Resolvent cubic

Suppose that P(x) is a depressed quartic—that is, that a₃ = 0. A possible definition of the resolvent cubic of P(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 applying Ferrari's method to find the roots of P(x). To be more precise:

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

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

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

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

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

But the equality

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

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 that R₁(y) = 0.

If y₀ is a root of R₁(y), then it is a consequence of the computations made above that the roots of P(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 if y₀ = 0, but since the constant term of R₁(y) is –a₁², 0 is a root of R₁(y) if and only if a₁ = 0, and in this case the roots of P(x) can be found using the quadratic formula.

Another possible definition[1] (still supposing that P(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:

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

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

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

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 that P(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, namely Descartes' method. If you try to find the roots of P(x) by expressing it as a product of two monic quadratic polynomials x² + αx + β and x² – αx + γ, then

${\displaystyle P(x)=(x^{2}+\alpha x+\beta )(x^{2}-\alpha x+\gamma )\Longleftrightarrow \left\{{\begin{array}{l}\beta +\gamma -\alpha ^{2}=a_{2}\\\alpha (-\beta +\gamma )=a_{1}\\\beta \gamma =a_{0}.\end{array}}\right.}$

If there is a solution of this system with α ≠ 0 (note that if a₁ ≠ 0, then this is automatically true for any solution), the previous system is equivalent to

${\displaystyle \left\{{\begin{array}{l}\beta +\gamma =a_{2}+\alpha ^{2}\\-\beta +\gamma ={\frac {a_{1}}{\alpha }}\\\beta \gamma =a_{0}.\end{array}}\right.}$

It is a consequence of the first two equations that then

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

and

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

After replacing, in the third equation, β and γ 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 that α² is a root of R₃(y). So, again, knowing the roots of R₃(y) helps to determine the roots of P(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 of P(x) are α₁, α₂, α₃, and α₄, then

${\displaystyle R_{4}(y)={\bigl (}y-(\alpha _{1}\alpha _{2}+\alpha _{3}\alpha _{4}){\bigr )}{\bigl (}y-(\alpha _{1}\alpha _{3}+\alpha _{2}\alpha _{4}){\bigr )}{\bigl (}y-(\alpha _{1}\alpha _{4}+\alpha _{2}\alpha _{3}){\bigr )}{\text{,}}}$

a fact the follows from Vieta's formulas. In other words, R₄(y) is the monic polynomial whose roots are α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, and α₁α₄ + 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 a multiple root if and only if R₄(y) has a multiple root. More precisely, P(x) and R₄(y) have the same discriminant.

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

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

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 of P(x) are α₁, α₂, α₃, and α₄, then

${\displaystyle R_{5}(y)={\bigl (}y-(\alpha _{1}+\alpha _{2})(\alpha _{3}+\alpha _{4}){\bigr )}{\bigl (}y-(\alpha _{1}+\alpha _{3})(\alpha _{2}+\alpha _{4}){\bigr )}{\bigl (}y-(\alpha _{1}+\alpha _{4})(\alpha _{2}+\alpha _{3}){\bigr )}{\text{,}}}$

again as a consequence of Vieta'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 with R₄(y), P(x) has a multiple root if and only if R₅(y) has a multiple root. More precisely, P(x) and R₅(y) have the same discriminant. This is also a consequence of the fact that R₅(y + a₂) = -R₄(-y).

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

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

It was explained above how R₁(y), R₂(y), and R₃(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 − a₃/4). For each root x₀ of this polynomial, x₀ − a₃/4 is a root of P(x).

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 R₃(y) has a non-null root of the form α², 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 − a₃/4) can and this polynomial is a reduced one. Then R₃(y) = y³ + 2a₂y² + (a₂² − 4a₀)y − a₁². There are two cases:

• If a₁ ≠ 0 then R₃(0) = −a₁² < 0. Since R₃(y) > 0 if y is large enough, then, by the intermediate value theorem, R₃(y) has a root y₀ with y₀ > 0. So, we can take α = √y₀.
• If a₁ = 0, then R₃(y) = y³ + 2a₂y² + (a₂² − 4a₀)y. The roots of this polynomial are 0 and the roots of the quadratic polynomial y² + 2a₂y + a₂² − 4a₀. If a₂² − 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 take α 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, 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 R₃(y) has a root of the form α², for some non-null rational number α (again, this can be determined using the rational root theorem).
• The number a₂² − 4a₀ is the square of a rational number and a₁ = 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 α² is a root of R₃(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 δ² = a₂² − 4a₀, then P(x) = (x² + (a₂ + δ)/2)(x² + (a₂ − δ)/2).

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 R₄(y) or R₅(y); they have the same splitting field). Then the group G is a subgroup of the symmetric group S₄. 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 S₄ generated by any of its six 4-cycles. If it is still irreducible, then G is one of the three subgroups of S₄ of order 8, each of which is isomorphic to the dihedral group of order 8.
• If m = 3, then G is the alternating group A₄.
• If m = 6, then G is the whole group S₄.