Resolvent cubic

Graph of the polynomial function Template:Math (in green) together with the graph of its resolvent cubic Template:Math (in red). The roots of both polynomials are visible too.

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 Template:Math using only sums, subtractions and multiplications.
• Knowing the roots of the resolvent cubic of Template:Math is useful for finding the roots of Template:Math itself. Hence the name “resolvent cubic”.
• The polynomial Template:Math has a multiple root if and only if its resolvent cubic has a multiple root.

Definitions

Suppose that the coefficients of Template:Math belong to a field Template:Math whose characteristic is different from Template:Math. Whenever roots of Template:Math are mentioned, they belong to some extension Template:Math of Template:Math such that Template:Math factors into linear factors in Template:Math. If Template:Math is the field Template:Math of rational numbers, then Template:Mvar can be the field Template:Math of complex numbers or the field Template:Overline of algebraic numbers.

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

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

First definition

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

$R_1(y)=8y^3+8a_2{y}^2+(2{a_2}^2-8a_0)y-{a_1}^2.$

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

$\begin{array}{cc}P(x)=0& ⟺x^4+a_2{x}^2=-a_{1}x-a_0\\ & ⟺{(x^2+\frac{a_2}{2})}^2=-a_{1}x-a_0+\frac{{a_2}^2}{4}.\end{array}$

Add a new unknown, Template:Mvar, to Template:Math. Now you have:

$\begin{array}{cc}{(x^2+\frac{a_2}{2}+y)}^2& =-a_{1}x-a_0+\frac{{a_2}^2}{4}+2x^{2}y+a_{2}y+y^2\\ & =2y{x}^2-a_{1}x-a_0+\frac{{a_2}^2}{4}+a_{2}y+y^2.\end{array}$

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

${(\sqrt{2y}x-\frac{a_1}{2\sqrt{2y}})}^2=2y{x}^2-a_{1}x-a_0+\frac{{a_2}^2}{4}+a_{2}y+y^2$

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 Template:Math = 0.

If Template:Math is a root of Template:Math, then it is a consequence of the computations made above that the roots of Template:Math 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 Template:Math, but since the constant term of Template:Math is Template:Math, Template:Math is a root of Template:Math if and only if Template:Math, and in this case the roots of Template:Math can be found using the quadratic formula.

Second definition

Another possible definition (still supposing that Template:Math 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:

$\begin{array}{cc}P(x)=0& ⟺x^4=-a_2{x}^2-a_{1}x-a_0\\ & ⟺(x^2+y{)}^2=-a_2{x}^2-a_{1}x-a_0+2y{x}^2+y^2\end{array}$

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

$R_2(y+\frac{a_2}{2})=R_1(y).$

Third definition

Another possible definition (again, supposing that Template:Math 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 Template:Math by expressing it as a product of two monic quadratic polynomials Template:Math and Template:Math, then

$P(x)=(x^2+\alpha x+\beta )(x^2-\alpha x+\gamma )⟺\{\begin{array}{c}\beta +\gamma -{\alpha }^2=a_2\\ \alpha (-\beta +\gamma )=a_1\\ \beta \gamma =a_0.\end{array}$

If there is a solution of this system with Template:Math (note that if Template:Math, then this is automatically true for any solution), the previous system is equivalent to

$\{\begin{array}{c}\beta +\gamma =a_2+{\alpha }^2\\ -\beta +\gamma =\frac{a_1}{\alpha }\\ \beta \gamma =a_0.\end{array}$

It is a consequence of the first two equations that then

$\beta =\frac{1}{2}(a_2+{\alpha }^2-\frac{a_1}{\alpha })$

and

$\gamma =\frac{1}{2}(a_2+{\alpha }^2+\frac{a_1}{\alpha }).$

After replacing, in the third equation, Template:Math and Template:Math by these values one gets that

${(a_2+{\alpha }^2)}^2-\frac{{a_1}^2}{{\alpha }^2}=4a_0\text{,}$

and this is equivalent to the assertion that Template:Math is a root of Template:Math. So, again, knowing the roots of Template:Math helps to determine the roots of Template:Math.

Note that

$R_3(y)=R_1\left(\frac{y}{2}\right)\text{.}$

Fourth definition

Still another possible definition is

$R_4(y)=y^3-a_2{y}^2+(a_1{a}_3-4a_0)y+4a_0{a}_2-{a_1}^2-a_0{a_3}^2.$

In fact, if the roots of Template:Math are Template:Math, and Template:Math, then

$R_4(y)=(y-({\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4))(y-({\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4))(y-({\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3))\text{,}$

a fact the follows from Vieta's formulas. In other words, R₄(y) is the monic polynomial whose roots are Template:Math, Template:Math, and Template:Math.

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, Template:Math has a multiple root if and only if Template:Math has a multiple root. More precisely, Template:Math and Template:Math have the same discriminant.

If Template:Math is a depressed polynomial, then

$\begin{array}{cc}{R}_4(y)& =y^3-a_2{y}^2-4a_{0}y+4a_0{a}_2-{a_1}^2\\ & =R_2\left(\frac{y}{2}\right)\text{.}\end{array}$

Fifth definition

Yet another definition is

$R_5(y)=y^3-2a_2{y}^2+({a_2}^2+a_3{a}_1-4a_0)y+{a_1}^2-a_3{a}_2{a}_1+{a_3}^2{a}_0\text{.}$

If, as above, the roots of Template:Math are Template:Math, and Template:Math, then

$R_5(y)=(y-({\alpha }_1+{\alpha }_2)({\alpha }_3+{\alpha }_4))(y-({\alpha }_1+{\alpha }_3)({\alpha }_2+{\alpha }_4))(y-({\alpha }_1+{\alpha }_4)({\alpha }_2+{\alpha }_3))\text{,}$

again as a consequence of Vieta's formulas. In other words, Template:Math is the monic polynomial whose roots are Template:Math, Template:Math, and Template:Math.

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 Template:Math, Template:Math has a multiple root if and only if Template:Math has a multiple root. More precisely, Template:Math and Template:Math have the same discriminant. This is also a consequence of the fact that Template:Math = Template:Math.

Note that if Template:Math is a depressed polynomial, then

$\begin{array}{cc}{R}_5(y)& =y^3-2a_2{y}^2+({a_2}^2-4a_0)y+{a_1}^2\\ & =-R_3(-y)\\ & =-R_1(-\frac{y}{2})\text{.}\end{array}$

Applications

Solving quartic equations

It was explained above how Template:Math, Template:Math, and Template:Math can be used to find the roots of Template:Math if this polynomial is depressed. In the general case, one simply has to find the roots of the depressed polynomial Template:Math. For each root Template:Math of this polynomial, Template:Math is a root of Template:Math.

Factoring quartic polynomials

If a quartic polynomial Template:Math is reducible in Template:Math, 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 Template:Math has a root in Template:Math. In order to determine whether or not Template:Math can be expressed as the product of two quadratic polynomials, let us assume, for simplicity, that Template:Math is a depressed polynomial. Then it was seen above that if the resolvent cubic Template:Math has a non-null root of the form Template:Math, for some Template:Math, then such a decomposition exists.

This can be used to prove that, in Template:Math, every quartic polynomial without real roots can be expressed as the product of two quadratic polynomials. Let Template:Math be such a polynomial. We can assume without loss of generality that Template:Math is monic. We can also assume without loss of generality that it is a reduced polynomial, because Template:Math can be expressed as the product of two quadratic polynomials if and only if Template:Math can and this polynomial is a reduced one. Then Template:Math = Template:Math. There are two cases:

• If Template:Math then Template:Math = Template:Math. Since Template:Math if Template:Math is large enough, then, by the intermediate value theorem, Template:Math has a root Template:Math with Template:Math. So, we can take Template:Math = Template:Math.
• If Template:Math = Template:Math, then Template:Math = Template:Math. The roots of this polynomial are Template:Math and the roots of the quadratic polynomial Template:Math. If Template:Math, then the product of the two roots of this polynomial is smaller than Template:Math and therefore it has a root greater than Template:Math (which happens to be Template:Math) and we can take Template:Math as the square root of that root. Otherwise, Template:Math and then,

$P(x)=(x^2+\frac{a_2+\sqrt{{a_2}^2-4a_0}}{2})(x^2+\frac{a_2-\sqrt{{a_2}^2-4a_0}}{2})\text{.}$

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

A similar approach can be used to get an algorithm to determine whether or not a quartic polynomial Template:Math is reducible and, if it is, how to express it as a product of polynomials of smaller degree. Again, we will suppose that Template:Math is monic and depressed. Then Template:Math is reducible if and only if at least one of the following conditions holds:

• The polynomial Template:Math has a rational root (this can be determined using the rational root theorem).
• The resolvent cubic Template:Math has a root of the form Template:Math, for some non-null rational number Template:Math (again, this can be determined using the rational root theorem).
• The number Template:Math is the square of a rational number and Template:Math = Template:Math.

Indeed:

• If Template:Math has a rational root Template:Math, then Template:Math is the product of Template:Math by a cubic polynomial in Template:Math, which can be determined by polynomial long division or by Ruffini's rule.
• If there is a rational number Template:Math such that Template:Math is a root of Template:Math, it was shown above how to express Template:Math as the product of two quadratic polynomials in Template:Math.
• Finally, if the third condition holds and if Template:Math is such that Template:Math=Template:Math, then Template:Math = Template:Math.

Galois groups of irreducible quartic polynomials

The resolvent cubic of an irreducible quartic polynomial Template:Math can be used to determine its Galois group Template:Math; that is, the Galois group of the splitting field of Template:Math. Let Template:Mvar be the degree over Template:Mvar of the splitting field of the resolvent cubic (it can be either Template:Math or Template:Math; they have the same splitting field). Then the group Template:Mvar is a subgroup of the symmetric group Template:Math. More precisely:

• If Template:Math (that is, if the resolvent cubic factors into linear factors in Template:Mvar), then Template:Mvar is the group Template:Math}.
• If Template:Math (that is, if the resolvent cubic has one and, up to multiplicity, only one root in Template:Math), then, in order to determine Template:Mvar, one can determine whether or not Template:Math is still irreducible after adjoining to the field Template:Mvar the roots of the resolvent cubic. If not, then Template:Mvar is a cyclic group of order 4; more precisely, it is one of the three cyclic subgroups of Template:Math generated by any of its six Template:Math-cycles. If it is still irreducible, then Template:Mvar is one of the three subgroups of Template:Math of order Template:Math, each of which is isomorphic to the dihedral group of order Template:Math.
• If Template:Math, then Template:Mvar is the alternating group Template:Math.
• If Template:Math, then Template:Mvar is the whole group Template:Math.

See also

• Resolvent (Galois theory)