Proposition 2

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

Lie Groups and Lie Algebras, chapter V, § 5. Invariants in the symmetric algebra

Retain the preceding notation and assume that S and R are graded polynomial algebras.

(i) R and S have the same transcendence degree r over K. (ii) Let $p_1, \ldots, p_r$ (resp. $q_1, \ldots, q_r$ ) be the characteristic degrees of S (resp. R). Then

\prod_{i=1}^r (1 - T^{q_i}) = \left( \sum_{j=1}^N T^{f_j} \right) \prod_{i=1}^r (1 - T^{p_i}).

(iii) $N p_1 p_2 \ldots p_r = q_1 q_2 \ldots q_r$ .

Formula (4) shows first of all that the multiplicity of the root 1 is the same in the polynomials $P_S(T)^{-1}$ and $P_R(T)^{-1}$ and, taking (3) into account, proves both (i) and (ii).

It follows from (ii) that

\prod_{i=1}^r (1 + T + T^2 + \cdots + T^{q_i-1}) = \left( \sum_{j=1}^N T^{f_j} \right) \prod_{i=1}^r (1 + T + T^2 + \cdots + T^{p_i-1}).

Putting $T = 1$ in this relation gives (iii).

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

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