Lemma 4 (« Newton's relations »)

Tag 06EP, permanent. Cite it as [Bourbaki, Tag 06EP].

Algebra, chapter IV, § 6. Symmetric functions

— 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.

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

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Lemma 4 (« Newton's relations »)

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