CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Vieta's formulas identify the coefficients of a split monic polynomial with elementary symmetric functions of its roots

Statement

Let $f (t) = t^{n} + a_{1} t^{n - 1} + \dots + a_{n} \in R [t]$ be monic and suppose that in a commutative $R$ -algebra $S$ it splits as

$f (t) = \prod_{i = 1}^{n} (t - α_{i}),$

with roots repeated according to multiplicity. Then

$a_{k} = (- 1)^{k} e_{k} (α_{1}, \dots, α_{n}) (1 \leq k \leq n) .$

The assertion includes the monic constant polynomial, for which there are no coefficient equations.

Facts & Assumptions

Given: A split monic polynomial $f$ as in the Statement.

[L1]

The universal Vieta expansion is $\prod_{i = 1}^{n} (t - x_{i}) = \sum_{k = 0}^{n} (- 1)^{k} e_{k} t^{n - k}$ (Vieta expansion: $\prod_{i = 1}^{n} (t - x_{i}) = \sum_{k = 0}^{n} (- 1)^{k} e_{k} t^{n - k}$ ).

[L2]

A polynomial splits over an extension when it is a product of linear factors there, with roots listed with multiplicity (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

Proof

technique · direct

1.1

Substitute $x_{i} = α_{i}$ in [L1] inside $S [t]$ to obtain $f (t) = \sum_{k = 0}^{n} (- 1)^{k} e_{k} (α_{1}, \dots, α_{n}) t^{n - k}$ .

givenL1L2

2.1

Equality of polynomials is coefficientwise, so comparison with $f (t) = t^{n} + a_{1} t^{n - 1} + \dots + a_{n}$ gives the displayed formula for every $k$ . If $n = 0$ , the comparison has no positive index.

step 1.1algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 26 results over 6 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 1 (standard reference, not scraped)