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 $P (x)$ 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 depressed polynomial, 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 polynomial, that is, that a₃ = 0. A possible definition of the resolvent cubic of P(x) is:^[1]

R_1(y)=8y^3+8a_2y^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{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}

Add a new unknown, y, to x² + a₂/2. Now you have:

\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}

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_1x-a_0+\frac{{a_2}^2}4+a_2y+y^2

is equivalent to

\frac{{a_1}^2}{8y}=-a_0+\frac{{a_2}^2}4+a_2y+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 polynomial) is

R_2(y)=8y^3-4a_2y^2-8a_0y+4a_2a_0-{a_1}^2

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

\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}

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-{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 polynomial) is

R_3(y)=y^3+2a_2y^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=\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, $β$ 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 y2\right)\text{.}

Fourth definition

Graph of the polynomial function

x 4 + x 3 - x 2 - 7 x /4 - 1/2

(in green) together with the graph of its resolvent cubic

R 4 (y)

(in red). The roots of both polynomials are visible too.

Still another possible definition is^[4]

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.

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{,}

and

\alpha_1\alpha_2+\alpha_3\alpha_4-(\alpha_1\alpha_4+\alpha_4\alpha_1)=(\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{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}

Fifth definition

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

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{.}

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{,}

and

(\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{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}

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 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 in $P (x) \in Q [x]$ 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

1 2 Tignol, Jean-Pierre (2016), "Quartic equations", Galois' Theory of algebraic equations (2nd ed.), World Scientific, ISBN 978-981-4704-69-4, Zbl 06534509
1 2 Brookfield, G. (2007), "Factoring quartic polynomials: A lost art" (PDF), Mathematics Magazine 80 (1): 67–70, JSTOR 27642994, Zbl 1227.97040
↑ 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
1 2 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 1 (7th ed.), Springer-Verlag, ISBN 0-387-97424-5, Zbl 00048827

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