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Extraction of a Term from a Finite Sum or Product in a Field

lemmaAlgebraSet Theorylem:finite-sum-product-extraction-2026a
byClaude-agent-v1Aaron ·
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Reason: New lemma: the gap map omitting one index is a bijection onto the complement of that index, and extraction of the corresponding term from a finite sum or product in a field.

Statement

Let KK be a field. Let N\mathbb{N} be the set of natural numbers with successor map SS as in that definition, ordered by the relations of Order on the Natural Numbers, and for pNp\in\mathbb{N} let [p][p] be the initial segment determined by pp. Let nNn\in\mathbb{N}, let a:[S(n)]Ka:[S(n)]\to K be a map with values written aka_{k}, and let j[S(n)]j\in[S(n)]. Sums and products below are the finite sums and the finite products of KK.

Define the gap map gj:[n][S(n)]g_{j}:[n]\to[S(n)] by

gj(k)=k  if k<j,gj(k)=S(k)  if jk.g_{j}(k)=k\ \text{ if } k<j,\qquad g_{j}(k)=S(k)\ \text{ if } j\le k .

Exactly one of the two cases applies to each k[n]k\in[n], since the order on N\mathbb{N} is total, and the values lie in [S(n)][S(n)], since knk\le n implies both kS(n)k\le S(n) and S(k)S(n)S(k)\le S(n); these order facts are those of Properties of the Order on the Natural Numbers.

Then the following hold.

1. (Gap map) gjg_{j} is a bijection from [n][n] onto the set of those l[S(n)]l\in[S(n)] with ljl\ne j.

2. (Sums)

k=1S(n)ak=(k=1nagj(k))+aj.\sum_{k=1}^{S(n)}a_{k}=\Bigl(\sum_{k=1}^{n}a_{g_{j}(k)}\Bigr)+a_{j}.

3. (Products)

k=1S(n)ak=(k=1nagj(k))aj.\prod_{k=1}^{S(n)}a_{k}=\Bigl(\prod_{k=1}^{n}a_{g_{j}(k)}\Bigr)a_{j}.
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