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Newton's identities through $p_{4}$ in three variables

Example

For three variables, Newton's identities give

$p_{1} = e_{1},$

$p_{2} = e_{1}^{2} - 2 e_{2},$

$p_{3} = e_{1}^{3} - 3 e_{1} e_{2} + 3 e_{3},$

$p_{4} = e_{1}^{4} - 4 e_{1}^{2} e_{2} + 2 e_{2}^{2} + 4 e_{1} e_{3} .$

If $6$ is invertible in the coefficient ring, the first three equations can be solved recursively for $e_{1}, e_{2}, e_{3}$ .

Facts & Assumptions

Given: Three variables over a commutative ring.

[L1]

Newton's identities are $k e_{k} = \sum_{i = 1}^{k} (- 1)^{i - 1} e_{k - i} p_{i}$ , with $e_{0} = 1$ and $e_{k} = 0$ for $k > 3$ (Newton's identities: $k e_{k} = \sum_{i = 1}^{k} (- 1)^{i - 1} e_{k - i} p_{i}$ ).

[L2]

If $3!$ is invertible, then $p_{1}, p_{2}, p_{3}$ freely generate the symmetric-polynomial ring (If $n!$ is invertible, then $p_{1}, \dots, p_{n}$ freely generate the symmetric-polynomial ring).

Verification

technique · direct

1.1

At $k = 1$ , [L1] gives $e_{1} = p_{1}$ . At $k = 2$ , it gives $2 e_{2} = e_{1} p_{1} - p_{2}$ , hence $p_{2} = e_{1}^{2} - 2 e_{2}$ .

L1algebra

2.1

At $k = 3$ , [L1] gives $3 e_{3} = e_{2} p_{1} - e_{1} p_{2} + p_{3}$ ; substituting step 1.1 yields $p_{3} = e_{1}^{3} - 3 e_{1} e_{2} + 3 e_{3}$ .

step 1.1L1algebra

3.1

At $k = 4$ , $e_{4} = 0$ , so [L1] gives $0 = e_{3} p_{1} - e_{2} p_{2} + e_{1} p_{3} - p_{4}$ . Substitution from steps 1.1 and 2.1 gives the displayed formula for $p_{4}$ .

step 1.1step 2.1L1algebra

4.1

When $6$ is invertible, so are $1, 2, 3$ , and the first three Newton identities recursively solve for the $e_{i}$ , as asserted by [L2].

L2algebra∎

Depends on

Used by

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Sources

K. Conrad, Symmetric Polynomials, Section 3 (standard reference, not scraped)

Newton's identities through p4p4​ in three variables