LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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The leading multidegree of a symmetric polynomial is weakly decreasing

Statement

If $0 \neq f \in R [x_{1}, \dots, x_{n}]$ is symmetric and has leading multidegree $(a_{1}, \dots, a_{n})$ in lexicographic order, then

$a_{1} \geq a_{2} \geq \dots \geq a_{n} .$

Facts & Assumptions

Given: A nonzero symmetric polynomial $f$ with leading multidegree $a = (a_{1}, \dots, a_{n})$ .

[L1]

Every permutation of the variables fixes a symmetric polynomial (Symmetric polynomials as the invariants of variable permutations).

[L2]

The leading multidegree is the lexicographically greatest exponent tuple carrying a nonzero coefficient (Lexicographic order on exponent tuples and the multidegree of a nonzero polynomial).

Proof

technique · contradiction

1.1

Suppose, for contradiction, that $a_{i} < a_{i + 1}$ for some $i < n$ .

assume-contra

1.2

Interchanging $x_{i}$ and $x_{i + 1}$ fixes $f$ , so the tuple $a^{'} = (a_{1}, \dots, a_{i - 1}, a_{i + 1}, a_{i}, a_{i + 2}, \dots, a_{n})$ occurs in $f$ with the same nonzero coefficient as $a$ .

givenL1

2.1

The tuples agree before coordinate $i$ and $a_{i}^{'} = a_{i + 1} > a_{i}$ , so $a^{'} >_{l e x} a$ , contradicting the maximality in [L2]. Therefore no such $i$ exists and the tuple is weakly decreasing.

step 1.1step 1.2L2discharge-contradiction∎

Depends on

Used by

Every symmetric polynomial over a commutative ring is a polynomial in the elementary symmetric polynomials Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 11 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

K. Conrad, Symmetric Polynomials, Section 2 (standard reference, not scraped)