# Resolvent cubic

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

- [math]\displaystyle{ P(x)=x^4+a_3x^3+a_2x^2+a_1x+a_0. }[/math]

In each case:

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

## Definitions

Suppose that the coefficients of *P*(*x*) belong to a field *k* whose characteristic is different from 2. In other words, we are working in a field in which 1 + 1 ≠ 0. Whenever roots of *P*(*x*) are mentioned, they belong to some extension *K* of *k* such that *P*(*x*) factors into linear factors in *K*[*x*]. If *k* is the field **Q** of rational numbers, then K can be the field **C** of complex numbers or the field **Q** of algebraic numbers.

In some cases, the concept of resolvent cubic is defined only when *P*(*x*) is a quartic in depressed form—that is, when *a*_{3} = 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.

### First definition

Suppose that *P*(*x*) is a depressed quartic—that is, that *a*_{3} = 0. A possible definition of the resolvent cubic of *P*(*x*) is:^{[1]}

- [math]\displaystyle{ R_1(y)=8y^3+8a_2y^2+(2{a_2}^2-8a_0)y-{a_1}^2. }[/math]

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

- [math]\displaystyle{ \begin{align}P(x)=0&\Longleftrightarrow x^4+a_2x^2=-a_1x-a_0\\ &\Longleftrightarrow \left(x^2+\frac{a_2}2\right)^2=-a_1x-a_0+\frac{{a_2}^2}4.\end{align} }[/math]

Add a new unknown, y, to *x*^{2} + *a*_{2}/2. Now you have:

- [math]\displaystyle{ \begin{align}\left(x^2+\frac{a_2}2+y\right)^2&=-a_1x-a_0+\frac{{a_2}^2}4+2x^2y+a_2y+y^2\\ &=2yx^2-a_1x-a_0+\frac{{a_2}^2}4+a_2y+y^2.\end{align} }[/math]

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

- [math]\displaystyle{ \sqrt{2y}\,x-\frac{a_1}{2\sqrt{2y}}. }[/math]

But the equality

- [math]\displaystyle{ \left(\sqrt{2y}\,x-\frac{a_1}{2\sqrt{2y}}\right)^2=2yx^2-a_1x-a_0+\frac{{a_2}^2}4+a_2y+y^2 }[/math]

is equivalent to

- [math]\displaystyle{ \frac{{a_1}^2}{8y}=-a_0+\frac{{a_2}^2}4+a_2y+y^2\text{,} }[/math]

and this is the same thing as the assertion that *R*_{1}(*y*) = 0.

If *y*_{0} is a root of *R*_{1}(*y*), then it is a consequence of the computations made above that the roots of *P*(*x*) are the roots of the polynomial

- [math]\displaystyle{ x^2-\sqrt{2y_0}\,x+\frac{a_2}2+y_0+\frac{a_1}{2\sqrt{2y_0}} }[/math]

together with the roots of the polynomial

- [math]\displaystyle{ x^2+\sqrt{2y_0}\,x+\frac{a_2}2+y_0-\frac{a_1}{2\sqrt{2y_0}}. }[/math]

Of course, this makes no sense if *y*_{0} = 0, but since the constant term of *R*_{1}(*y*) is –*a*_{1}^{2}, 0 is a root of *R*_{1}(*y*) if and only if *a*_{1} = 0, and in this case the roots of *P*(*x*) can be found using the quadratic formula.

### Second definition

Another possible definition^{[1]} (still supposing that *P*(*x*) is a depressed quartic) is

- [math]\displaystyle{ R_2(y)=8y^3-4a_2y^2-8a_0y+4a_2a_0-{a_1}^2 }[/math]

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

- [math]\displaystyle{ \begin{align}P(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{align} }[/math]

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

- [math]\displaystyle{ 8y^3-4a_2y^2-8a_0y+4a_2a_0-{a_1}^2=0\text{.} }[/math]

A simple computation shows that

- [math]\displaystyle{ R_2\left(y+\frac{a_2}2\right)=R_1(y). }[/math]

### Third definition

Another possible definition^{[2]}^{[3]} (again, supposing that *P*(*x*) is a depressed quartic) is

- [math]\displaystyle{ R_3(y)=y^3+2a_2y^2+({a_2}^2-4a_0)y-{a_1}^2\text{.} }[/math]

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 *x*^{2} + *αx* + *β* and *x*^{2} – *αx* + *γ*, then

- [math]\displaystyle{ 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. }[/math]

If there is a solution of this system with *α* ≠ 0 (note that if *a*_{1} ≠ 0, then this is automatically true for any solution), the previous system is equivalent to

- [math]\displaystyle{ \left\{\begin{array}{l}\beta+\gamma=a_2+\alpha^2\\-\beta+\gamma=\frac{a_1}{\alpha}\\ \beta\gamma=a_0.\end{array}\right. }[/math]

It is a consequence of the first two equations that then

- [math]\displaystyle{ \beta=\frac12\left(a_2+\alpha^2-\frac{a_1}{\alpha}\right) }[/math]

and

- [math]\displaystyle{ \gamma=\frac12\left(a_2+\alpha^2+\frac{a_1}{\alpha}\right). }[/math]

After replacing, in the third equation, *β* and *γ* by these values one gets that

- [math]\displaystyle{ \left(a_2+\alpha^2\right)^2-\frac{{a_1}^2}{\alpha^2}=4a_0\text{,} }[/math]

and this is equivalent to the assertion that *α*^{2} is a root of *R*_{3}(*y*). So, again, knowing the roots of *R*_{3}(*y*) helps to determine the roots of *P*(*x*).

Note that

- [math]\displaystyle{ R_3(y)=R_1\left(\frac y2\right)\text{.} }[/math]

### Fourth definition

Still another possible definition is^{[4]}

- [math]\displaystyle{ R_4(y)=y^3-a_2y^2+(a_1a_3-4a_0)y+4a_0a_2-{a_1}^2-a_0{a_3}^2. }[/math]

In fact, if the roots of *P*(*x*) are *α*_{1}, *α*_{2}, *α*_{3}, and *α*_{4}, then

- [math]\displaystyle{ R_4(y)=\bigl(y-(\alpha_1\alpha_2+\alpha_3\alpha_4)\bigr)\bigl(y-(\alpha_1\alpha_3+\alpha_2\alpha_4)\bigr)\bigl(y-(\alpha_1\alpha_4+\alpha_2\alpha_3)\bigr)\text{,} }[/math]

a fact the follows from Vieta's formulas. In other words, *R*_{4}(*y*) is the monic polynomial whose roots are
*α*_{1}*α*_{2} + *α*_{3}*α*_{4},
*α*_{1}*α*_{3} + *α*_{2}*α*_{4}, and
*α*_{1}*α*_{4} + *α*_{2}*α*_{3}.

It is easy to see that

- [math]\displaystyle{ \alpha_1\alpha_2+\alpha_3\alpha_4-(\alpha_1\alpha_3+\alpha_2\alpha_4)=(\alpha_1-\alpha_4)(\alpha_2-\alpha_3)\text{,} }[/math]
- [math]\displaystyle{ \alpha_1\alpha_3+\alpha_2\alpha_4-(\alpha_1\alpha_4+\alpha_2\alpha_3)=(\alpha_1-\alpha_2)(\alpha_3-\alpha_4)\text{,} }[/math]
- [math]\displaystyle{ \alpha_1\alpha_2+\alpha_3\alpha_4-(\alpha_1\alpha_4+\alpha_2\alpha_3)=(\alpha_1-\alpha_3)(\alpha_2-\alpha_4)\text{.} }[/math]

Therefore, *P*(*x*) has a multiple root if and only if *R*_{4}(*y*) has a multiple root. More precisely, *P*(*x*) and *R*_{4}(*y*) have the same discriminant.

One should note that if *P*(*x*) is a depressed polynomial, then

- [math]\displaystyle{ \begin{align}R_4(y)&=y^3-a_2y^2-4a_0y+4a_0a_2-{a_1}^2\\ &=R_2\left(\frac y2\right)\text{.}\end{align} }[/math]

### Fifth definition

Yet another definition is^{[5]}^{[6]}

- [math]\displaystyle{ R_5(y)=y^3-2a_2y^2+({a_2}^2+a_3a_1-4a_0)y+{a_1}^2-a_3a_2a_1+{a_3}^2a_0\text{.} }[/math]

If, as above, the roots of *P*(*x*) are *α*_{1}, *α*_{2}, *α*_{3}, and *α*_{4}, then

- [math]\displaystyle{ R_5(y)=\bigl(y-(\alpha_1+\alpha_2)(\alpha_3+\alpha_4)\bigr)\bigl(y-(\alpha_1+\alpha_3)(\alpha_2+\alpha_4)\bigr)\bigl(y-(\alpha_1+\alpha_4)(\alpha_2+\alpha_3)\bigr)\text{,} }[/math]

again as a consequence of Vieta's formulas. In other words, *R*_{5}(*y*) is the monic polynomial whose roots are
(*α*_{1} + *α*_{2})(*α*_{3} + *α*_{4}),
(*α*_{1} + *α*_{3})(*α*_{2} + *α*_{4}), and
(*α*_{1} + *α*_{4})(*α*_{2} + *α*_{3}).

It is easy to see that

- [math]\displaystyle{ (\alpha_1+\alpha_2)(\alpha_3+\alpha_4)-(\alpha_1+\alpha_3)(\alpha_2+\alpha_4)=-(\alpha_1-\alpha_4)(\alpha_2-\alpha_3)\text{,} }[/math]
- [math]\displaystyle{ (\alpha_1+\alpha_2)(\alpha_3+\alpha_4)-(\alpha_1+\alpha_4)(\alpha_2+\alpha_3)=-(\alpha_1-\alpha_3)(\alpha_2-\alpha_4)\text{,} }[/math]
- [math]\displaystyle{ (\alpha_1+\alpha_3)(\alpha_2+\alpha_4)-(\alpha_1+\alpha_4)(\alpha_2+\alpha_3)=-(\alpha_1-\alpha_2)(\alpha_3-\alpha_4)\text{.} }[/math]

Therefore, as it happens with *R*_{4}(*y*), *P*(*x*) has a multiple root if and only if *R*_{5}(*y*) has a multiple root. More precisely, *P*(*x*) and *R*_{5}(*y*) have the same discriminant. This is also a consequence of the fact that *R*_{5}(*y* + *a*_{2}) = -*R*_{4}(-*y*).

Note that if *P*(*x*) is a depressed polynomial, then

- [math]\displaystyle{ \begin{align}R_5(y)&=y^3-2a_2y^2+({a_2}^2-4a_0)y+{a_1}^2\\ &=-R_3(-y)\\ &=-R_1\left(-\frac y2\right)\text{.}\end{align} }[/math]

## Applications

### Solving quartic equations

It was explained above how *R*_{1}(*y*), *R*_{2}(*y*), and *R*_{3}(*y*) can be used to find the roots of *P*(*x*) if this polynomial is depressed. In the general case, one simply has to find the roots of the depressed polynomial *P*(*x* − *a*_{3}/4). For each root *x*_{0} of this polynomial, *x*_{0} − *a*_{3}/4 is a root of *P*(*x*).

### Factoring quartic polynomials

If a quartic polynomial *P*(*x*) is reducible in *k*[*x*], then it is the product of two quadratic polynomials or the product of a linear polynomial by a cubic polynomial. This second possibility occurs if and only if *P*(*x*) has a root in *k*. In order to determine whether or not *P*(*x*) can be expressed as the product of two quadratic polynomials, let us assume, for simplicity, that *P*(*x*) is a depressed polynomial. Then it was seen above that if the resolvent cubic *R*_{3}(*y*) has a non-null root of the form *α*^{2}, for some *α* ∈ *k*, then such a decomposition exists.

This can be used to prove that, in **R**[*x*], every quartic polynomial without real roots can be expressed as the product of two quadratic polynomials. Let *P*(*x*) be such a polynomial. We can assume without loss of generality that *P*(*x*) is monic. We can also assume without loss of generality that it is a reduced polynomial, because *P*(*x*) can be expressed as the product of two quadratic polynomials if and only if *P*(*x* − *a*_{3}/4) can and this polynomial is a reduced one. Then *R*_{3}(*y*) = *y*^{3} + 2*a*_{2}*y*^{2} + (*a*_{2}^{2} − 4*a*_{0})*y* − *a*_{1}^{2}. There are two cases:

- If
*a*_{1}≠ 0 then*R*_{3}(0) = −*a*_{1}^{2}< 0. Since*R*_{3}(*y*) > 0 if*y*is large enough, then, by the intermediate value theorem,*R*_{3}(*y*) has a root*y*_{0}with*y*_{0}> 0. So, we can take*α*= √*y*_{0}. - If
*a*_{1}= 0, then*R*_{3}(*y*) =*y*^{3}+ 2*a*_{2}*y*^{2}+ (*a*_{2}^{2}− 4*a*_{0})*y*. The roots of this polynomial are 0 and the roots of the quadratic polynomial*y*^{2}+ 2*a*_{2}*y*+*a*_{2}^{2}− 4*a*_{0}. If*a*_{2}^{2}− 4*a*_{0}< 0, then the product of the two roots of this polynomial is smaller than 0 and therefore it has a root greater than 0 (which happens to be −*a*_{2}+ 2√*a*_{0}) and we can take*α*as the square root of that root. Otherwise,*a*_{2}^{2}− 4*a*_{0}≥ 0 and then,

- [math]\displaystyle{ P(x)=\left(x^2+\frac{a_2+\sqrt{{a_2}^2-4a_0}}2\right)\left(x^2+\frac{a_2-\sqrt{{a_2}^2-4a_0}}2\right)\text{.} }[/math]

More generally, if *k* is a real closed field, then every quartic polynomial without roots in *k* can be expressed as the product of two quadratic polynomials in *k*[*x*]. Indeed, this statement can be expressed in first-order logic and any such statement that holds for **R** also holds for any real closed field.

A similar approach can be used to get an algorithm^{[2]} to determine whether or not a quartic polynomial *P*(*x*) ∈ **Q**[*x*] is reducible and, if it is, how to express it as a product of polynomials of smaller degree. Again, we will suppose that *P*(*x*) is monic and depressed. Then *P*(*x*) is reducible if and only if at least one of the following conditions holds:

- The polynomial
*P*(*x*) has a rational root (this can be determined using the rational root theorem). - The resolvent cubic
*R*_{3}(*y*) has a root of the form*α*^{2}, for some non-null rational number*α*(again, this can be determined using the rational root theorem). - The number
*a*_{2}^{2}− 4*a*_{0}is the square of a rational number and*a*_{1}= 0.

Indeed:

- If
*P*(*x*) has a rational root*r*, then*P*(*x*) is the product of*x*−*r*by a cubic polynomial in**Q**[*x*], which can be determined by polynomial long division or by Ruffini's rule. - If there is a rational number
*α*≠ 0 such that*α*^{2}is a root of*R*_{3}(*y*), it was shown above how to express*P*(*x*) as the product of two quadratic polynomials in**Q**[*x*]. - Finally, if the third condition holds and if
*δ*∈**Q**is such that*δ*^{2}=*a*_{2}^{2}− 4*a*_{0}, then*P*(*x*) = (*x*^{2}+ (*a*_{2}+*δ*)/2)(*x*^{2}+ (*a*_{2}−*δ*)/2).

### Galois groups of irreducible quartic polynomials

The resolvent cubic of an irreducible quartic polynomial *P*(*x*) can be used to determine its Galois group *G*; that is, the Galois group of the splitting field of *P*(*x*). Let m be the degree over k of the splitting field of the resolvent cubic (it can be either *R*_{4}(*y*) or *R*_{5}(*y*); they have the same splitting field). Then the group G is a subgroup of the symmetric group *S*_{4}. More precisely:^{[4]}

- If
*m*= 1 (that is, if the resolvent cubic factors into linear factors in k), then G is the group {*e*, (12)(34), (13)(24), (14)(23)}. - If
*m*= 2 (that is, if the resolvent cubic has one and, up to multiplicity, only one root in*k*), then, in order to determine G, one can determine whether or not*P*(*x*) is still irreducible after adjoining to the field k the roots of the resolvent cubic. If not, then G is a cyclic group of order 4; more precisely, it is one of the three cyclic subgroups of*S*_{4}generated by any of its six 4-cycles. If it is still irreducible, then G is one of the three subgroups of*S*_{4}of order 8, each of which is isomorphic to the dihedral group of order 8. - If
*m*= 3, then G is the alternating group*A*_{4}. - If
*m*= 6, then G is the whole group*S*_{4}.

## See also

## References

- ↑
^{1.0}^{1.1}Tignol, Jean-Pierre (2016), "Quartic equations",*Galois' Theory of algebraic equations*(2nd ed.), World Scientific, ISBN 978-981-4704-69-4 - ↑
^{2.0}^{2.1}Brookfield, G. (2007), "Factoring quartic polynomials: A lost art",*Mathematics Magazine***80**(1): 67–70, doi:10.1080/0025570X.2007.11953453, http://web.calstatela.edu/faculty/gbrookf/pubs/quartic.pdf - ↑ Hartshorne, Robin (1997),
*Geometry: Euclid and Beyond*, Springer-Verlag, ISBN 0-387-98650-2 - ↑
^{4.0}^{4.1}Kaplansky, Irving (1972),*Fields and Rings*, Chicago Lectures in Mathematics (2nd ed.), University of Chicago Press, ISBN 0-226-42451-0 - ↑ Rotman, Joseph (1998), "Galois groups of quadratics, cubics, and quartics",
*Galois Theory*(2nd ed.), Springer-Verlag, ISBN 0-387-98541-7 - ↑ van der Waerden, Bartel Leendert (1991),
*Algebra*,**1**(7th ed.), Springer-Verlag, ISBN 0-387-97424-5

Original source: https://en.wikipedia.org/wiki/Resolvent cubic.
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