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For monic $f$ of degree $n$ , $Res (f, f^{'}) = (- 1)^{n (n - 1) / 2} Disc (f)$

Statement

Let $F$ be a field and let $f \in F [t]$ be monic of degree $n$ . Then

$Res (f, f^{'}) = (- 1)^{n (n - 1) / 2} Disc (f) .$

Facts & Assumptions

Given: A monic polynomial $f \in F [t]$ of degree $n$ , split as $f (t) = \prod_{i = 1}^{n} (t - α_{i})$ in a splitting field.

[L1]

The root-product formula gives $Res (f, f^{'}) = \prod_{i} f^{'} (α_{i})$ (For monic $f$ , $Res (f, g) = \prod_{i} g (α_{i})$ and it vanishes exactly when $f$ and $g$ have a common root).

[L2]

The discriminant root formula is $Disc (f) = \prod_{i < j} (α_{i} - α_{j})^{2}$ (The discriminant is $\prod_{i < j} (α_{i} - α_{j})^{2}$ and vanishes exactly when a monic polynomial has a repeated root).

[L3]

The formal derivative of $f = \sum_{i} a_{i} x^{i}$ is $f^{'} = \sum_{i \geq 1} i a_{i} x^{i - 1}$ (The formal derivative of a polynomial).

[L4]

Over a commutative ring, formal differentiation is additive and $F$ -linear and satisfies $(f g)^{'} = f^{'} g + f g^{'}$ (Linearity, power rule, Leibniz rule and the degree bound for the formal derivative).

Proof

technique · direct

1.1

Iterating the Leibniz rule of [L4] over the $n$ factors of $f = \prod_{i = 1}^{n} (t - α_{i})$ gives $f^{'} = \sum_{i = 1}^{n} \prod_{j \neq i} (t - α_{j})$ , since each $(t - α_{i})^{'} = 1$ by [L3]. At $t = α_{i}$ , every summand except the $i$ -th contains the zero factor $α_{i} - α_{i}$ , so $f^{'} (α_{i}) = \prod_{j \neq i} (α_{i} - α_{j})$ .

givenL3L4algebra

2.1

By [L1], $Res (f, f^{'}) = \prod_{i} \prod_{j \neq i} (α_{i} - α_{j})$ . Group the two ordered factors belonging to each unordered pair $i < j$ .

step 1.1L1algebra

3.1

For each $i < j$ , $(α_{i} - α_{j}) (α_{j} - α_{i}) = - (α_{i} - α_{j})^{2}$ . There are $n (n - 1) / 2$ unordered pairs, so step 2.1 becomes $(- 1)^{n (n - 1) / 2} \prod_{i < j} (α_{i} - α_{j})^{2}$ .

step 2.1algebra

4.1

Apply [L2] to identify the remaining product with $Disc (f)$ . The empty-degree cases $n = 0, 1$ give $1 = 1$ .

step 3.1L2∎

Depends on

Used by

The discriminant of x³+px+q is -4p³-27q² Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 31 results over 10 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

J. S. Milne, Fields and Galois Theory, Proposition 4.35 and discriminant discussion (standard reference, not scraped)

For monic ff of degree nn, Res⁡(f,f′)=(−1)n(n−1)/2Disc⁡(f)Res(f,f′)=(−1)n(n−1)/2Disc(f)