Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-30
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Over an infinite field, a triangular change makes a nonzero polynomial monic

Statement

Let k be an infinite field, let n≥1, and let f∈k[x1,…,xn] be nonzero. Then there exist a1,…,an−1∈k and c∈k× such that g(x1,…,xn)=c f(x1+a1xn,…,xn−1+an−1xn,xn) is monic as a polynomial in xn with coefficients in k[x1,…,xn−1].

Facts & Assumptions

Given: An infinite field k, an integer n≥1, and a nonzero polynomial f∈k[x1,…,xn].

[L1]

A nonzero polynomial over an integral domain does not vanish on every tuple from an infinite subring (A polynomial vanishing at every tuple from an infinite subdomain is the zero polynomial).

Proof

technique · direct
1.1

Let d be the total degree of f, and let H be the homogeneous degree d part of f. Then H is nonzero.

givenalgebra
2.1

The polynomial H(X1,…,Xn−1,1) in k[X1,…,Xn−1] is nonzero, so [L1] yields a1,…,an−1∈k with H(a1,…,an−1,1)≠0.

L1step 1.1choose
3.1

Substitute xi+aixn for xi when i<n. Every degree-d monomial of f contributes to the coefficient of xnd, and the lower-degree part of f contributes only lower powers of xn. Therefore the coefficient of xnd in the transformed polynomial is exactly H(a1,…,an−1,1)∈k×.

step 2.1algebra
4.1

Multiplying by the inverse scalar c=H(a1,…,an−1,1)−1 makes the transformed polynomial monic in xn.

step 3.1algebra∎

Depends on

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Sources