Corollary 1

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

Algebra, chapter IV, § 6. Symmetric functions

— 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).

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

Cited by

Corollary 2, § 6

Cites

Nothing.