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)

from which we obtain by induction on $k$

(6)

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)

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

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

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

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

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

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

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

We have

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

hence finally

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

and finally

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

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

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

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

is commutative.