Inalgebra, aresolvent cubic is one of several distinct, although related,cubic polynomials defined from amonicpolynomial of degree four:

${\displaystyle P(x)=x^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0.}$

In each case:

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

Suppose that the coefficients ofP(x) belong to afieldk whosecharacteristic is different from 2. In other words, we are working in a field in which1 + 1 ≠ 0. Whenever roots ofP(x) are mentioned, they belong to someextensionK ofk such thatP(x) factors into linear factors inK[x]. Ifk is the fieldQ of rational numbers, thenK can be the fieldC of complex numbers or the fieldQ ofalgebraic numbers.

In some cases, the concept of resolvent cubic is defined only whenP(x) is a quartic in depressed form—that is, whena₃ = 0.

Note that thefourth andfifth definitions below also make sense and that the relationship between these resolvent cubics andP(x) are still valid if the characteristic ofk is equal to 2.

Suppose thatP(x) is a depressed quartic—that is, thata₃ = 0. A possible definition of the resolvent cubic ofP(x) is:^[1]

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

The origin of this definition lies in applyingFerrari's method to find the roots ofP(x). To be more precise:

Add a new unknown,y, tox² + a₂/2. Now you have:

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

${\displaystyle \sqrt{2y}x-\frac{a_1}{2\sqrt{2y}}.}$

But the equality

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

is equivalent to

${\displaystyle \frac{{a_1}^2}{8y}=-a_0+\frac{{a_2}^2}{4}+a_{2}y+y^2\text{,}}$

and this is the same thing as the assertion thatR₁(y) = 0.

Ify₀ is a root ofR₁(y), then it is a consequence of the computations made above that the roots ofP(x) are the roots of the polynomial

${\displaystyle x^2-\sqrt{2y_0}x+\frac{a_2}{2}+y_0+\frac{a_1}{2\sqrt{2y_0}}}$

together with the roots of the polynomial

${\displaystyle x^2+\sqrt{2y_0}x+\frac{a_2}{2}+y_0-\frac{a_1}{2\sqrt{2y_0}}.}$

Of course, this makes no sense ify₀ = 0, but since theconstant term ofR₁(y) is–a₁²,0 is a root ofR₁(y) if and only ifa₁ = 0, and in this case the roots ofP(x) can be found using thequadratic formula.

Another possible definition^[1] (still supposing thatP(x) is a depressed quartic) is

${\displaystyle R_2(y)=8y^3-4a_2{y}^2-8a_{0}y+4a_2{a}_0-{a_1}^2}$

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

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

${\displaystyle 8y^3-4a_2{y}^2-8a_{0}y+4a_2{a}_0-{a_1}^2=0\text{.}}$

A simple computation shows that

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

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

${\displaystyle R_3(y)=y^3+2a_2{y}^2+({a_2}^2-4a_0)y-{a_1}^2\text{.}}$

The origin of this definition lies in another method of solving quartic equations, namelyDescartes' method. If you try to find the roots ofP(x) by expressing it as a product of two monic quadratic polynomialsx² + αx + β andx² – αx + γ, then

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

If there is a solution of this system withα ≠ 0 (note that ifa₁ ≠ 0, then this is automatically true for any solution), the previous system is equivalent to

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

It is a consequence of the first two equations that then

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

and

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

After replacing, in the third equation,β andγ by these values one gets that

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

and this is equivalent to the assertion thatα² is a root ofR₃(y). So, again, knowing the roots ofR₃(y) helps to determine the roots ofP(x).

Note that

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

Still another possible definition is^[4]

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

In fact, if the roots ofP(x) areα₁,α₂,α₃, andα₄, then

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

a fact the follows fromVieta's formulas. In other words,R₄(y) is the monic polynomial whose roots areα₁α₂ +α₃α₄,α₁α₃ +α₂α₄, andα₁α₄ +α₂α₃.

It is easy to see that

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

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

${\displaystyle {\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4-({\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3)=({\alpha }_1-{\alpha }_3)({\alpha }_2-{\alpha }_4)\text{.}}$

Therefore,P(x) has amultiple root if and only ifR₄(y) has a multiple root. More precisely,P(x) andR₄(y) have the samediscriminant.

One should note that ifP(x) is a depressed polynomial, then

Yet another definition is^[5][6]

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

If, as above, the roots ofP(x) areα₁,α₂,α₃, andα₄, then

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

again as a consequence ofVieta's formulas. In other words,R₅(y) is the monic polynomial whose roots are(α₁ +α₂)(α₃ +α₄),(α₁ +α₃)(α₂ +α₄), and(α₁ +α₄)(α₂ +α₃).

It is easy to see that

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

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

${\displaystyle ({\alpha }_1+{\alpha }_3)({\alpha }_2+{\alpha }_4)-({\alpha }_1+{\alpha }_4)({\alpha }_2+{\alpha }_3)=-({\alpha }_1-{\alpha }_2)({\alpha }_3-{\alpha }_4)\text{.}}$

Therefore, as it happens withR₄(y),P(x) has a multiple root if and only ifR₅(y) has a multiple root. More precisely,P(x) andR₅(y) have the same discriminant. This is also a consequence of the fact thatR₅(y + a₂) = -R₄(-y).

Note that ifP(x) is a depressed polynomial, then

It was explained above howR₁(y),R₂(y), andR₃(y) can be used to find the roots ofP(x) if this polynomial is depressed. In the general case, one simply has to find the roots of the depressed polynomialP(x − a₃/4). For each root x₀ of this polynomial,x₀ − a₃/4 is a root of P(x).

If a quartic polynomialP(x) isreducible ink[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 ifP(x) has a root in k. In order to determine whether or notP(x) can be expressed as the product of two quadratic polynomials, let us assume, for simplicity, thatP(x) is a depressed polynomial. Then it was seenabove that if the resolvent cubicR₃(y) has a non-null root of the formα², for someα ∈ k, then such a decomposition exists.

This can be used to prove that, inR[x], every quartic polynomial without real roots can be expressed as the product of two quadratic polynomials. LetP(x) be such a polynomial. We can assumewithout loss of generality thatP(x) is monic. We can also assume without loss of generality that it is a reduced polynomial, becauseP(x) can be expressed as the product of two quadratic polynomials if and only ifP(x − a₃/4) can and this polynomial is a reduced one. ThenR₃(y) = y³ + 2a₂y² + (a₂² − 4a₀)y − a₁². There are two cases:

• Ifa₁ ≠ 0 thenR₃(0) = −a₁² < 0. SinceR₃(y) > 0 ify is large enough, then, by theintermediate value theorem,R₃(y) has a rooty₀ withy₀ > 0. So, we can takeα = √y₀.
• Ifa₁ = 0, thenR₃(y) = y³ + 2a₂y² + (a₂² − 4a₀)y. The roots of this polynomial are 0 and the roots of the quadratic polynomial y² + 2a₂y + a₂² − 4a₀. Ifa₂² − 4a₀ < 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−a₂ + 2√a₀) and we can takeα as the square root of that root. Otherwise,a₂² − 4a₀ ≥ 0 and then,

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

More generally, ifk is areal closed field, then every quartic polynomial without roots ink can be expressed as the product of two quadratic polynomials ink[x]. Indeed, this statement can be expressed infirst-order logic and any such statement that holds forR 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 polynomialP(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 polynomialP(x) has a rational root (this can be determined using therational root theorem).
• The resolvent cubic R₃(y) has a root of the formα², for some non-null rational number α (again, this can be determined using therational root theorem).
• The numbera₂² − 4a₀ is the square of arational number anda₁ = 0.

Indeed:

• IfP(x) has a rational rootr, thenP(x) is the product ofx − r by a cubic polynomial inQ[x], which can be determined bypolynomial long division or byRuffini's rule.
• If there is a rational number α ≠ 0 such thatα² is a root of R₃(y), it was shownabove how to express P(x) as the product of two quadratic polynomials inQ[x].
• Finally, if the third condition holds and ifδ ∈ Q is such thatδ² = a₂² − 4a₀, thenP(x) = (x² + (a₂ + δ)/2)(x² + (a₂ − δ)/2).

The resolvent cubic of anirreducible quartic polynomialP(x) can be used to determine itsGalois groupG; that is, the Galois group of thesplitting field ofP(x). Let m be thedegree overk of the splitting field of the resolvent cubic (it can be eitherR₄(y) orR₅(y); they have the same splitting field). Then the group G is a subgroup of thesymmetric groupS₄. More precisely:^[4]

• Ifm = 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)}.
• Ifm = 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 notP(x) is still irreducible after adjoining to the fieldk the roots of the resolvent cubic. If not, thenG is acyclic group oforder 4; more precisely, it is one of the three cyclic subgroups of S₄ generated by any of its six4-cycles. If it is still irreducible, thenG is one of the three subgroups of S₄ of order 8, each of which is isomorphic to thedihedral group of order 8.
• Ifm = 3, thenG is thealternating groupA₄.
• Ifm = 6, thenG is the whole groupS₄.

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