under the action of Sn, and KS Œx1; : : : ; xn is the direct sum of the vector subspaces KS Œx
d
1; : : : ; xn D Kd Œx1; : : : ; xn \ KS Œx1; : : : ; xn for d 0. See Exercise 9.6.3.
Lemma 9.6.1.
(a) The action of Sn on KŒx1; : : : ; xn is an action by ring automorphisms; the action of of Sn on K.x1; : : : ; xn/ is an action by by field automorphisms.
(b) KS Œx1; : : : ; xn is a
subring of KŒx1; : : : ; xn
and
KS .x1; : : : ; xn/ is a subfield of K.x1; : : : ; xn/.
(c) The field of symmetric rational functions is the field of fractions of the ring of symmetric polynomials in n-variables.
Proof. Exercise 9.6.2.
n
Proposition 9.6.2. The field K.x1; : : : ; xn/ of rational functions is Galois over the field KS .x1; : : : ; xn/ of symmetric rational functions, and the Galois group AutKS.x1;:::;xn/.K.x1; : : : ; xn// is Sn.
Proof. By Exercise 9.6.1, Sn acts on K.x1; : : : ; xn/ by field automorphisms and KS .x1; : : : ; xn/ is the fixed field. Therefore, by Proposition 9.5.5 the extension is Galois, with Galois group is Sn.
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9.6. SYMMETRIC FUNCTIONS
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We define a distinguished family of symmetric polynomials, the ele mentary symmetric functions as follows: 0.x1; : : : ; xn/
D
1
1.x1; : : : ; xn/
D x1 C x2 C C xn
X 2.x1; : : : ; xn/
D
xi xj 1i
X
k.x1; : : : ; xn/
D
xi x
1
ik
1i1
D
x1x2 xn
We put j .x1; : : : xn/ D 0 if j > n.
Lemma 9.6.3.
.x
x1/.x
x2/. /.x
xn/
D xn 1xn 1 C 2xn 2 C . 1/nn
n
X
D . 1/kkxn k;
kD0 where k is short for k.x1; : : : ; xn/.
Proof. Exercise 9.6.4.
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Corollary 9.6.4.
(a) Let f .x/ D xn C an 1xn 1 C C a0 be a monic polynomial in KŒx and let ˛1; : : : ; ˛n be the roots of f in a splitting field.
Then ai D . 1/n i n i .˛1; : : : ; ˛n/.
(b) Let f .x/ D anxn C an 1xn 1 C C a0 2 KŒx be of degree n, and let ˛1; : : : ; ˛n be the roots of f in a splitting field. Then ai =an D . 1/n i n i .˛1; : : : ; ˛n/.
Proof. For part (a), f .x/ D .x
˛1/.x
˛2/ .x ˛n/
D xn 1.˛1; : : : ; ˛n/xn 1 C 2.˛1; : : : ; ˛n/xn 2 C . 1/nn.˛1; : : : ; ˛n/:
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9. FIELD EXTENSIONS – SECOND LOOK
For part (b), apply part (a) to
X
.x
˛1/.x
˛2/ .x
˛n/ D .ai =an/xi :
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n
Definition 9.6.5. Let K be a field and fu1; : : : ; ung a set of elements in an extension field. We say that fu1; : : : ; ung is algebraically independent over K if there is no polynomial f 2 KŒx1; : : : ; xn such that f .u1; : : : ; un/ D 0.
The following is called the fundamental theorem of symmetric func tions:
Theorem 9.6.6. The set of elementary symmetric functions f1; : : : ; ng in KŒx1; : : : ; xn is algebraically independent over K, and generates KS Œx1; : : : ; xn as a ring.
Consequently, K.1; : : : ; n/
D KS .x1; : : : ; xn/.
The algebraic independence of the i is the same as linear indepen dence of the monic monomials in the i . First, we establish an indexing system for the monic monomials:
A partition is a finite decreasing se quence of nonnegative integers, D .1; : : : ; k/. We can picture a partition by means of an M -by-N matrix of zeroes and ones, where M k and N 1; rs D 1 if r k and s r , and rs D 0 otherwise. Here is a matrix representing the partition D .5; 4; 4; 2/: 21 1 1 1 1 03
61
1 1 1 0 07 61 1 1 1 0 07
D 6
7
6
7 :
61
1 0 0 0 07 60 0 0 0 0 07
4
5
0 0 0 0 0 0
The size of a partition is jj D P i . The nonzero entries in are referred to as the parts of . The number of parts is called the length of . The conjugate partition is that represented by the transposed matrix . Note that r D jfi W i rgj, and ./ D ; see Exercise 9.6.6. For D .5; 4; 4; 2/, we have D .4; 4; 3; 3; 1/, corresponding to the matrix
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21 1 1 1 0 03
61
1 1 1 0 07 61 1 1 0 0 07
D 6
7
6
7 :
61
1 1 0 0 07 61 0 0 0 0 07
4
5
0 0 0 0 0 0 For D .1; 2; : : : ; s/, define
Y
.x1; : : : ; xn/ D
.x
i 1; : : : ; xn/:
i
For example, .5;4;4;2/ D .5/.4/2.2/. Note that .x1; : : : ; xn/ D 0 if 1 > n.
We will show that the set of with jj D d and 1 n is a basis of KS Œx
d 1; : : : ; xn. In order to do this, we first produce a more obvious basis.
For a partition D .1; : : : n/ with no more than n nonzero parts, define the monomial symmetric function
X
m.x1; : : : ; xn/ D .1=f / .x/;
2Sn
where f is the size of the stabilizer of x under the action of the symmetric group. (Thus, m is a sum of monic monomials, each occurring exactly once.) For example, m.5;4;4;2/.x1; : : : ; x4/ D x51x42x43x24 C x51x42x23x44 C x51x22x43x44 C : : : ; a sum of 12 monic monomials.
In the Exercises, you are asked to check that the monomial symmetric functions m with D .1; : : : ; n/ and jj D d form a linear basis of KS Œx
d 1; : : : ; xn.
Define a total order on n-tuples of nonnegative integers and, in particu lar, on partitions by ˛ > ˇ if the first nonzero difference ˛i ˇi is positive.
Note that ./ for any permutation . This total order on n-tuples ˛ induces a total order on monomials x˛ called lexicographic order.
Lemma 9.6.7.
(a) The leading (i.e., lexicographically highest) monomial of is x.
(b)
X
.x1; : : : ; xn/ D Tm.x1; : : : ; xn/;
jjDjj
where T is a nonnegative integer, T D 1, and T D 0 if > .
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9. FIELD EXTENSIONS – SECOND LOOK (c)
X m D S.x1; : : : ; xn/;
jjDjj
where S is a nonnegative integer, S D 1, and S D 0 if > .
Proof.
.x1; : : : ; xn/ D .x1 x /.x / .x
/ C : : :
1
1 x2
1 xn
D x1x2
1
2
xn
n
C ; where the omitted monomials are lexicographically lower, and j is the number of k such that k j ; but then j D , and, therefore, the
j leading monomial of is x. Since is symmetric, we have D mC a sum of m, where jj D jj and