Examples

Tag 0DHL, permanent. Cite it as [Bourbaki, Tag 0DHL].

Algebra, chapter IV, § 6. Symmetric functions

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.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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