Resolvent cubic

In algebra, a resolvent cubic is one of several distinct, although related, cubic polynomials defined from a monic polynomial of degree four: In each case: Suppose that the coefficients of P(x) belong to a field k whose characteristic is different from 2.

In some cases, the concept of resolvent cubic is defined only when P(x) is a quartic in depressed form—that is, when a3 = 0.

Note that the fourth and fifth definitions below also make sense and that the relationship between these resolvent cubics and P(x) are still valid if the characteristic of k is equal to 2.

This time, we start by doing: and a computation similar to the previous one shows that this last expression is a square if and only if A simple computation shows that Another possible definition[2][3] (again, supposing that P(x) is a depressed quartic) is The origin of this definition lies in another method of solving quartic equations, namely Descartes' method.

If you try to find the roots of P(x) by expressing it as a product of two monic quadratic polynomials x2 + αx + β and x2 – αx + γ, then If there is a solution of this system with α ≠ 0 (note that if a1 ≠ 0, then this is automatically true for any solution), the previous system is equivalent to It is a consequence of the first two equations that then and After replacing, in the third equation, β and γ by these values one gets that and this is equivalent to the assertion that α2 is a root of R3(y).

One should note that if P(x) is a depressed polynomial, then Yet another definition is[5][6] If, as above, the roots of P(x) are α1, α2, α3, and α4, then again as a consequence of Vieta's formulas.

In the general case, one simply has to find the roots of the depressed polynomial P(x − a3/4).

Then it was seen above that if the resolvent cubic R3(y) has a non-null root of the form α2, for some α ∈ k, then such a decomposition exists.