Bourbaki

Proposition 1

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

Algebra, chapter IV, § 6. Symmetric functions

*For every pair of positive integers k, n, let Sk(n)S_k^{(n)} be the A-module consisting of all symmetric polynomials in X1,…,XnX_1, \ldots, X_n, which are homogeneous of degree k. If the integer m satisfies 0≤k≤m≤n0 \leq k \leq m \leq n, then the mapping f↦f(X1,…,Xm,0,…,0)f \mapsto f (X_1, \ldots, X_m, 0, \ldots, 0) is an isomorphism of Sk(n)S_k^{(n)} onto Sk(m)S_k^{(m)}.

By the commutativity of the diagram (10) it is enough to show that every isobaric polynomial of weight k in S1,…,SnS_1, \ldots, S_n depends only on S1,…,SmS_1, \ldots, S_m under the hypotheses 0≤k≤m≤n0 \leq k \leq m \leq n. Now the weight of a monomial S1α(1)⋯Snα(n)S_1^{\alpha(1)} \cdots S_n^{\alpha(n)} is equal to the integer α(1)+2α(2)+…+nα(n)\alpha(1) + 2\alpha(2) + \ldots + n\alpha(n); since the integers α(1),…,α(n)\alpha(1), \ldots, \alpha(n) are positive, the relation

α(1)+2α(2)+⋯+nα(n)=k≤n \alpha(1) + 2\alpha(2) + \cdots + n\alpha(n) = k \leq n
implies α(j)=0\alpha(j) = 0 for k<j≤nk < j \leq n, whence the assertion.

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

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Cites

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Transcribed from the printed edition, 2003, Springer. Not translated.