LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-30

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Over an infinite field, a triangular change makes a nonzero polynomial monic

Statement

Let $k$ be an infinite field, let $n \geq 1$ , and let $f \in k [x_{1}, \dots, x_{n}]$ be nonzero. Then there exist $a_{1}, \dots, a_{n - 1} \in k$ and $c \in k^{\times}$ such that $g (x_{1}, \dots, x_{n}) = c f (x_{1} + a_{1} x_{n}, \dots, x_{n - 1} + a_{n - 1} x_{n}, x_{n})$ is monic as a polynomial in $x_{n}$ with coefficients in $k [x_{1}, \dots, x_{n - 1}]$ .

Facts & Assumptions

Given: An infinite field $k$ , an integer $n \geq 1$ , and a nonzero polynomial $f \in k [x_{1}, \dots, x_{n}]$ .

[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

1.1givenalgebra

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

2.1L1step 1.1choose

The polynomial $H (X_{1}, \dots, X_{n - 1}, 1)$ in $k [X_{1}, \dots, X_{n - 1}]$ is nonzero, so [L1] yields $a_{1}, \dots, a_{n - 1} \in k$ with $H (a_{1}, \dots, a_{n - 1}, 1) \neq 0$ .

3.1step 2.1algebra

Substitute $x_{i} + a_{i} x_{n}$ for $x_{i}$ when $i < n$ . Every degree- $d$ monomial of $f$ contributes to the coefficient of $x_{n}^{d}$ , and the lower-degree part of $f$ contributes only lower powers of $x_{n}$ . Therefore the coefficient of $x_{n}^{d}$ in the transformed polynomial is exactly $H (a_{1}, \dots, a_{n - 1}, 1) \in k^{\times}$ .

4.1step 3.1algebra∎

Multiplying by the inverse scalar $c = H (a_{1}, \dots, a_{n - 1}, 1)^{- 1}$ makes the transformed polynomial monic in $x_{n}$ .

Depends on

A polynomial vanishing at every tuple from an infinite subdomain is the zero polynomial

Over an infinite field, a triangular change makes a nonzero polynomial monic

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources