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_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}.

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 $1 + 1 \neq 0$ . 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]

R_{1}(y)=8y^{3}+8a_{2}y^{2}+(2{a_{2}}^{2}-8a_{0})y-{a_{1}}^{2}.

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

{\begin{aligned}P(x)=0&\Longleftrightarrow x^{4}+a_{2}x^{2}=-a_{1}x-a_{0}\\&\Longleftrightarrow \left(x^{2}+{\frac {a_{2}}{2}}\right)^{2}=-a_{1}x-a_{0}+{\frac {{a_{2}}^{2}}{4}}.\end{aligned}}

Add a new unknown, $y$ , to $x 2 + a 2 /2$ . Now you have:

{\begin{aligned}\left(x^{2}+{\frac {a_{2}}{2}}+y\right)^{2}&=-a_{1}x-a_{0}+{\frac {{a_{2}}^{2}}{4}}+2x^{2}y+a_{2}y+y^{2}\\&=2yx^{2}-a_{1}x-a_{0}+{\frac {{a_{2}}^{2}}{4}}+a_{2}y+y^{2}.\end{aligned}}

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

{\sqrt {2y}}\,x-{\frac {a_{1}}{2{\sqrt {2y}}}}.

But the equality

\left({\sqrt {2y}}\,x-{\frac {a_{1}}{2{\sqrt {2y}}}}\right)^{2}=2yx^{2}-a_{1}x-a_{0}+{\frac {{a_{2}}^{2}}{4}}+a_{2}y+y^{2}

is equivalent to

{\frac {{a_{1}}^{2}}{8y}}=-a_{0}+{\frac {{a_{2}}^{2}}{4}}+a_{2}y+y^{2}{\text{,}}

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

x^{2}-{\sqrt {2y_{0}}}\,x+{\frac {a_{2}}{2}}+y_{0}+{\frac {a_{1}}{2{\sqrt {2y_{0}}}}}

together with the roots of the polynomial

x^{2}+{\sqrt {2y_{0}}}\,x+{\frac {a_{2}}{2}}+y_{0}-{\frac {a_{1}}{2{\sqrt {2y_{0}}}}}.

Of course, this makes no sense if $y 0 = 0$ , but since the constant term of $R 1 (y)$ is $- a 12$ , $0$ 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

R_{2}(y)=8y^{3}-4a_{2}y^{2}-8a_{0}y+4a_{2}a_{0}-{a_{1}}^{2}

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

{\begin{aligned}P(x)=0&\Longleftrightarrow x^{4}=-a_{2}x^{2}-a_{1}x-a_{0}\\&\Longleftrightarrow (x^{2}+y)^{2}=-a_{2}x^{2}-a_{1}x-a_{0}+2yx^{2}+y^{2}\end{aligned}}

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

8y^{3}-4a_{2}y^{2}-8a_{0}y+4a_{2}a_{0}-{a_{1}}^{2}=0{\text{.}}

A simple computation shows that

R_{2}\left(y+{\frac {a_{2}}{2}}\right)=R_{1}(y).

Third definition

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

R_{3}(y)=y^{3}+2a_{2}y^{2}+({a_{2}}^{2}-4a_{0})y-{a_{1}}^{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 $P (x)$ by expressing it as a product of two monic quadratic polynomials $x 2 + αx + β$ and $x 2 - αx + γ$ , 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 $α \neq 0$ (note that if $a 1 \neq 0$ , 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 ={\frac {1}{2}}\left(a_{2}+\alpha ^{2}-{\frac {a_{1}}{\alpha }}\right)

and

\gamma ={\frac {1}{2}}\left(a_{2}+\alpha ^{2}+{\frac {a_{1}}{\alpha }}\right).

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

\left(a_{2}+\alpha ^{2}\right)^{2}-{\frac {{a_{1}}^{2}}{\alpha ^{2}}}=4a_{0}{\text{,}}

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

R_{3}(y)=R_{1}\left({\frac {y}{2}}\right){\text{.}}

Fourth definition

Still another possible definition is^[4]

R_{4}(y)=y^{3}-a_{2}y^{2}+(a_{1}a_{3}-4a_{0})y+4a_{0}a_{2}-{a_{1}}^{2}-a_{0}{a_{3}}^{2}.

In fact, if the roots of $P (x)$ are $α 1, α 2, α 3$ , and $α 4$ , then

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{,}}

a fact the follows from Vieta's formulas. In other words, R₄(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

\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{,}}

\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{,}}

\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{.}}

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

{\begin{aligned}R_{4}(y)&=y^{3}-a_{2}y^{2}-4a_{0}y+4a_{0}a_{2}-{a_{1}}^{2}\\&=R_{2}\left({\frac {y}{2}}\right){\text{.}}\end{aligned}}

Fifth definition

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

R_{5}(y)=y^{3}-2a_{2}y^{2}+({a_{2}}^{2}+a_{3}a_{1}-4a_{0})y+{a_{1}}^{2}-a_{3}a_{2}a_{1}+{a_{3}}^{2}a_{0}{\text{.}}

If, as above, the roots of $P (x)$ are $α 1, α 2, α 3$ , and $α 4$ , then

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{,}}

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

(\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{,}}

(\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{,}}

(\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{.}}

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

{\begin{aligned}R_{5}(y)&=y^{3}-2a_{2}y^{2}+({a_{2}}^{2}-4a_{0})y+{a_{1}}^{2}\\&=-R_{3}(-y)\\&=-R_{1}\left(-{\frac {y}{2}}\right){\text{.}}\end{aligned}}

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 $α \in k$ , 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 $R 3 (y)$ = $y 3 + 2 a 2 y 2 + (a 22 - 4 a 0) y - a 12$ . There are two cases:

If $a 1 \neq 0$ then $R 3 (0)$ = $- a 12 < 0$ . 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 $α$ = $\sqrt y 0$ .
If $a 1$ = $0$ , then $R 3 (y)$ = $y 3 + 2 a 2 y 2 + (a 22 - 4 a 0) y$ . The roots of this polynomial are $0$ and the roots of the quadratic polynomial $y 2 + 2 a 2 y + a 22 - 4 a 0$ . If $a 22 - 4 a 0 < 0$ , 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 $- a 2 + 2 \sqrt a 0$ ) and we can take $α$ as the square root of that root. Otherwise, $a 22 - 4 a 0 \geq 0$ and then,

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{.}}

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 $P (x) \in Q [x]$ 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 $a 22 - 4 a 0$ 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 $α \neq 0$ 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 $δ \in Q$ is such that $δ 2$ = $a 22 - 4 a 0$ , 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 $S 4$ .

References

^ ^a ^b Tignol, Jean-Pierre (2016), "Quartic equations", Galois' Theory of algebraic equations (2nd ed.), World Scientific, ISBN 978-981-4704-69-4, Zbl 1333.12001
^ ^a ^b Brookfield, G. (2007), "Factoring quartic polynomials: A lost art" (PDF), Mathematics Magazine, 80 (1): 67–70, doi:10.1080/0025570X.2007.11953453, JSTOR 27642994, S2CID 53375377, Zbl 1227.97040, archived from the original (PDF) on 2015-02-21
^ Hartshorne, Robin (1997), "Construction problems and field extensions: Cubic and quartic equations", Geometry: Euclid and Beyond, Springer-Verlag, ISBN 0-387-98650-2, Zbl 0954.51001
^ ^a ^b Kaplansky, Irving (1972), "Fields: Cubic and quartic equations", Fields and Rings, Chicago Lectures in Mathematics (2nd ed.), University of Chicago Press, ISBN 0-226-42451-0, Zbl 1001.16500
^ Rotman, Joseph (1998), "Galois groups of quadratics, cubics, and quartics", Galois Theory (2nd ed.), Springer-Verlag, ISBN 0-387-98541-7, Zbl 0924.12001
^ van der Waerden, Bartel Leendert (1991), "The Galois theory: Equations of the second, third, and fourth degrees", Algebra, vol. 1 (7th ed.), Springer-Verlag, ISBN 0-387-97424-5, Zbl 0724.12001

[Tignol-1] Tignol, Jean-Pierre (2016), "Quartic equations", Galois' Theory of algebraic equations (2nd ed.), World Scientific, ISBN 978-981-4704-69-4, Zbl 1333.12001

[Brookfield-2] Brookfield, G. (2007), "Factoring quartic polynomials: A lost art" (PDF), Mathematics Magazine, 80 (1): 67–70, doi:10.1080/0025570X.2007.11953453, JSTOR 27642994, S2CID 53375377, Zbl 1227.97040, archived from the original (PDF) on 2015-02-21

[3] Hartshorne, Robin (1997), "Construction problems and field extensions: Cubic and quartic equations", Geometry: Euclid and Beyond, Springer-Verlag, ISBN 0-387-98650-2, Zbl 0954.51001

[Kaplanski-4] Kaplansky, Irving (1972), "Fields: Cubic and quartic equations", Fields and Rings, Chicago Lectures in Mathematics (2nd ed.), University of Chicago Press, ISBN 0-226-42451-0, Zbl 1001.16500

[5] Rotman, Joseph (1998), "Galois groups of quadratics, cubics, and quartics", Galois Theory (2nd ed.), Springer-Verlag, ISBN 0-387-98541-7, Zbl 0924.12001

[6] van der Waerden, Bartel Leendert (1991), "The Galois theory: Equations of the second, third, and fourth degrees", Algebra, vol. 1 (7th ed.), Springer-Verlag, ISBN 0-387-97424-5, Zbl 0724.12001

[1]

[2]

[3]

[4]

[5]

[6]