§ 6. Symmetric functions

1. Symmetric polynomials

Let $n$ be a positive integer. For every permutation $\sigma \in \mathfrak{S}_n$ let $\varphi_\sigma$ be the automorphism of the A-algebra $A[X_1, \ldots, X_n]$ which maps $X_i$ to $X_{\sigma(i)}$ for $1 \leq i \leq n$ . It is clear that $\sigma \mapsto \varphi_\sigma$ is a homomorphism of $\mathfrak{S}_n$ into the group of automorphisms of $A[X_1, \ldots, X_n]$ . We shall put $\sigma f = \varphi_\sigma(f)$ for $\sigma \in \mathfrak{S}_n$ and $f \in A[X_1, \ldots, X_n]$ . The polynomial f is said to be symmetric if $\sigma f = f$ for all $\sigma \in \mathfrak{S}_n$ ; the symmetric polynomials form a unital graded subalgebra of $A[X_1, \ldots, X_n]$ ; we shall denote it by $A[X_1, \ldots, X_n]^{\mathrm{sym}}$ in the rest of this paragraph. For every positive integer k we denote by $\mathfrak{P}_k$ the set of k-element subsets of the set $\{1, 2, \ldots, n\}$ and put

(1)

s_k = \sum_{H \in \mathfrak{P}_k} \prod_{i \in H} X_i .

When we wish to specify the integer n we shall write $s_{k,n}$ instead of $s_k$ . We have in particular

\begin{aligned} s_0 &= 1 \\ s_1 &= \sum_{1 \leq i \leq n} X_i \\ s_2 &= \sum_{1 \leq i < j \leq n} X_i X_j \\ &\ \ \cdots \cdots \cdots \cdots \cdots \\ s_n &= X_1 \ldots X_n \end{aligned}

and $s_k = 0$ for $k > n$ . It is clear that $s_k$ is a homogeneous symmetric polynomial of degree k ; we shall call it the elementary symmetric polynomial of degree k. In the ring $A[X_1, \ldots, X_n, U, V]$ we have the relation

(2)

\prod_{i=1}^{n} (U + V X_i) = \sum_{k=0}^{n} U^{n-k} V^k s_k ;

by appropriate substitutions we deduce the relations

(3)

\prod_{i=1}^{n} (1 + T X_i) = \sum_{k=0}^{n} s_k T^k ,

(4)

\prod_{i=1}^{n} (\mathbf{X} - \mathbf{X}_i) = \sum_{k=0}^{n} (-1)^{n-k} s_{n-k} \mathbf{X}^k .

Theorem 1

— Let us put $E = A[\mathbf{X}_1, \ldots, \mathbf{X}_n]$ and $S = A[\mathbf{X}_1, \ldots, \mathbf{X}_n]^{\text{sym}}$ .

a) The $A$ -algebra $S$ of symmetric polynomials is generated by $s_1, \ldots, s_n$ .

b) The elements $s_1, \ldots, s_n$ of $E$ are algebraically independent over $A$ (IV, p. 4).

c) The family of monomials $X^\nu = X_1^{\nu(1)} \ldots X_n^{\nu(n)}$ such that $0 \leq \nu(i) < i$ for $1 \leq i \leq n$ is a basis of the $S$ -module $E$ . In particular, $E$ is a free $S$ -module of rank $n!$ .

We shall prove the theorem by induction on $n$ , the case $n = 0$ being trivial. Let us write $B = A[X_n]$ and denote by $s_k'$ the elementary symmetric polynomial of degree $k$ in $X_1, \ldots, X_{n-1}$ ; we thus have $B[X_1, \ldots, X_{n-1}] = A[X_1, \ldots, X_n]$ . If we replace $n$ by $n-1$ and $A$ by $B$ in the statement of Th. 1, we can formulate the induction hypothesis as follows:

(A) The $B$ -algebra $S'$ of polynomials $f \in A[\mathbf{X}_1, \ldots, \mathbf{X}_n]$ invariant under all permutations of $X_1, \ldots, X_{n-1}$ is generated by $s_1', \ldots, s_{n-1}'$ .

(B) The elements $s_1', \ldots, s_{n-1}'$ , of $E$ are algebraically independent over $B$ .

(C) The family of monomials $X_1^{\nu(1)} \ldots X_{n-1}^{\nu(n-1)}$ such that $0 \leq \nu(i) < i$ for $1 \leq i \leq n-1$ is a basis of the $S'$ -module $E$ .

We have the obvious relation

(5)

s_k = s_k' + s_{k-1}' X_n \quad (1 \leq k \leq n-1),

from which we obtain by induction on

k

(6)

s_k' = (-1)^k X_n^k + \sum_{i=1}^k (-1)^{k-i} s_i X_n^{k-i} \quad (1 \leq k \leq n-1).

We have $S \subset S'$ , hence $s_1, \ldots, s_n$ belong to $S'$ ; by (A) and formula (6) the B-algebra $S'$ is therefore generated by $s_1, \ldots, s_{n-1}'$ .

By (B) there exists an endomorphism $\mu$ of the B-algebra $S'$ such that

(7)

u(s_k') = (-1)^k X_n^k + \sum_{i=1}^k (-1)^{k-i} s_i' X_n^{k-i} \quad (1 \leq k \leq n-1).

By (5), we have $u(s_k) = u(s_k') + u(s_{k-1}') X_n$ , whence $u(s_k) = s_k'$ by an easy calculation. Let $P \in B[Y_1, \ldots, Y_{n-1}]$ ; then from $P(s_1, \ldots, s_{n-1}) = 0$ we deduce

0 = u(P(s_1, \ldots, s_{n-1})) = P(s_1', \ldots, s_{n-1}'),

whence

P = 0

by (B). It follows that the B-algebra

S'

is generated by the algebraically independent elements

s_1, \ldots, s_{n-1}

. This property may be reformulated as follows: (D) The

A

-algebra

S'

is generated by the algebraically independent elements

s_1, \ldots, s_{n-1}, X_n

We can thus identify S' with the polynomial ring $C[X_n]$ , where $C$ is the sub-A-algebra of E generated by $s_1, ..., s_{n-1}$ .

To prove a), let $f \in S$ be a homogeneous symmetric polynomial of degree $m$ . We have $f \in S' = C[X_n]$ , so there exists an element $g = P(s_1, ..., s_{n-1})$ of C, homogeneous of degree $m$ in $X_1, ..., X_n$ , such that $f - g$ is divisible by $X_n$ . Since $f - g$ is symmetric, each of its terms is also divisible by $X_1, ..., X_{n-1}$ , hence $f - g$ is divisible by $s_n = X_1 ... X_n$ . In other words, there exists $h \in S$ such that $f = g + hs_n$ , whence $\deg h < m$ . By induction on $m$ it follows that $f$ belongs to

C[s_n]_E = A[s_1, ..., s_{n-1}, s_n]_E .

Thus we have shown that the A-algebra S is generated by $s_1, ..., s_n$ .

Next we prove b). If we substitute $X$ , for $X$ in (4), we find

(-1)^{n+1}s_n = X_n^n + \sum_{k=1}^{n-1} (-1)^{n-k} s_{n-k} X_n^k ;

in other words, $(-1)^{n+1}s_n$ is a monic polynomial in $X$ , of degree $n$ and with coefficients in $C$ . By IV, p. 11 we thus have the following property : (E) The C-algebra homomorphism $\varphi$ of $C[T]$ into $C[X_n] = S'$ which maps T to $s_n$ is injective, and $S'$ is a free module over the image of $\varphi$ , with basis $(1, X_n, ..., X_n^{n-1})$ .

The elements $s_1, ..., s_{n-1}$ of $C$ are algebraically independent over A, by (D); thus the injectivity of $\varphi$ means that $s_1, ..., s_{n-1}, s_n$ are algebraically independent over A, whence b).

To prove c), the image of $\varphi$ equals $C[s_n]_E = S$ , hence by (E), $S'$ is a free module over S on the basis $(1, X_1, ..., X_n^{n-1})$ . Now the assertion c) follows from the induction hypothesis (C) and Prop. 25 of II, p. 222. This completes the proof.

Let $f$ be a symmetric polynomial in $X_1, ..., X_n$ , homogeneous of degree $m$ . By Th. 1 (IV, p. 62) there exists a polynomial $Q \in A[Y_1, ..., Y_n]$ such that $f = Q(s_1, ..., s_n)$ . The preceding proof provides a means of explicit calculation of $Q$ , by a double induction on $n$ and $m$ . For as we have seen, there exists a polynomial $P \in A[Y_1, ..., Y_{n-1}]$ and a polynomial h symmetric in $X_1, ..., X_n$ which is homogeneous of degree $m-n$ , such that

f = P(s_1, ..., s_{n-1}) + s_n h .

For every polynomial $u \in A[X_1, ..., X_n]$ we put

u'(X_1, ..., X_{n-1}) = u(X_1, ..., X_{n-1}, 0)

Then $s'_1, ..., s'_{n-1}$ are the elementary symmetric polynomials in $X_1, ..., X_{n-1}$ and formula (8) implies

f' = P(s'_1, ..., s'_{n-1})

The determination of $P$ is thus reduced to a calculation of symmetric polynomials in $n - 1$ indeterminates, and $h$ is obtained from (8).

We illustrate the method by two examples.

Examples. — 1) Let $n = 3$ and

f = X_1^2(X_2 + X_3) + X_2^2(X_3 + X_1) + X_3^2(X_1 + X_2).

We have

f' = X_1^2X_2 + X_1X_2^2 = X_1X_2(X_1 + X_2) = s'_1s'_2.

Let us write $g = f - s_1s_2$ ; we have

g = f - (X_1 + X_2 + X_3)(X_1X_2 + X_1X_3 + X_2X_3) = -3X_1X_2X_3,

hence finally

f = s_1s_2 - 3s_3.

2) Again let $n = 3$ and put $p = X_1^3 + X_2^3 + X_3^3$ . We have $p(X_1, 0, 0) = X_1^3 = s_1(X_1, 0, 0)^3$ . Writing $q = p - s_1^3$ , we obtain

q = -3f - 6X_1X_2X_3 = -3s_1s_2 + 3s_3

and finally

p = s_1^3 - 3s_1s_2 + 3s_3.

Let $S_1, ..., S_n$ be indeterminates. We shall equip the polynomial algebra $A[S_1, ..., S_n]$ with the graduation of type $N$ for which $S_k$ is of weight $k$ for $1 \leq k \leq n$ (IV, p. 3), and we equip $A[X_1, ..., X_n]$ with the usual graduation. For $1 \leq k \leq n$ the elementary symmetric polynomial $s_{k,n}$ in $X_1, ..., X_n$ is homogeneous of degree $k$ . By Th. 1 (IV, p. 62) the mapping $g \mapsto g(s_{1,n}, ..., s_{n,n})$ is thus an isomorphism of graded algebras

\varphi_n : A[S_1, ..., S_n] \to A[X_1, ..., X_n]^{\text{sym}}

Let $m$ be an integer such that $0 \leq m \leq n$ . For every integer $k$ such that $1 \leq k \leq m$ we have

s_{k,m}(X_1, ..., X_m) = s_{k,n}(X_1, ..., X_m, 0, ..., 0)

by definition (IV, p. 61, formula (1)) of $s_k$ . Hence the diagram

\begin{array}{ccc} A[S_1, ..., S_m] & \xrightarrow{j} & A[S_1, ..., S_n] \\ \varphi_m \downarrow & & \varphi_n \downarrow \\ A[X_1, ..., X_m]^{\text{sym}} & \xleftarrow{p} & A[X_1, ..., X_n]^{\text{sym}} \end{array}

(where j denotes the canonical inclusion and $p$ the homomorphism

g \mapsto g (X_1, \ldots, X_m, 0, \ldots, 0)

is commutative.

Proposition 1

For every pair of positive integers k, n, let $S_k^{(n)}$ be the A-module consisting of all symmetric polynomials in $X_1, \ldots, X_n$ , which are homogeneous of degree k. If the integer m satisfies $0 \leq k \leq m \leq n$ , then the mapping $f \mapsto f (X_1, \ldots, X_m, 0, \ldots, 0)$ is an isomorphism of $S_k^{(n)}$ onto $S_k^{(m)}$ .

By the commutativity of the diagram (10) it is enough to show that every isobaric polynomial of weight k in $S_1, \ldots, S_n$ depends only on $S_1, \ldots, S_m$ under the hypotheses $0 \leq k \leq m \leq n$ . Now the weight of a monomial $S_1^{\alpha(1)} \cdots S_n^{\alpha(n)}$ is equal to the integer $\alpha(1) + 2\alpha(2) + \ldots + n\alpha(n)$ ; since the integers $\alpha(1), \ldots, \alpha(n)$ are positive, the relation

\alpha(1) + 2\alpha(2) + \cdots + n\alpha(n) = k \leq n

implies

\alpha(j) = 0

for

k < j \leq n

, whence the assertion.

Example 3

By example 2 of IV, p. 64 and Prop. 1 above we thus have

\sum_{i=1}^n X_i^3 = s_{1,n}^3 - 3s_{1,n}s_{2,n} + 3s_{3,n}

for every integer

n \geq 3

. The commutativity of the diagram (10) moreover gives the formulae

X_1^3 + X_2^3 = s_{1,2}^3 - 3s_{1,2}s_{2,2},

X_1^3 = s_{1,1}^3.

Remark

— Let n and k be two positive integers. We denote by $\Delta_{k,n}$ the set of elements of length k in $N^n$ and further equip $A$ , , with the order relation, written $a \leq \beta$ , induced by the lexicographic order on $N^n$ (Set Theory, III, p. 157), and define an action of the group $\mathfrak{S}_n$ on $N^n$ by $(\sigma a)(i) = a(\sigma^{-1}(i))$ for $\sigma \in \mathfrak{S}_n, a \in N^n$ and $l \leq i \leq n$ . Moreover, we denote by $D_k$ the set of elements $a = (\alpha(1), \ldots, \alpha(k))$ of $\mathbf{N}^k$ such that

\alpha(1) \geq \alpha(2) \geq \cdots \geq \alpha(k), \quad \alpha(1) + \cdots + \alpha(k) = k.

Suppose that $k \leq n$ and let us identify $\mathbf{N}^k$ with a subset of $N^n$ by the mapping $(a(1), \ldots, a(k)) \mapsto (a(1), \ldots, a(k), 0, \ldots, 0)$ . Then $D_k$ consists of the elements $a$ of $\Delta_{k,n}$ such that $\sigma a \leq a$ for all $\sigma \in \mathfrak{S}_n$ . It follows that every orbit of the group $\mathfrak{S}_n$ in $A$ , , contains a unique element of $D_k$ . For every $a \in D_k$ let $O(a)$ be the orbit of $a$ in $\Delta_{k,n}$ under the operation of $\mathfrak{S}_n$ ; put

M(\alpha) = \sum_{\beta \in O(\alpha)} X^\beta.

It follows from Lemma 1 of IV, p. 47, that the family $(\mathbf{M}(\alpha))_{\alpha \in D_k}$ is a basis of the A-module $S_k^{(n)}$ .

For each $\alpha \in D_k$ put

(12)\quad S(\alpha) = \prod_{i=1}^k s_i^{\alpha(i) - \alpha(i+1)} \quad \text{(where it is understood that } a\ (k+1) = 0 \text{)} ;

since we have $\sum_{i=1}^k i \cdot (a(i) - a(i+1)) = \sum_{i=1}^k a(i) = k$ , the symmetric polynomial $S(\alpha)$ is homogeneous of degree $k$ . It follows directly from Th. 1 (IV, p. 62) that the family $(S(\alpha))_{\alpha \in D_k}$ is a basis of the A-module $S_k^{(n)}$ .

Let $\alpha, \beta$ be in $D_k$ , and let $c_{\alpha\beta}$ be the coefficient of the monomial $X^\beta$ in the polynomial $S(\alpha)$ defined by (12) in the case $A = \mathbf{Z}$ and $k = n$ . This is then a positive integer, independent of the ring A and of the integer n. By formula (9) (IV, p. 64) we then have

(13)\quad S(\alpha) = \sum_{\beta \in D_k} c_{\alpha\beta} \cdot \mathbf{M}(\beta) \quad (\alpha \in D_k) .

It may be shown (cf. IV, p. 101, Exercise 13) that the matrix $C = (c_{\alpha\beta})_{\alpha,\beta \in D_k}$ is such that $c_{\alpha\alpha} = 1$ and $c_{\alpha\beta} = 0$ for $\alpha < \beta$ ; generalizing the terminology introduced in II, p. 351, we shall say that $C$ belongs to the lower total triangular group. The same is true of the matrix D inverse to C. The values of the matrices C and D when $2 \leq k \leq 5$ , are given in the table (IV, p. 103-105).

Suppose now that $n < k$ and let us identify $N^n$ with a subset of $N^k$ by the mapping $(a(1), \ldots, a(n)) \mapsto (a(1), \ldots, a(n), 0, \ldots, 0)$ . For every $a$ in $D_k \cap N^n$ we again denote by $O(\alpha)$ the orbit of $a$ in $\Delta_{k,n}$ under the operation of $\mathfrak{S}_n$ and define $\mathbf{M}(\alpha)$ by (11). Further we define $S(\alpha)$ by (12), then the families $(\mathbf{M}(a))_{a \in D_k \cap N^n}$ and $(S(a))_{a \in D_k \cap N^n}$ are bases of the A-module $S_k^{(n)}$ . By formula (9) of IV, p. 64, we have a formula analogous to (13) where $D_k$ is replaced by $D_k \cap N^n$ , with the same integers $c_{\alpha\beta}$ .

Example 4

By Example 3 of IV, p. 65, we have

\mathbf{M}(3,0,0) = S(3,0,0) - 3S(2,1,0) + 3S(1,1,1)

for every integer $n \geq 3$ , and so

\mathbf{M}(3,0) = S(3,0) - 3S(2,1)

for $n = 2$ .

2. Symmetric rational fractions

Let $K$ be a commutative field and $X_1, X_2, \ldots, X_n$ indeterminates. For every $\sigma \in S_n$ we have defined in No. 1 (IV, p. 61) an automorphism $\varphi_\sigma$ of $K[X_1, X_2, \ldots, X_n]$ . This automorphism has a unique extension $\psi_\sigma$ to the field $K(X_1, \ldots, X_n)$ , and $\sigma \mapsto \psi_\sigma$ is an injective homomorphism of $S_n$ into the automorphism group of $K(X_1, \ldots, X_n)$ . For each $f \in K(X_1, \ldots, X_n)$ we have $(\psi_\sigma f)(X_1, \ldots, X_n) = f(X_{\sigma(1)}, \ldots, X_{\sigma(n)})$ . The rational fractions $f$ such that $\psi_\sigma(f) = f$ for all $\sigma \in S_n$ are called symmetric rational fractions. The set of all symmetric rational fractions in $X_1, \ldots, X_n$ is a subfield of $K(X_1, \ldots, X_n)$ .

Proposition 2

— The field of symmetric rational fractions in $X_1, \ldots, X_n$ is the field of fractions of the ring of symmetric polynomials in $X_1, \ldots, X_n$ ..

Let $f \in K(X_1, \ldots, X_n)$ be a symmetric rational fraction, and let $u_1, v_1$ be two elements of $K[X_1, \ldots, X_n]$ such that $f = \frac{u_1}{v_1}$ . We put $v = \prod_{\sigma \in S_n} \psi_\sigma(v_1) \in K[X_1, \ldots, X_n]$ and $u = fv \in K[X_1, \ldots, X_n]$ ; then $v$ is symmetric, hence $u$ is symmetric, because $f$ is, and $f = \frac{u}{v}$ , whence the result.

Corollary

— *Let $s_1, s_2, \ldots, s_n$ be the elementary symmetric polynomials in $X_1, \ldots, X_n$ . For every rational fraction $g \in K(S_1, S_2, \ldots, S_n)$ the sequence $(s_1, s_2, \ldots, s_n)$ is substitutable in $g$ and the mapping $g \mapsto g(s_1, s_2, \ldots, s_n)$ is an isomorphism of $K(S_1, S_2, \ldots, S_n)$ onto the field of symmetric rational fractions in $X_1, \ldots, X_n$ .

This follows from Prop. 2 and Th. 1 of IV, p. 62.

3. Symmetric formal power series

Let $I$ be a set, $X = (X_i)_{i \in I}$ a family of indeterminates and $A[[X]]$ the algebra of formal power series in the $X_i$ . For every permutation $a \in S_I$ there exists a unique continuous automorphism $\varphi_a$ of the algebra $A[[X]]$ which maps $X_i$ to $X_{a(i)}$ for each $i \in I$ (IV, p. 28, Prop. 4); it is clear that $\sigma \mapsto \varphi_\sigma$ is a homomorphism of $S_I$ into the group of continuous automorphisms of the algebra $A[[X]]$ . Let $f \in A[[X]]$ be a formal power series; we put $af = \varphi_a(f)$ and we shall say that the formal power series $f$ is symmetric if $\sigma f = f$ for each $\sigma \in S_I$ . The symmetric formal power series form a closed subalgebra of $A[[X]]$ which is denoted by $A[[X]]^{\text{sym}}$ and is equipped with the topology induced by that of $A[[X]]$ .

Let T be an indeterminate. In the ring of formal power series $A[[X, T]]$ the family $(X_i T)_{i \in I}$ is summable, hence the family $(1 + X_i T)_{i \in I}$ , is multipliable (IV, p. 27, Prop. 2); moreover we have

\prod_{i \in I} (1 + X_i T) = 1 + \sum_{k \geq 1} s_k T^k,

where the formal power series $s_k \in A[[X]]$ is defined by

s_k = \sum_{H \in \mathscr{P}_k} \left( \prod_{i \in H} X_i \right) \quad (k \geq 1)

(we denote by $\mathscr{P}_k$ the set of all k-element subsets of I). In particular we have $s_1 = \sum_{i \in I} X_i$ . When I is finite with $n$ elements we have $s_k = 0$ for $k > n$ ; more precisely, when $I = \{1, \ldots, n\}$ then the formal power series $s_k$ is nothing other than the elementary symmetric polynomial of degree k in $X_1, \ldots, X_n$ .

Let $S = (S_k)_{k \geq 1}$ be a sequence of indeterminates. Since the formal power series $s_k$ is of order $\geq k$ , and belongs to $A[[X]]^{\text{sym}}$ , conditions a) and b) of Prop. 4 of IV, p. 28 are satisfied with $E = A[[X]]^{\text{sym}}$ ; there exists thus a unique continuous A-algebra homomorphism

\varphi_I : A[[S]] \to A[[X]]^{\text{sym}}

such that $\varphi_I(S_k) = s_k$ for each integer $k \geq 1$ .

Theorem 2. — a) If I is a finite set of n elements, then $\varphi_I$ induces a bicontinuous isomorphism of $A[[S_1, \ldots, S_n]]$ onto $A[[X]]^{\text{sym}}$ .

b) If I is infinite, $\varphi_I$ is a bicontinuous isomorphism of $A[[S]]$ onto $A[[X]]^{\text{sym}}$ .

In case a) we put $B = A[[S_1, \ldots, S_n]]$ and equip this algebra with the topology induced by that of $A[[S]]$ ; we also denote by $\psi_I$ the restriction of $\varphi_I$ to B. In case b) we put $B = A[[S]]$ and $\psi_I = \varphi_I$ . We shall equip the polynomial algebra $A[S]$ with the graduation of type N for which $S_k$ is of weight k for every integer $k \geq 1$ .

Lemma 1. — Let J be a finite subset of I, r an integer such that Card $J \geq r$ and fa symmetric formal power series which is homogeneous of degree r in the $X_i$ ( $i \in I$ ). Let $\bar{f}$ be the formal power series obtained by substituting 0 for $X_i$ for each i in $I - J$ . If $\bar{f} = 0$ , then we have $f = 0$ .

Let us put $f = \sum_{|\alpha| = r} a_\alpha X^\alpha$ (where $|\alpha|$ is the length $\sum_{i \in I} \alpha_i$ of the multiindex $a = (\alpha_i)_{i \in I}$ ). If $a$ is a multiindex of length $r$ , and $J'$ is the support of $a$ (the set of $i \in I$ such that $\alpha_i \neq 0$ ), then Card $J' \leq r$ , hence there exists a permutation $\sigma \in S_I$ such that $\sigma(J') \subset J$ . Put $\beta_i = \alpha_{\sigma^{-1}(i)}$ for $i \in I$ , then the monomial $X^\beta = \prod_{i \in I} X_{\sigma(i)}^{\alpha_i}$ depends only on the indeterminates $X_j$ ( $j \in J$ ), whence $a_\beta = 0$ by the hypothesis $\overline{f} = 0$ . Since $f$ is symmetric, we have $a_\alpha = a_\beta$ and since $\alpha$ was arbitrary, we conclude that $f = 0$ .

Lemma 2

Let $f$ be a symmetric formal power series of degree $r$ in the $X_i (i \in I)$ . There exists a unique polynomial $P \in B \cap A[S]$ , isobaric of weight $r$ , such that $f = \psi_1(P)$ .

The case when $I$ is finite follows from Th. 1 (IV, p. 62).

Assume that $I$ is infinite and choose a finite subset $J$ of $I$ , containing $r$ elements. We keep the notation of Lemma 1; we remark that every isobaric polynomial of weight $r$ in the $S_n \ (n \geq 1)$ depends only on $S_1, \ldots, S_r$ and that $\overline{s}_1, \ldots, \overline{s}_r$ are the elementary symmetric polynomials in the $r$ indeterminates $X_j \ (j \in J)$ . If $P$ is an isobaric polynomial of weight $r$ in the $S_n$ and $h = f - \psi_1(P)$ , then $\overline{h} = \overline{f} - P(\overline{s}_1, \ldots, S_r)$ and Lemma 1 shows the relation $f = \psi_1(P)$ to be equivalent to $\overline{f} = P(\overline{s}_1, \ldots, \overline{s}_r)$ . By Th. 1 (IV, p. 62) there exists a unique polynomial $P \in A[S]$ , isobaric of weight $r$ and such that $\overline{f} = P(\overline{s}_1, \ldots, \overline{s}_r)$ , whence the result.

Lemma 3

For every integer $m \geq 0$ let $c_m$ be the ideal of the algebra $A[[X]]^{sym}$ consisting of all symmetric formal power series of order $\geq m$ . The sequence $(c_m)_{m \geq 0}$ is a fundamental system of neighbourhoods of $0$ in $A[[X]]^{sym}$ .

The lemma clearly holds when $I$ is finite, so let us assume that $I$ is infinite. For every finite subset $J$ of $I$ , consisting of $m$ elements, denote by $\tilde{J}$ the set of elements of $N^{(1)}$ of length $< m$ and with support in $J$ . Further denote by $a_J'$ the set of formal power series containing no term $aX^\alpha$ with $a \in \tilde{J}$ . Since $\tilde{J}$ is a finite subset of $N^{(1)}$ and every finite subset of $N^{(1)}$ is contained in a set of the form $\tilde{J}$ , the family $(a_J')$ is a base of neighbourhoods of $0$ in $A[[X]]$ (IV, p. 26). Now Lemma 1 implies the relation $a_J' \cap A[[X]]^{sym} = c_m$ for every subset $J$ with $m$ elements, and this proves Lemma 3.

Since there are only a finite number of monomials of a given weight in the $S_k$ , every formal power series $f \in B$ may be written in a unique way as $f = \sum_{r \geq 0} P_r$ , where $P_r$ is an isobaric polynomial of weight $r$ in $B \cap A[S]$ . For every integer $m \geq 0$ let $b_m$ be the ideal of $B$ consisting of all formal power series of the above type such that $P_r = 0$ for $0 \leq r < m$ . The sequence $(b_m)_{m \geq 0}$ is a base of neighbourhoods of $0$ in $B$ .

With the above notation $\psi_1(P_r)$ is a symmetric formal power series in the $X_i$ , homogeneous of degree $r$ , and so this is the homogeneous component of degree $r$ of $\psi_1(f)$ . Lemma 2 shows that $\psi_1$ is an algebra homomorphism of $B$ onto $A[[X]]^{sym}$ , mapping $b_m$ to $c_m$ for every integer $m \geq 0$ ; now Lemma 3 shows $\psi_1$ to be bicontinuous, and this completes the proof of Theorem 2.

4. Sums of powers

We again write $X = (X_i)_{i \in I}$ for a family of indeterminates. The symmetric formal power series $s_k$ are defined as before by

s_k = \sum_{H \in \mathfrak{P}_k} \prod_{i \in H} X_i \quad (k \geq 1),

where $\mathfrak{P}_k$ is the set of all k-element subsets of $I$ . We also write

p_k = \sum_{i \in I} X_i^k \quad (k \geq 1)

This is a symmetric formal power series which is homogeneous of degree k.

Lemma 4 (« Newton's relations »). — For every integer $d \geq 1$ we have

p_d = \sum_{k=1}^{d-1} (-1)^{k-1} s_k p_{d-k} + (-1)^{d+1} d s_d

Let us define a continuous derivation $A$ in $\mathbf{A}[[\mathbf{X}]]$ by $A(u) = \sum_{n \geq 0} n u_n$ , where for every $u$ in $\mathbf{A}[[\mathbf{X}]]$ , $u_n$ is the homogeneous component of degree $n$ of $u$ . By (16) and Prop. 2 of IV, p. 27 we have

1 + \sum_{k \geq 1} s_k = \prod_{i \in I} (1 + X_i)

By Prop. 9 of IV, p. 34 we thus have

\left( \sum_{k \geq 1} k s_k \right) \cdot \left( 1 + \sum_{k \geq 1} s_k \right)^{-1} = \sum_{i \in I} \Delta(X_i)/(1 + X_i).

We have $A(X_i) = X_i$ and $X_i/(1 + X_i) = \sum_{k \geq 1} (-1)^k \ X_i^k$ . The right-hand side of (20) is thus equal to $\sum_{k \geq 1} (-1)^{k-1} p_k$ . Hence by (20) we have

\sum_{k \geq 1} k s_k = \left( 1 + \sum_{k \geq 1} s_k \right) \cdot \left( \sum_{k \geq 1} (-1)^{k-1} p_k \right)

and now Lemma 4 follows by a comparison of homogeneous components of degree d.

Remark

— With the notation of the above proof we have

\Delta u = \sum_{i \in I} X_i \cdot D_i(u)

(IV, p. 33, Cor. 1). In other words, Euler's relation (IV, p. 8, Prop 6) extends to formal power series : if $u \in A[X]$ is homogeneous of degree $n$ , then

(21)

n \cdot u = \sum_{i \in I} X_i \cdot D_i(u) .

When I is finite, consisting of $n$ elements, we have $s_k = 0$ for $k > n$ . Then Newton's relations may be written

\begin{aligned} p_2 &= s_1 p_1 - 2 s_2 \\ p_3 &= s_1 p_2 - s_2 p_1 + 3 s_3 \\ &\ \ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ p_{n-1} &= s_1 p_{n-2} - s_2 p_{n-3} + \cdots + (-1)^{n-1} s_{n-2} p_1 + (-1)^n (n-1) s_{n-1} \\ p_n &= s_1 p_{n-1} - s_2 p_{n-2} + \cdots + (-1)^n s_{n-1} p_1 + (-1)^{n+1} n s_n \end{aligned}

and

(22)

p_k = s_1 p_{k-1} - s_2 p_{k-2} + \cdots + (-1)^{n+1} s_n p_{k-n} \quad (\text{for } k > n) .

The first $n$ of the above relations hold for any $I$ ; we find for example

p_1 = s_1 , \quad p_2 = s_1^2 - 2 s_2 , \quad p_3 = s_1^3 - 3 s_1 s_2 + 3 s_3 .

More generally, let $S = (S_n)_{n \geq 1}$ be a family of indeterminates. Let us define the polynomials $P_d \in \mathbf{Z}[S_1, \ldots, S_d]$ recursively by $P_1 = S_1$ and

P_d = \sum_{k=1}^{d-1} (-1)^{k-1} S_k P_{d-k} + (-1)^{d+1} d S_d \quad (d \geq 2) .

Then we have the « universal formulae » $p_d = P_d(s_1, \ldots, s_d)$ holding over any ring A and family X of indeterminates. Let $P = (P_k)_{k \geq 1}$ be a sequence of indeterminates. Since $p_k$ is homogeneous of degree k in $A[[X]]$ , there exists precisely one continuous A-algebra homomorphism

\lambda_I : A[[P]] \to A[[X]]^{\mathrm{sym}}

such that $\lambda_I(P_k) = p_k$ for every integer $k \geq 1$ (IV, p. 28). If we assign to $P_k$ the weight k, then $\lambda_I$ transforms an isobaric polynomial of weight $n$ in the $P_k$ into a formal power series which is homogeneous of degree $n$ in the $X_i$ .

Proposition 3

— a) If I is a finite set with n elements and $n! \cdot 1$ is invertible in A, then $\lambda_I$ induces a bicontinuous isomorphism of $A[[P_1, \ldots, P_n]]$ onto $A[[X]]^{\mathrm{sym}}$ . b) If I is infinite and A is a Q-algebra, then $\lambda_I$ is a bicontinuous isomorphism of $A[[P]]$ onto $A[[X]]^{\mathrm{sym}}$ . We shall treat the case a) only, the case b) being quite similar.

By Th. 2 (IV, p. 68) we can identify $\mathbf{A}[[\mathbf{X}]]^{\text{sym}}$ with the algebra of formal power series $\mathbf{A}[[S_1, \ldots, S_n]]$ , with $S_k$ corresponding to $s_k$ . By Lemma 4 of IV, p. 70, there exist formal power series $g_1, \ldots, g_s$ of order $\geq 2$ in the indeterminates $s_1, \ldots, s_n$ such that

p_k = (-1)^{k+1} k s_k + g_k(s_1, \ldots, s_s) \quad (1 \leq k \leq n).

Since $k! . 1$ is invertible in $\mathbf{A}$ , Lemma 2 of IV, p. 35 proves the existence of an automorphism $T$ of the topological $\mathbf{A}$ -algebra $\mathbf{A}[[\mathbf{X}]]^{\text{sym}}$ which maps $s_k$ to $p_k$ for $l \leq k \leq n$ . Now Prop. 3, a) is an immediate consequence.

Corollary

— *Let $\xi_1, \ldots, \xi_n, \eta_1, \ldots, \eta_n$ be elements of $\mathbf{A}$ and suppose that $\mathbf{A}$ is an integral domain.

a) If $s_k(\xi_1, \ldots, \xi_n) = s_k(\eta_1, \ldots, \eta_n)$ for $1 \leq k \leq n$ , then there exists a permutation $\sigma \in \mathfrak{S}_n$ such that $\eta_i = \xi_{\sigma(i)}$ for $1 \leq i \leq n$ .

b) Suppose that $n! . 1 \neq 0$ in $\mathbf{A}$ and

\xi_1^k + \cdots + \xi_n^k = \eta_1^k + \cdots + \eta_n^k

for $1 \leq k \leq n$ . Then there exists a permutation $\sigma \in \mathfrak{S}_n$ such that $\eta_i = \xi_{\sigma(i)}$ for $1 \leq i \leq n$ .

Under the hypotheses a) we have $\prod_{i=1}^n (X - \xi_i) = \prod_{i=1}^n (X - \eta_i)$ . If we substitute $\eta_n$ for $X$ , we find that $\prod_{i=1}^n (\eta_n - \xi_i) = 0$ and since $\mathbf{A}$ is an integral domain, there is an integer $\sigma(n)$ such that $l \ s \ \sigma(n) \leq n$ and $\eta_n = \xi_{\sigma(n)}$ . Now the assertion a) follows easily by induction because $\mathbf{A}[X]$ is an integral domain.

Under the hypotheses of b) there exist by Prop. 3 polynomials $\Pi_1, \ldots, \Pi_n$ in $n$ indeterminates, with coefficients in the field of fractions of $\mathbf{A}$ such that $s_k = \Pi_k(p_1, \ldots, p_n)$ for $1 \leq k \leq n$ . Now the relation (23) implies

s_k(\xi_1, \ldots, \xi_n) = s_k(\eta_1, \ldots, \eta_n)

for $1 \leq k \leq n$ and so b) follows from a).

5. Symmetric functions in the roots of a polynomial

Consider a monic polynomial of degree $n$ , with coefficients in $\mathbf{A}$ :

f = X^n + a_1 X^{n-1} + \cdots + a_{n-1} X + a_n.

We define an associative, commutative and unital $\mathbf{A}$ -algebra $E_f$ with the generators $x_1, \ldots, x$ , and relations

\sum_{i_1 < \ldots < i_k} x_{i_1} \ldots x_{i_k} = (-1)^k a_k \quad (1 \leq k \leq n).

More precisely, we have

E_f = A[X_1, ..., X_n]/a

where the ideal $a$ is generated by the polynomials $s_k + (-1)^{k+1} a_k$ for $1 \leq k \leq n$ , and $x_i$ is the residue class of $X_i$ mod $a$ for $1 \leq i \in n$ . The relation (24) is also equivalent to $f(X) = \prod_{i=1}^n (X - x_i)$ . When there is any risk of ambiguity, we write $x_{1,f}, ..., x_{n,f}$ in place of $x_1, ..., x_n$ .

Proposition 4

Let B be a commutative ring, $\rho$ a homomorphism of A into B and $\xi_1, ..., 5$ , elements of B. Suppose that the relation $^\rho f(X) = \prod_{i=1}^n (X - \xi_i)$ holds in B[X]. Then there exists one and only one ring homomorphism $u : E_f \to B$ such that $\rho(a) = u(a . 1)$ for all $a \in A$ and $u(x_i) = \xi_i$ for $1 \leq i \leq n$ .

We consider B as associative, commutative and unital A-algebra by means of $\rho$ . Then the relation $^\rho f(X) = \prod_{i=1}^n (X - \xi_i)$ may also be written in the form

\sum_{i_1 < ... < i_k} \xi_{i_1} ... \xi_{i_k} = (-1)^k a_k . 1 \quad (1 \leq k \leq n)

in B. Since the relations (24) define a presentation of $E_f$ , Prop. 4 follows.

Prop. 4 justifies the name « universal decomposition algebra of $f$ » for $E_f$ . The relation $f(X) = \prod_{i=1}^n (X - x_{i,f})$ is called the « universal decomposition of $f$ ». Let $\sigma \in S_n$ be a permutation ; since $f(X) = \prod_{i=1}^n (X - x_{\sigma(i),f})$ , there exists an automorphism $t_\sigma$ of the A-algebra $E_f$ characterized by $t_\sigma(x_{i,f}) = x_{\sigma(i),f}$ for $1 \leq i \leq n$ . We have $t_{\sigma \tau} = t_\sigma \circ t_\tau$ for $\sigma, \tau$ in $S_n$ , and hence obtain an action of the group $\mathscr{G}$ , an the A-algebra $E_f$ .

Proposition 5

In the universal decomposition algebra $E_f$ the family of monomials $x_1^{\nu(1)} ... x_n^{\nu(n)}$ such that $0 \leq \nu(i) < i$ for $1 \leq i \leq n$ is a basis of the A-module $E_f$ . In particular $E_f$ is a free A-module of rank $n!$ .

Write $B = A[X_1, ..., X_n]$ and $C = A[X_1, ..., X_n]^{\text{sym}}$ . By Th. 1 (IV, p. 62) we have $C = A[s_1, ..., s_n]$ and $s_1, ..., s_n$ are algebraically independent over A. The polynomials without constant term in $s_1, ..., s_n$ form an ideal $C^+$ in C, supplementary to A and generated by $s_1, ..., s_n$ . Let c be the ideal of C generated by $s_1 + a_1, s_2 - a_2, ..., s_n + (-1)^{n+1} a_n$ . There exists an A-algebra automorphism of C which maps $s_k$ to $s_k + (-1)^{k+1} a_k$ for $1 \leq k \leq n$ , and hence maps $C^+$ onto c ; therefore we have $C = A \oplus c$ . Moreover Th. 1, c) of IV, p. 62 shows that

B = \bigoplus_{\nu \in S} CX^\nu

where S is the set of all $\nu \in \mathbf{N}^n$ such that $0 \leq \nu(i) < i$ for $1 \leq i \leq n$ . The ideal a of B is generated by c, whence $a = Bc = \bigoplus_{\nu \in S} c \cdot X^\nu$ . Since $C = A \oplus c$ , we find

B = a \oplus \bigoplus_{\nu \in S} AX^\nu,

whence Prop. 5, because

E_f = B/a

Corollary

— *The canonical homomorphism of A into the universal decomposition algebra of the monic polynomial $f \in A[X]$ is injective.

For the unit element of $E_f$ forms part of a basis of the A-module $E_f$ .

Proposition 6

— *Let $f \in A[X]$ be a monic polynomial of degree n, and let P be a symmetric polynomial in $X_1, ..., X_n$ with coefficients in A. Then there exists precisely one element a of A with the following property: (FS) For any ring homomorphism $\rho : A \to B$ and decomposition $^0f(X) = \prod_{i=1}^n (X - \xi_i)$ in B[X] we have $\rho(a) = P(\xi_1, ..., \xi_n)$ .

Write $f = X^n + \sum_{k=1}^n a_k X^{n-k}$ , then by Th. 1, IV, p. 62 there exists a polynomial $\Pi$ in n indeterminates with coefficients in A, such that $P = \Pi(s_1, ..., s_n)$ . Put $a = \Pi(-a_1, a_2, ..., (-1)^n a_n)$ . Under the hypothesis (FS) we have

s_k(\xi_1, ..., \xi_n) = (-1)^k \rho(a_k)

whence

\begin{align*} \rho(a) &= \Pi(-\rho(a_1), \rho(a_2), ..., (-1)^n \rho(a_n)) \\ &= \Pi(s_1(\xi_1, ..., \xi_n), ..., s_n(\xi_1, ..., \xi_n)) \\ &= P(\xi_1, ..., \xi_n) . \end{align*}

This proves the existence of an element a satisfying (FS). The uniqueness of a follows from the Cor. to Prop. 5, because we have $a \cdot 1 = P(x_{1,f}, ..., x_{n,f})$ in the universal decomposition algebra $E_f$ .

With the notation of Prop. 6 one sometimes writes $a = P^*(f)$ . Here are some examples.

Examples. — 1) If $P = s_k$ , then $P^*(f) = (-1)^k a_k$ . \ 2) Let g be a polynomial in $A[X]$ and put

P(X_1, ..., X_n) = g(X_1) ... g(X_n) .

Then

P^*(f)

is just the resultant

\operatorname{res}(f, g)

, by Cor. 1 of IV, p. 80. 3) Put

A(X_1, ..., X_n) = \prod_{i < j} (X_i - X_j)^2

, then

\Delta^*(f)

is just the discriminant of the monic polynomial

f

(IV, p. 82, formula (46)). \*

4) Put $P(X_1, \ldots, X_n) = X_1^k + \ldots + X_n^k$ ; moreover, define the algebra $E = A[X]/(f)$ and write $x$ for the image of $X$ in $E$ . We recall that the $A$ -module $E$ is free, with basis $(1, x, \ldots, x^{n-1})$ (IV, p. 11, Cor.). Let us show

\text{Tr}_{E/A}(x^k) = P^*(f)

Put $\pi_k = \text{Tr}_{E/A}(x^k)$ for every integer $k \geq 1$ . Bearing in mind Newton's relations (IV, p. 70) we need only establish the relations

\begin{align*} &\pi_k + a_1 \pi_{k-1} + \cdots + a_{k-1} \pi_1 + k a_k = 0 \quad \text{for } 1 \leq k \leq n \\ &\pi_k + a_1 \pi_{k-1} + \cdots + a_{n-1} \pi_{k-n+1} + a_n \pi_{k-n} = 0 \quad \text{for } k > n \end{align*}

(which we shall also call « Newton's relations »). The relation (27) is clear, because the left-hand side is the trace of

x^{k-n}(x^n + a_1 x^{n-1} + \cdots + a_{n-1} x + a_n) = 0 .

Suppose that $1 \leq k \leq n$ and put

y = x^k + a_1 x^{k-1} + \cdots + a_{k-1} x + a_k \cdot 1 ;

let $M = (m_{ij})$ be the matrix of the linear mapping $u \mapsto yu$ in $E$ relative to the basis $(x')_{0,\ldots,n}$ . We easily obtain the relations

\begin{align*} &m_{ii} = a_k \quad \text{for } 0 \leq i < n - k \\ &m_{ii} = 0 \quad \text{for } n - k \leq i < n , \end{align*}

whence

\text{Tr}_{E/A}(y) = \sum_{i=0}^{n-1} m_{ii} = (n-k) a_k .

Moreover we have

\text{Tr}_{E/A}(y) = \pi_k + a_1 \pi_{k-1} + \cdots + a_{k-1} \pi_1 + n a_k ,

whence formula (26) follows.

6. The resultant

In this No. we assume as given two positive integers $p, q$ and two polynomials $f, g$ in $A[X]$ of the form

\begin{align*} f &= t_p X^p + t_{p-1} X^{p-1} + \cdots + t_0 \\ g &= u_q X^q + u_{q-1} X^{q-1} + \cdots + u_0 \end{align*}

such that $\deg f \leq p, \deg g \leq q$ . For every integer $n \geq 0$ we denote by $S$ , the sub- $A$ -module of $A[X]$ consisting of all polynomials of degree $< n$ ; it has the family $(X')_{0 \leq i < n}$ as basis, and hence is of rank $n$ .

We provide $S, \times S_p$ with the basis

B_1 = ((X^{q-1}, 0), ..., (X, 0), (1, 0), (0, X^{p-1}), ..., (0, X), (0, 1))

and

S_{p+q}

with the basis

B_2 = (X^{p+2-1}, ..., X, 1).

We define a linear mapping

\varphi : S, \times S_p \to S_{p+q}

\varphi(u, v) = uf + vg

and we denote by

M(f, g, p, q)

the matrix of

\varphi

with respect to the bases

B_1

and

B_2

. This is a square matrix of order

p + q

, indexed by the set

\{0, 1, ..., p + q - 1\}

. Its elements

a_{ij}

are given by the rules: a)

a_{ij} = t_{p-i+j}

for

0 \leq j \leq q-1

, b)

a_{ij} = u_{j-i}

for

q \leq j \leq p + q - 1

, where

t_k

is taken to be 0 if

k \notin (0, p)

and

u_k = 0

k \notin (0, q)

For example, when $p = 2$ and $q = 3$ , we have the matrix

\begin{pmatrix} t_2 & 0 & 0 & u_3 & 0 \\ t_1 & t_2 & 0 & u_2 & u_3 \\ t_0 & t_1 & t_2 & u_1 & u_2 \\ 0 & t_0 & t_1 & u_0 & u_1 \\ 0 & 0 & t_0 & 0 & u_0 \end{pmatrix}.

Definition 1

With the above notation the determinant of the matrix $M(f, g, p, q)$ is called the resultant of the pair $(f, g)$ for the degrees $p$ and $q$ , or simply the resultant of $f$ and $g$ if $p = \deg f$ and $q = \deg g$ .

The resultant is denoted by $\operatorname{res}_{p,q}(f, g)$ or simply $\operatorname{res}(f, g)$ when $p = \deg f$ , $q = \deg g$ .

Example 1

Given $\Lambda, \mu$ in $\Lambda$ , we have the formulae

\operatorname{res}_{p,0}(f, \lambda) = \lambda^p, \quad \operatorname{res}_{0,q}(\mu, g) = \mu^q

\operatorname{res}_{p,1}(f, \lambda) = \lambda^p t_p, \quad \operatorname{res}_{1,q}(\mu, g) = (-1)^q \mu^q u_q,

whose proof is immediate.

Example 2

When $p = q = 1$ , we have

\operatorname{res}_{1,1}(t_1 X + t_0, u_1 X + u_0) = t_1 u_0 - t_0 u_1

Remark 1

The matrix $M(g, f, q, p)$ is obtained from $M(f, g, p, q)$ by $pq$ transpositions of columns, whence

\operatorname{res}_{q,p}(g, f) = (-1)^{pq} \operatorname{res}_{p,q}(f, g)

Remark 2

Let $\rho : A \to B$ be a ring homomorphism. Def. / implies directly the formula

\operatorname{res}_{p,q}(\rho f, \rho g) = \rho(\operatorname{res}_{p,q}(f, g)) .

Remark 3

Given $\lambda, \mu$ in $A$ , we have

\operatorname{res}_{p,q}(\lambda f, \mu g) = \lambda^q \mu^p \operatorname{res}_{p,q}(f, g) .

(28)

Remark 4

Suppose that $p + q \geq 1$ . By III, p. 532, formula (28), the image of $\varphi$ contains the constant polynomial $\operatorname{res}_{p,q}(f, g)$ . Hence there exist a pair of polynomials $(u, v)$ , with $u \in S_q, v \in S_p$ such that

\operatorname{res}_{p,q}(f, g) = uf + vg ,

whence

\operatorname{res}_{p,q}(f, g) \in A \cap (f, g) .

This pair

(u, v)

is unique when

\operatorname{res}_{p,q}(f, g)

is cancellable in

A

: for then

\varphi

is injective (III, p. 524, Prop. 3).

Remark 5

Suppose that $p \geq q$ , and let $h \in A[X]$ be a polynomial of degree $\leq p - q$ . Let us show that

\operatorname{res}_{p,q}(f, g) = \operatorname{res}_{p,q}(f + gh, g) .

(29) For if we write

\omega(u, v) = (u, uh + v)

for

(u, v) \in S_q \times S_p

, then

w

is an automorphism of the

A

-module

S_q \times S_p

and we have

\omega^{-1}(u, v) = (u, -uh + v) .

The matrix of

w

with respect to the basis

B_1

is lower triangular and its diagonal elements are equal to

1

. On the other hand,

\varphi \circ w

maps

(u, v)

u(f + gh) + vg

. Now formula (29) means that the matrices representing

\varphi

and

\varphi \circ w

have the same determinant and this follows from the relation

\det w = 1

Assume that $f$ is monic of degree $p$ ; put $E = A[X]/(f)$ and denote by $x$ the canonical image of $X$ in $E$ . We know (IV, p. 11) that $E$ is a free $A$ -module with basis $(1, x, \ldots, x^{p-1})$ . We can thus define the norm $N_{E/A}(u)$ of any element $u$ of $E$ (III, p. 543, Def. 2).

Proposition 7

— Assume that f is monic of degree p; with the above notations we have¹

\operatorname{res}_{p,q}(f, g) = N_{E/A}(g(x)) .

(30)

¹ The resultant $\operatorname{res}_{p,q}(f, g)$ is thus independent of $q$ when $f$ is monic of degree $p$ . We then denote it simply by $\operatorname{res}(f, g)$ .

Let us define an A-linear mapping $0$ of $S_q \times S_p$ into $S_{p+q}$ by $\theta(u, v) = uf + v$ . Then $0$ maps the basis $B_1$ of $S_q \times S_s$ into the sequence

(fX^{q-1}, \ldots, fX, f, X^{p-1}, \ldots, X, 1)

of elements of

S_{p+q}

; hence the matrix

M

, of

0

with respect to the bases

B_1, B_2

is lower triangular and its diagonal elements are equal to 1, whence

\det M = 1

It follows that $\theta$ is bijective and $\operatorname{res}_{p,q}(f, g)$ is equal to the determinant of the endomorphism $\varphi' = \varphi \circ \theta^{-1}$ of $S_{p-q}$ . Explicitly, we have

\varphi'(uf + v) = uf + vg

for any pair

(u, v)

S_q \times S_s

. Now we have

A[X] = S_{p+q} + (f)

and

fS_q = S_s \cap (f)

, hence the canonical injection of

S_{p-s}

into

A[X]

defines by the passage to quotients an isomorphism

\gamma

S_{p+q}/fS_q

onto E. Let

\psi

be the multiplication by

g(x)

in E. Formula (31) shows that

\varphi'

induces the identity on

fS_q

and

\gamma^{-1}\psi\gamma

S_{p+q}/fS_q

. We thus have

\det \varphi' = \det \psi

, whence (30) follows, because

\det \varphi' = \operatorname{res}_{p,q}(f, g)

and

\det \psi = N_{E/A}(g(x))

, by definition.

Corollary 1

— Let $f \in A[X]$ be a monic polynomial; for every polynomial $g \in A[X]$ the following conditions are equivalent: (i) $\operatorname{res}(f, g)$ is invertible in $A$ ; (ii) there exist polynomials $u, v$ in $A[X]$ such that $uf + vg = 1$ ; (iii) $g(x)$ is invertible in the algebra $A[X]/(f)$ . The equivalence of (i) and (iii) follows from Prop. 3 of III, p. 545 and Prop. 7; that of (ii) and (iii) follows trivially.

Corollary 2

— Suppose that $A$ is a field and let $f, g \in A[X]$ , then the following conditions are equivalent when $f, g$ are non-zero: (i) $\operatorname{res}(f, g) \neq 0$ ; (ii) the polynomials $f$ and $g$ are relatively prime in $A[X]$ ; \* (iii) for any extension L of A the polynomials $f$ and $g$ have no common root in L. \

We can reduce at once to the case where $f$ is monic (IV, p. 77, remark 3). The equivalence of (i) and (ii) is just a translation of Cor. 1, by Def. 1 of IV, p. 12; the equivalence of (ii) and (iii) is just Cor. 7 of IV, p. 13.

Corollary 3

— For every $\lambda \in A$ we have

\operatorname{res}_{p,1}(f, \lambda - X) = f(\lambda), \quad \operatorname{res}_{1,q}(X - \lambda, g) = g(\lambda).

When

f(X) = X - \lambda

, the algebra E is equal to A and we have

x = A

; the second formula (32) now follows from Prop. 7 (IV, p. 77). By remarks 1 and 3 (IV, p. 76, 77) we conclude

\operatorname{res}_{p,1}(f, \lambda - X) = (-1)^p \operatorname{res}_{1,p}(\lambda - X, f) = (-1)^{p+p} \operatorname{res}_{1,p}(X - A, f) = f(A).

Suppose now that f and g are monic. We denote by F the A-algebra $A[X, Y]/(f(X), g(Y))$ and denote by x (resp. y) the canonical image of X (resp. Y) in F.

Proposition 8

— Suppose that f and g are monic of degrees p and q respectively. With the above notation, the A-module F is free with basis $(x^i y^j)_{0 \leq i < p,\ 0 \leq j < q}$ and we have

\text{res}(f, g) = N_{F/A}(x - y).

Put $E = A[X]/(f)$ and $E' = A[Y]/(g)$ . By II, p. 253, Cor. 1, the homomorphism $\sigma$ of $E @ E'$ into F derived from the canonical homomorphism $A[X] \otimes A[Y] \to A[X, Y]$ is bijective; this proves the assertion about the basis of F. We now identify E with its image in F under $\sigma$ . Then the E-algebra homomorphism of $E[Y]/(g(Y))$ into F which maps Y to y is an isomorphism.

By the transitivity of the norm (III, p. 546) we have

N_{F/A}(x - y) = N_{E/A}(N_{F/E}(x - y)).

By Prop. 7 (IV, p. 77), $N_{F/E}(x - y)$ is the resultant of the polynomials g(Y) and x − Y in E[Y], hence equal to g(x) (IV, p. 78, Cor. 3). By formula (34) and Prop. 7 (IV, p. 77) we thus have

N_{F/A}(x - y) = N_{E/A}(g(x)) = \text{res}(f, g)

Proposition 9

— Let $p_1$ and $q_1$ be positive integers and $f_1, g_1$ polynomials in $A[X]$ such that $\deg f_1 \leq p_1, \deg g_1 \leq q_1$ ; then we have

\text{res}_{p, q + q_1}(f, gg_1) = \text{res}_{p, q}(f, g) \cdot \text{res}_{p, q_1}(f, g_1)

\text{res}_{p + p_1, q}(ff_1, g) = \text{res}_{p, q}(f, g) \cdot \text{res}_{p_1, q}(f_1, g).

We have $\text{res}_{p, q}(f, g) = (-1)^{pq} \text{res}_{q, p}(g, f)$ (IV, p. 76); so it is enough to prove (35). Likewise, Remark 3 (loc. cit.) shows that if formula (35) is established for a polynomial $f$ , then it holds for all polynomials of the form Af, where $A \in A$ . Finally, when f is monic of degree p, formula (35) follows from Prop. 7 (IV, p. 77) by the formula $N_{E/A}(ab) = N_{E/A}(a) \cdot N_{E/A}(b)$ . In all we conclude that (35) holds when the coefficient $t_p$ of $X^p$ in f is invertible.

Lemma 5

Let t be an element of A. There exists a commutative ring C containing A as subring, a subring B of C containing A, an element $\tau$ of B invertible in C and a ring homomorphism $p : B \to A$ such that $p(\tau) = t$ and the restriction of p to A is equal to Id.

It suffices to take for B the algebra $A^{(\mathbf{N})}$ of the monoid $\mathbf{N}$ , that is, the polynomial algebra $A[\tau]$ in one indeterminate $\tau$ , for C the algebra $A^{(\mathbf{Z})}$ of the group $\mathbf{Z}$ and for p the homomorphism $P \mapsto P(t)$ of $A(\tau)$ into A.

With the notation of Lemma 5, where we have taken t = t_p, let us put

F = \tau X^p + t_{p-1} X^{p-1} + \cdots + t_1 X + t_0

in B[X]. The coefficient of

X^p

in F is invertible in C; if we consider F, g,

g_1

as polynomials of C[X] we thus have

\operatorname{res}_{p,q+q_1}(F, gg_1) = \operatorname{res}_{p,q}(F, g) \cdot \operatorname{res}_{p,q_1}(F, g_1)

by what has been said. The resultants are unchanged if we consider F, g and

g_1

as polynomials in B[X]. Since

^\rho F = f

^\rho g = g

^\rho g_1 = g_1

, formula (35) follows from (38) and Remark 2 (IV, p. 77).

Corollary 1

— (i) Let $\lambda, \alpha_1, \ldots, \alpha_p$ be elements of A and suppose that $f(X) = \lambda (X - \alpha_1) \ldots (X - \alpha_p)$ . We have

\operatorname{res}_{p,q}(f, g) = \lambda^q g(\alpha_1) \ldots g(\alpha_p) .

(ii) Let

\mu, \beta_1, \ldots, \beta_q

be elements of A and suppose further that

g(X) = \mu (X - \beta_1) \ldots (X - \beta_q)

. Then we have

\operatorname{res}_{p,q}(f, g) = \lambda^q \mu^p \prod_{\substack{1 \leq i \leq p \\ 1 \leq j \leq q}} (\alpha_i - \beta_j) .

Assertion (i) follows directly from the formulae (28), (32) and (36). Now Assertion (ii) follows from (i).

Corollary 2

— For every integer $r \geq 0$ we have

\operatorname{res}_{p,q+r}(f, g) = t_p^r \cdot \operatorname{res}_{p,q}(f, g) .

Bearing in mind Example 1 (IV, p. 76) we need only take

q_1 = r, g_1 = 1

in (35).

Assume that f is monic, and let $\rho : A \to B$ be a ring homomorphism and $\xi_1, \ldots, \xi_p$ elements of B such that we have the decomposition

^\rho f(X) = (X - \xi_1) \ldots (X - \xi_p)

By Remark 2 (IV, p. 77) and Cor. 1 above we have

\rho(\operatorname{res}(f, g)) = g(\xi_1) \ldots g(\xi_p)

This remark applies in particular to the universal decomposition off (IV, p. 73) and since

\rho

is then injective, this provides a means of calculating

\operatorname{res}(f, g)

Example 3

Let us prove the formula

\operatorname{res}_{2,2}(aX^2 + bX + c, a'X^2 + b'X + c') = (ac' - ca')^2 + (bc' - cb')(ba' - ab')

Arguing as for Prop. 9 (IV, p. 79) we see that it is enough to prove the formula when $a$ is invertible. There exists then a decomposition of the form

aX^2 + bX + c = a(X - x)(X - y)

in $B[X]$ , where $B$ is an appropriate ring containing $A$ as subring. By Cor. 1 above, the required resultant is equal to

R = a^2(a'x^2 + b'x + c')(a'y^2 + b'y + c') .

Now we have

ax + ay = -b , \quad axy = c

by (43), whence $(ax)^2 + (ay)^2 = b^2 - 2ac$ .

By (44) we have

R = {a'}^2(axy)^2 + a{b'}^2(axy) + a^2{c'}^2 + a'b'(axy)(ax + ay) +

+ a'c'((ax)^2 + (ay)^2) + ab'c'(ax + ay)

= {a'}^2c^2 + a{b'}^2c + a^2{c'}^2 - a'b'cb + a'c'(b^2 - 2ac) - ab'c'b

= (ac' - ca')^2 + (ab' - a'b)(b'c - c'b) ,

from which the desired result follows.

7. The discriminant

Definition 2

Let $f$ be a monic polynomial of $A[X]$ of degree $m$ , and denote by $E$ the $A$ -algebra $A[X]/(f)$ and by $x$ the canonical image of $X$ in $E$ . We define the discriminant of $f$ , written $\operatorname{dis}(f)$ , as the discriminant $D_{E/A}(1, x, \ldots, x^{m-1})$ of the basis $(1, x, \ldots, x^{m-1})$ of the $A$ -algebra $E$ .

For every positive integer $k$ we write $\rho_k = \operatorname{Tr}_{E/A}(x^k)$ . By III, p. 549, Def. 2 amounts to the formula

\operatorname{dis}(f) = \det(p_{i+j})_{0 \leq i,j < m}

*Examples. — 1) If $f$ is a monic polynomial of degree 0 or 1, we have $\operatorname{dis}(f) = 1$ , by (45). 2) Let $f(X) = X^2 + \alpha X + \beta$ be a monic polynomial of degree 2. Newton's relations may be written in the form (IV, p. 75)

p_0 = 2 \\ p_1 + \alpha = 0 \\ p_2 + \alpha p_1 + 2\beta = 0 ,

whence $p_1 = -a$ , $p_2 = a^2 - 2p$ . It follows that we have

\operatorname{dis}(f) = \det \begin{pmatrix} 2 & -\alpha \\ -a & \alpha^2 - 2p \end{pmatrix} = \alpha^2 - 4\beta

Let B be a commutative ring, $\rho$ a homomorphism from A to B and $\xi_1, \ldots, \xi_m$ elements of B such that

\rho f(X) = (X - \xi_1) \cdots (X - \xi_m).

Let us write M for the matrix

(\rho(p_{i + j}))_{0 \leq i, j < m}

and D for the Van der Monde matrix

(\xi_i^{j+1})_{0 \leq i, j < m}

. By Example 4 of IV, p. 75 we have

\rho(p_k) = \xi_1^k + \cdots + \xi_m^k,

whence

M = 'D . D ;

by III, p. 532 we have

D = \prod_{i > j} (\xi_i - \xi_j)

and

\det M = (\det D)^2

, that is

\rho(\operatorname{dis}(f)) = \prod_{i < j} (\xi_i - \xi_j)^2

Moreover (on writing D for the derivation

\frac{d}{dX}

) we have

D(\rho f)(\xi_i) = (\xi_i - \xi_1) \cdots (\xi_i - \xi_{i-1})(\xi_i - \xi_{i+1}) \cdots (\xi_i - \xi_m)

for

1 \leq i \leq m

, whence

\rho(\operatorname{dis}(f)) = (-1)^{m(m-1)/2} \prod_{i \neq j} (\xi_i - \xi_j) = (-1)^{m(m-1)/2} \prod_{i=1}^m D(\rho f)(\xi_i).

By Cor. 1 of IV, p. 80 applied to the universal decomposition of

f

, we finally obtain

\operatorname{res}(f, Df) = \operatorname{res}(Df, f) = (-1)^{m(m-1)/2} \operatorname{dis}(f).

Proposition 10

Let $m \geq 1$ . There exists a unique polynomial $A \in \mathbf{Z}[A_1, \ldots, A_m]$ with the following property : for any commutative ring A and monic polynomial $f = X^m + \sum_{i=1}^m a_i X^{m-i}$ in $A[X]$ we have

\operatorname{dis}(f) = \Delta(a_1, \ldots, a_m).

Moreover A is of degree

\leq 2m-2

and if we assign weight i to

A_i

then A is isobaric of weight

m(m-1)

. a) Uniqueness of A : if A satisfies (48), we have in particular

A = \operatorname{dis}(F)

, where F is the polynomial

X^m + \sum_{i=1}^m A_i X^{m-i}

with coefficients in

\mathbf{Z}[A_1, \ldots, A_m]

. b) Existence of A : let

s_1, \ldots, s_m

be the elementary symmetric polynomials in the indeterminates

X_1, \ldots, X_m

. There exists a polynomial

A \in \mathbf{Z}[A_1, \ldots, A_m]

, isobaric of weight

m(m-1)

, such that

\Delta(-s_1, s_2, \ldots, (-1)^m s_m) = \prod_{i < j} (X_i - X_j)^2;

for the right-hand side is a symmetric polynomial P, homogeneous of degree m(m − 1) in $Z[X_1, ..., X_m]$ (IV, p. 62, Th. 1). Now formula (46) means that dis(f) = P (f ), in the notation of IV, p. 74, and (48) follows directly from this.

c) Degree of A : the relation (47) and the definition of the resultant (IV, p. 76) imply the formula

(-1)^{m(m-1)/2} \Delta = \det(a_{ij})_{0 \leq i,j \leq 2m-2}

with the following values of the $a_{ij}$

\begin{align*} a_{00} &= 1 , \quad a_{0,m-1} = m , \\ a_{ij} &= A_{i-j} & \text{for } 1 \leq i \leq 2m-2 , \quad 0 \leq j \leq m-2 \\ a_{ij} &= (j-i+1) A_{m+i-j-1} & \text{for } 1 \leq i \leq 2m-2 , \quad m-1 \leq j \leq 2m-2 . \end{align*}

$a_{0j} = 0$ if $j \neq 0 , \ j \neq m-1$

In these formulae it is understood that $A_i = 1$ and $A_i = 0$ for $i < 0$ or $i > m$ . Now (50) shows at once that A is of degree $\leq 2m-2$ , as we had to prove.

Prop. 10 allows us to extend the definition of the discriminant to non-monic polynomials. Let $m \geq 1$ be an integer, then there exists a unique homogeneous polynomial of degree $2m-2$ , $\tilde{\Delta}$ say, in $Z[A_0, A_1, ..., A_m]$ such that

\Delta(A_1, ..., A_r) = \tilde{\Delta}(1, A_1, ..., A_r)

for, since A is of degree $\leq 2m-2$ , the rational fraction

A_0^{2m-2} \Delta(A_1/A_0, ..., A_m/A_0)

belongs to the subring $Z[A_0, A_1, ..., A_m]$ of $Q(A_0, A_1, ..., A_m)$ . If $A_i$ has weight i for $0 \leq i \leq m$ , then $\tilde{\Delta}$ is isobaric of weight $m(m-1)$ . Iff is a polynomial of degree $\leq m$ , say

f = a_0 X^m + a_1 X^{m-1} + \cdots + a_{m-1} X + a_m ,

we shall put

\operatorname{dis}_m(f) = \tilde{\Delta}(a_0, a_1, ..., a_r).

When $m = \deg f$ , we shall simply write dis(f) for $\operatorname{dis}_m(f)$ ; if f is monic, dis(f) agrees with the discriminant defined in Def. 2, by (48), (51) and (52).

Proposition 11

— Let f in $A[X]$ be of degree $\leq m$ .

(i) If $\rho : A \to B$ is a ring homomorphism, we have $\operatorname{dis}_m(\rho_f) = \rho(\operatorname{dis}_m(f))$ .

(ii) Let $\lambda, \alpha_1, ..., a_r$ be elements of A. If $f = X(X-\alpha_1) ... (X-a_r)$ ; then we have

\operatorname{dis}_m(f) = \lambda^{2m-2} \prod_{i < j} (\alpha_i - \alpha_j)^2 .

(iii) Let $a_0$ be the coefficient of $X^m$ in $f$ ; then we have

\text{res}_{m,m-1}(f, Df) = \text{res}_{m-1,m}(Df, f) = (-1)^{m(m-1)/2} a_0 \text{ dis}_m(f) .

Assertion (i) is clear.

Since $\tilde{\Delta}$ is homogeneous of degree $2m - 2$ , we have

\text{dis}_m(\lambda f) = \lambda^{2m-2} \text{dis}_m(f)

for every polynomial $f \in A[X]$ of degree $\leq m$ . Assertion (ii) now follows from formulae (46) and (55).

When $f$ is monic of degree $m$ , we have $a_0 = 1$ and Assertion (iii) is reduced to formula (47). Bearing in mind (55) and the relation

\text{res}_{m,n}(\lambda f, \mu g) = \lambda^n \mu^m \text{res}_{m,n}(f, g),

(IV, p. 77), we can pass from there to the case where $a_0$ is invertible in $A$ . Now the general case follows by Prop. 11, (i) and Lemma 5 (IV, p. 79).

Corollary 1

— Let $g \in A[X]$ and let $n$ be a positive integer such that $\deg g \leq n$ . We have

\text{dis}_{m+n}(fg) = \text{dis}_m(f) \cdot \text{dis}_n(g) \cdot \text{res}_{m,n}(f,g)^2 .

By reasoning as before we are reduced to the case where $f$ and $g$ are monic of degrees $m$ and $n$ respectively. Now put $B = E_f \otimes E_g$ , where $E_f$ (resp. $E_g$ ) is the universal decomposition algebra of $f$ (resp. $g$ ) (IV, p. 73). Then $A$ is a subring of $B$ , and in $B[X]$ we have the decompositions

f = \prod_{i=1}^m (X - \alpha_i), \quad g = \prod_{j=1}^n (X - \beta_j) .

Hence we have

fg = \prod_{k=1}^{m+n} (X - \gamma_k) ,

with $\gamma_i = \alpha_i$ for $1 \leq i \leq m$ and $\gamma_{m+j} = \beta_j$ for $1 \leq j \leq n$ . We have the obvious identity

\prod_{k < k'} (\gamma_k - \gamma_{k'}) = \prod_{i < i'} (\alpha_i - \alpha_{i'}) \cdot \prod_{j < j'} (\beta_j - \beta_{j'}) \cdot \prod_{i,j} (\alpha_i - \beta_j)

On squaring this relation we obtain (57), by (40) and (46).

Corollary 2

— If $a_0$ is the coefficient of $X^m$ in $f$ , we have

\text{dis}_{m+1}(f) = a_0^2 \text{dis}_m(f)

This follows from Cor. 1 on taking $n = 1,\ g = 1$ , by the formula $\operatorname{res}_{m,1}(f,1) = a_0$ (IV, p. 76, Example 1).

Corollary 3

— Let $A$ be a field, and let $f$ be a non-constant polynomial in $A[X]$ . For $f$ and $Df$ to be relatively prime, it is necessary and sufficient that $\mathrm{dis}(f) \neq 0$ . This follows from Prop. 11, (iii) and Cor. 2 of IV, p. 78.

Remark

Adouble application of Cor. 2 above shows that we have $\mathrm{dis}_m(f) = 0$ for every polynomial $f$ of degree $\leq m - 2$ .

Example 3

Let $m = 2$ . By Example2 (IV, p. 81) we have $\Delta(A_1, A_2) = A_1^2 - 4A_2$ , whence $\tilde{\Delta}(A_0, A_1, A_2) = A_1^2 - 4A_0A_2$ . In other words, we have

\mathrm{dis}_2(a_0X^2 + a_1X + a_2) = a_1^2 - 4a_0a_2 .

Example 4

Consider the polynomial

F = A_0X^3 + 3A_1X^2 + 3A_2X + A,

with coefficients in

Q[A_0, A_1, A_2, A_3]

. We have

DF = 3(A_0X^2 + 2A_1X + A_2) ,

F - 1/3X \cdot DF = A_1X^2 + 2A_2X + A_3 .

By the formulae (54) (IV, p. 84) and (29) (IV, p. 77) we have

A_0 \cdot \mathrm{dis}_3(F) = - \mathrm{res}_{2,3}(DF, F) = - \mathrm{res}_{2,3}(DF, F - 1/3X \cdot DF)

Applying Cor. 2 of IV, p. 80 we finally obtain

\mathrm{dis}_3(F) = - 27 \mathrm{res}_{2,2}(A_0X^2 + 2A_1X + A_2,\ A_1X^2 + 2A_2X + A_3) .

By Example 3 of IV, p. 80 we thus have

\tilde{\Delta}(A_0, 3A_1, 3A_2, A_3) = - 27 (A_0A_3 - A_1A_2)^2 - 108 (A_1A_3 - A_2^2)(A_1^2 - A_0A_2) .

After some calculations we find, that if

f = a_0X^3 + a_1X^2 + a_2X + a_3

, we have

\mathrm{dis}_3(f) = a_1^2a_2^2 + 18a_0a_1a_2a_3 - 4a_1^3a_3 - 4a_0a_2^3 - 27a_0^2a_3^2 .

In particular, we have

\mathrm{dis}(X^3 + pX + q) = - (4p^3 + 27q^2) .

Exercises

See the exercises for § 6.