Resolvent Cubic

In algebra, a resolvent cubic is one of several distinct, although related, cubic polynomials defined from a monic polynomial of degree four:

:$P(x)=x^4+a_3x^3+a_2x^2+a_1x+a_0.$

In each case:
* The coefficients of the resolvent cubic can be obtained from the coefficients of using only sums, subtractions and multiplications.
* Knowing the roots of the resolvent cubic of is useful for finding the roots of itself. Hence the name “resolvent cubic”.
* The polynomial has a multiple root if and only if its resolvent cubic has a multiple root.

Definitions

Suppose that the coefficients of belong to a field whose characteristic is different from . In other words, we are working in a field in which . Whenever roots of are mentioned, they belong to some extension of such that factors into linear factors in . If is the field of rational numbers, then can be the field of complex numbers or the field of algebraic numbers.

In some cases, the concept of resolvent cubic is defined only when is a quartic in depressed form—that is, when .

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

First definition

Suppose that is a depressed quartic—that is, that . A possible definition of the resolvent cubic of is:

:$R_1(y)=8y^3+8a_2y^2+(2^2-8a_0)y-^2.$

The origin of this definition lies in applying Ferrari's method to find the roots of . To be more precise:

:$\beginP(x)=0&\Longleftrightarrow x^4+a_2x^2=-a_1x-a_0\\ &\Longleftrightarrow \left(x^2+\frac2\right)^2=-a_1x-a_0+\frac4.\end$

Add a new unknown, , to . Now you have:

:$\begin\left(x^2+\frac2+y\right)^2&=-a_1x-a_0+\frac4+2x^2y+a_2y+y^2\\ &=2yx^2-a_1x-a_0+\frac4+a_2y+y^2.\end$

If this expression is a square, it can only be the square of

:$\sqrt\,x-\frac.$

But the equality

:$\left(\sqrt\,x-\frac\right)^2=2yx^2-a_1x-a_0+\frac4+a_2y+y^2$

is equivalent to

:$\frac=-a_0+\frac4+a_2y+y^2\text$

and this is the same thing as the assertion that  = 0.

If is a root of , then it is a consequence of the computations made above that the roots of are the roots of the polynomial

:$x^2-\sqrt\,x+\frac2+y_0+\frac$

together with the roots of the polynomial

:$x^2+\sqrt\,x+\frac2+y_0-\frac.$

Of course, this makes no sense if , but since the constant term of is , is a root of if and only if , and in this case the roots of can be found using the quadratic formula.

Second definition

Another possible definition (still supposing that is a depressed quartic) is

:$R_2(y)=8y^3-4a_2y^2-8a_0y+4a_2a_0-^2$

The origin of this definition is similar to the previous one. This time, we start by doing:

:$\beginP(x)=0&\Longleftrightarrow x^4=-a_2x^2-a_1x-a_0\\ &\Longleftrightarrow(x^2+y)^2=-a_2x^2-a_1x-a_0+2yx^2+y^2\end$

and a computation similar to the previous one shows that this last expression is a square if and only if

:$8y^3-4a_2y^2-8a_0y+4a_2a_0-^2=0\text$

A simple computation shows that

:$R_2\left(y+\frac2\right)=R_1(y).$

Third definition

Another possible definition (again, supposing that is a depressed quartic) is

:$R_3(y)=y^3+2a_2y^2+(^2-4a_0)y-^2\text$

The origin of this definition lies in another method of solving quartic equations, namely Descartes' method. If you try to find the roots of by expressing it as a product of two monic quadratic polynomials and , then
:$P(x)=(x^2+\alpha x+\beta)(x^2-\alpha x+\gamma)\Longleftrightarrow\left\{\begin{array}{l}\beta+\gamma-\alpha^2=a_2\\ \alpha(-\beta+\gamma)=a_1\\ \beta\gamma=a_0.\end{array}\right.$

If there is a solution of this system with (note that if , then this is automatically true for any solution), the previous system is equivalent to
:$\left\{\begin{array}{l}\beta+\gamma=a_2+\alpha^2\\-\beta+\gamma=\frac{a_1}{\alpha}\\ \beta\gamma=a_0.\end{array}\right.$

It is a consequence of the first two equations that then
:$\beta=\frac12\left(a_2+\alpha^2-\frac{a_1}{\alpha}\right)$
and
:$\gamma=\frac12\left(a_2+\alpha^2+\frac{a_1}{\alpha}\right).$
After replacing, in the third equation, and by these values one gets that
:\left(a_2+\alpha^2\right)^2-\frac.
* If (that is, if the resolvent cubic has one and, up to multiplicity, only one root in ), then, in order to determine , one can determine whether or not is still irreducible after adjoining to the field the roots of the resolvent cubic. If not, then is a cyclic group of order 4; more precisely, it is one of the three cyclic subgroups of  generated by any of its six -cycles. If it is still irreducible, then is one of the three subgroups of  of order , each of which is isomorphic to the dihedral group of order .
* If , then is the alternating group .
* If , then is the whole group .

See also

*Resolvent (Galois theory)

References

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Algebra
Equations
Polynomials