resolvent cubic

Suppose that the coefficients of {{math|P(x)}} belong to a field {{math|k}} whose characteristic is different from {{math|2}}. In other words, we are working in a field in which {{math|1 + 1 ≠ 0}}. Whenever roots of {{math|P(x)}} are mentioned, they belong to some extension {{math|K}} of {{math|k}} such that {{math|P(x)}} factors into linear factors in {{math|K[x]}}. If {{math|k}} is the field {{math|Q}} of rational numbers, then {{mvar|K}} can be the field {{math|C}} of complex numbers or the field {{overline|{{math|Q}}}} of algebraic numbers.

In some cases, the concept of resolvent cubic is defined only when {{math|P(x)}} is a quartic in depressed form—that is, when {{math|a₃ {{=}} 0}}.

Note that the fourth and fifth definitions below also make sense and that the relationship between these resolvent cubics and {{math|P(x)}} are still valid if the characteristic of {{mvar|k}} is equal to {{math|2}}.

Suppose that {{math|P(x)}} is a depressed quartic—that is, that {{math|a₃ {{=}} 0}}. A possible definition of the resolvent cubic of {{math|P(x)}} is:{{Citation|last = Tignol|first = Jean-Pierre| author-link=Jean-Pierre Tignol |title=Galois' Theory of algebraic equations|edition = 2nd|publisher = World Scientific|year = 2016|chapter = Quartic equations|isbn = 978-981-4704-69-4|zbl = 1333.12001}}

:$R\_1(y)=8y^3+8a\_2y^2+(2\{a\_2\}^2-8a\_0)y-\{a\_1\}^2.$

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

:

\left(x^2+\frac{a_2}2\right)^2=-a_1x-a_0+\frac{{a_2}^2}4.\end{align}

Add a new unknown, {{mvar|y}}, to {{math|x² + a₂/2}}. Now you have:

:

&=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

:$\backslash sqrt\{2y\}\backslash ,x-\backslash frac\{a\_1\}\{2\backslash sqrt\{2y\}\}.$

But the equality

:$\backslash left(\backslash sqrt\{2y\}\backslash ,x-\backslash frac\{a\_1\}\{2\backslash sqrt\{2y\}\}\backslash right)^2=2yx^2-a\_1x-a\_0+\backslash frac\{\{a\_2\}^2\}4+a\_2y+y^2$

is equivalent to

:$\backslash frac\{\{a\_1\}^2\}\{8y\}=-a\_0+\backslash frac\{\{a\_2\}^2\}4+a\_2y+y^2\backslash text\{,\}$

and this is the same thing as the assertion that {{math|R₁(y)}} = 0.

If {{math|y₀}} is a root of {{math|R₁(y)}}, then it is a consequence of the computations made above that the roots of {{math|P(x)}} are the roots of the polynomial

:$x^2-\backslash sqrt\{2y\_0\}\backslash ,x+\backslash frac\{a\_2\}2+y\_0+\backslash frac\{a\_1\}\{2\backslash sqrt\{2y\_0\}\}$

together with the roots of the polynomial

:$x^2+\backslash sqrt\{2y\_0\}\backslash ,x+\backslash frac\{a\_2\}2+y\_0-\backslash frac\{a\_1\}\{2\backslash sqrt\{2y\_0\}\}.$

Of course, this makes no sense if {{math|y₀ {{=}} 0}}, but since the constant term of {{math|R₁(y)}} is {{math|–a₁²}}, {{math|0}} is a root of {{math|R₁(y)}} if and only if {{math|a₁ {{=}} 0}}, and in this case the roots of {{math|P(x)}} can be found using the quadratic formula.

Another possible definition (still supposing that {{math|P(x)}} is a depressed quartic) 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:

:

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

:$8y^3-4a\_2y^2-8a\_0y+4a\_2a\_0-\{a\_1\}^2=0\backslash text\{.\}$

A simple computation shows that

:$R\_2\backslash left(y+\backslash frac\{a\_2\}2\backslash right)=R\_1(y).$

Another possible definition{{Citation | author=Brookfield, G. | title=Factoring quartic polynomials: A lost art | journal=Mathematics Magazine | volume=80 | issue=1 | year=2007 | pages=67–70 | doi=10.1080/0025570X.2007.11953453 | jstor=27642994 | url=http://web.calstatela.edu/faculty/gbrookf/pubs/quartic.pdf | zbl=1227.97040 | s2cid=53375377 | url-status=dead | archive-url=https://web.archive.org/web/20150221005838/http://web.calstatela.edu/faculty/gbrookf/pubs/quartic.pdf | archive-date=2015-02-21 }}{{Citation|last = Hartshorne|first = Robin|author-link = Robin Hartshorne|title = Geometry: Euclid and Beyond|year = 1997|publisher=Springer-Verlag|section = Construction problems and field extensions: Cubic and quartic equations|isbn = 0-387-98650-2|zbl = 0954.51001}} (again, supposing that {{math|P(x)}} is a depressed quartic) is

:$R\_3(y)=y^3+2a\_2y^2+(\{a\_2\}^2-4a\_0)y-\{a\_1\}^2\backslash text\{.\}$

The origin of this definition lies in another method of solving quartic equations, namely Descartes' method. If you try to find the roots of {{math|P(x)}} by expressing it as a product of two monic quadratic polynomials {{math|x² + αx + β}} and {{math|x² – αx + γ}}, then

:$P(x)=(x^2+\backslash alpha\; x+\backslash beta)(x^2-\backslash alpha\; x+\backslash gamma)\backslash Longleftrightarrow\backslash left\backslash \{\backslash begin\{array\}\{l\}\backslash beta+\backslash gamma-\backslash alpha^2=a\_2\backslash \backslash \; \backslash alpha(-\backslash beta+\backslash gamma)=a\_1\backslash \backslash \; \backslash beta\backslash gamma=a\_0.\backslash end\{array\}\backslash right.$

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

:$\backslash left\backslash \{\backslash begin\{array\}\{l\}\backslash beta+\backslash gamma=a\_2+\backslash alpha^2\backslash \backslash -\backslash beta+\backslash gamma=\backslash frac\{a\_1\}\{\backslash alpha\}\backslash \backslash \; \backslash beta\backslash gamma=a\_0.\backslash end\{array\}\backslash right.$

It is a consequence of the first two equations that then

:$\backslash beta=\backslash frac12\backslash left(a\_2+\backslash alpha^2-\backslash frac\{a\_1\}\{\backslash alpha\}\backslash right)$

and

:$\backslash gamma=\backslash frac12\backslash left(a\_2+\backslash alpha^2+\backslash frac\{a\_1\}\{\backslash alpha\}\backslash right).$

After replacing, in the third equation, {{math|β}} and {{math|γ}} by these values one gets that

:$\backslash left(a\_2+\backslash alpha^2\backslash right)^2-\backslash frac\{\{a\_1\}^2\}\{\backslash alpha^2\}=4a\_0\backslash text\{,\}$

and this is equivalent to the assertion that {{math|α²}} is a root of {{math|R₃(y)}}. So, again, knowing the roots of {{math|R₃(y)}} helps to determine the roots of {{math|P(x)}}.

Note that

:$R\_3(y)=R\_1\backslash left(\backslash frac\; y2\backslash right)\backslash text\{.\}$

Still another possible definition is{{Citation | last=Kaplansky | first=Irving | author-link=Irving Kaplansky | title = Fields and Rings | edition=2nd | zbl=1001.16500 | series=Chicago Lectures in Mathematics | publisher = University of Chicago Press | year = 1972 | isbn = 0-226-42451-0 | section = Fields: Cubic and quartic equations }}

:$R\_4(y)=y^3-a\_2y^2+(a\_1a\_3-4a\_0)y+4a\_0a\_2-\{a\_1\}^2-a\_0\{a\_3\}^2.$

In fact, if the roots of {{math|P(x)}} are {{math|α₁, α₂, α₃}}, and {{math|α₄}}, then

:$R\_4(y)=\backslash bigl(y-(\backslash alpha\_1\backslash alpha\_2+\backslash alpha\_3\backslash alpha\_4)\backslash bigr)\backslash bigl(y-(\backslash alpha\_1\backslash alpha\_3+\backslash alpha\_2\backslash alpha\_4)\backslash bigr)\backslash bigl(y-(\backslash alpha\_1\backslash alpha\_4+\backslash alpha\_2\backslash alpha\_3)\backslash bigr)\backslash text\{,\}$

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

{{math|α₁α₂ + α₃α₄}},

{{math|α₁α₃ + α₂α₄}}, and

{{math|α₁α₄ + α₂α₃}}.

It is easy to see that

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

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

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

Therefore, {{math|P(x)}} has a multiple root if and only if {{math|R₄(y)}} has a multiple root. More precisely, {{math|P(x)}} and {{math|R₄(y)}} have the same discriminant.

One should note that if {{math|P(x)}} is a depressed polynomial, then

:

Yet another definition is{{Citation|last = Rotman|first=Joseph|title = Galois Theory|publisher=Springer-Verlag|edition = 2nd|isbn = 0-387-98541-7|year = 1998|chapter = Galois groups of quadratics, cubics, and quartics|zbl = 0924.12001}}{{Citation|last = van der Waerden|first=Bartel Leendert|author-link = Bartel Leendert van der Waerden|title = Algebra|volume = 1|publisher=Springer-Verlag|edition = 7th|isbn = 0-387-97424-5|year = 1991|section = The Galois theory: Equations of the second, third, and fourth degrees|zbl = 0724.12001}}

:$R\_5(y)=y^3-2a\_2y^2+(\{a\_2\}^2+a\_3a\_1-4a\_0)y+\{a\_1\}^2-a\_3a\_2a\_1+\{a\_3\}^2a\_0\backslash text\{.\}$

If, as above, the roots of {{math|P(x)}} are {{math|α₁, α₂, α₃}}, and {{math|α₄}}, then

:$R\_5(y)=\backslash bigl(y-(\backslash alpha\_1+\backslash alpha\_2)(\backslash alpha\_3+\backslash alpha\_4)\backslash bigr)\backslash bigl(y-(\backslash alpha\_1+\backslash alpha\_3)(\backslash alpha\_2+\backslash alpha\_4)\backslash bigr)\backslash bigl(y-(\backslash alpha\_1+\backslash alpha\_4)(\backslash alpha\_2+\backslash alpha\_3)\backslash bigr)\backslash text\{,\}$

again as a consequence of Vieta's formulas. In other words, {{math|R₅(y)}} is the monic polynomial whose roots are

{{math|(α₁ + α₂)(α₃ + α₄)}},

{{math|(α₁ + α₃)(α₂ + α₄)}}, and

{{math|(α₁ + α₄)(α₂ + α₃)}}.

It is easy to see that

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

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

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

Therefore, as it happens with {{math|R₄(y)}}, {{math|P(x)}} has a multiple root if and only if {{math|R₅(y)}} has a multiple root. More precisely, {{math|P(x)}} and {{math|R₅(y)}} have the same discriminant. This is also a consequence of the fact that {{math|R₅(y + a₂)}} = {{math|-R₄(-y)}}.

Note that if {{math|P(x)}} is a depressed polynomial, then

:

It was explained above how {{math|R₁(y)}}, {{math|R₂(y)}}, and {{math|R₃(y)}} can be used to find the roots of {{math|P(x)}} if this polynomial is depressed. In the general case, one simply has to find the roots of the depressed polynomial {{math|P(x − a₃/4)}}. For each root {{math|x₀}} of this polynomial, {{math|x₀ − a₃/4}} is a root of {{math|P(x)}}.

If a quartic polynomial {{math|P(x)}} is reducible in {{math|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 {{math|P(x)}} has a root in {{math|k}}. In order to determine whether or not {{math|P(x)}} can be expressed as the product of two quadratic polynomials, let us assume, for simplicity, that {{math|P(x)}} is a depressed polynomial. Then it was seen above that if the resolvent cubic {{math|R₃(y)}} has a non-null root of the form {{math|α²}}, for some {{math|α ∈ k}}, then such a decomposition exists.

This can be used to prove that, in {{math|R[x]}}, every quartic polynomial without real roots can be expressed as the product of two quadratic polynomials. Let {{math|P(x)}} be such a polynomial. We can assume without loss of generality that {{math|P(x)}} is monic. We can also assume without loss of generality that it is a reduced polynomial, because {{math|P(x)}} can be expressed as the product of two quadratic polynomials if and only if {{math|P(x − a₃/4)}} can and this polynomial is a reduced one. Then {{math|R₃(y)}} = {{math|y³ + 2a₂y² + (a₂² − 4a₀)y − a₁²}}. There are two cases:

• If {{math|a₁ ≠ 0}} then {{math|R₃(0)}} = {{math|−a₁² < 0}}. Since {{math|R₃(y) > 0}} if {{math|y}} is large enough, then, by the intermediate value theorem, {{math|R₃(y)}} has a root {{math|y₀}} with {{math|y₀ > 0}}. So, we can take {{math|α}} = {{math|{{sqrt|y₀}}}}.
• If {{math|a₁}} = {{math|0}}, then {{math|R₃(y)}} = {{math|y³ + 2a₂y² + (a₂² − 4a₀)y}}. The roots of this polynomial are {{math|0}} and the roots of the quadratic polynomial {{math|y² + 2a₂y + a₂² − 4a₀}}. If {{math|a₂² − 4a₀ < 0}}, then the product of the two roots of this polynomial is smaller than {{math|0}} and therefore it has a root greater than {{math|0}} (which happens to be {{math|−a₂ + 2{{sqrt|a₀}}}}) and we can take {{math|α}} as the square root of that root. Otherwise, {{math|a₂² − 4a₀ ≥ 0}} and then,

::$P(x)=\backslash left(x^2+\backslash frac\{a\_2+\backslash sqrt\{\{a\_2\}^2-4a\_0\}\}2\backslash right)\backslash left(x^2+\backslash frac\{a\_2-\backslash sqrt\{\{a\_2\}^2-4a\_0\}\}2\backslash right)\backslash text\{.\}$

More generally, if {{math|k}} is a real closed field, then every quartic polynomial without roots in {{math|k}} can be expressed as the product of two quadratic polynomials in {{math|k[x]}}. Indeed, this statement can be expressed in first-order logic and any such statement that holds for {{math|R}} 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 {{math|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 {{math|P(x)}} is monic and depressed. Then {{math|P(x)}} is reducible if and only if at least one of the following conditions holds:

• The polynomial {{math|P(x)}} has a rational root (this can be determined using the rational root theorem).
• The resolvent cubic {{math|R₃(y)}} has a root of the form {{math|α²}}, for some non-null rational number {{math|α}} (again, this can be determined using the rational root theorem).
• The number {{math|a₂² − 4a₀}} is the square of a rational number and {{math|a₁}} = {{math|0}}.

Indeed:

• If {{math|P(x)}} has a rational root {{math|r}}, then {{math|P(x)}} is the product of {{math|x − r}} by a cubic polynomial in {{math|Q[x]}}, which can be determined by polynomial long division or by Ruffini's rule.
• If there is a rational number {{math|α ≠ 0}} such that {{math|α²}} is a root of {{math|R₃(y)}}, it was shown above how to express {{math|P(x)}} as the product of two quadratic polynomials in {{math|Q[x]}}.
• Finally, if the third condition holds and if {{math|δ ∈ Q}} is such that {{math|δ² }}={{math| a₂² − 4a₀}}, then {{math|P(x)}} = {{math|(x² + (a₂ + δ)/2)(x² + (a₂ − δ)/2)}}.

The resolvent cubic of an irreducible quartic polynomial {{math|P(x)}} can be used to determine its Galois group {{math|G}}; that is, the Galois group of the splitting field of {{math|P(x)}}. Let {{mvar|m}} be the degree over {{mvar|k}} of the splitting field of the resolvent cubic (it can be either {{math|R₄(y)}} or {{math|R₅(y)}}; they have the same splitting field). Then the group {{mvar|G}} is a subgroup of the symmetric group {{math|S₄}}. More precisely:

• If {{math|m {{=}} 1}} (that is, if the resolvent cubic factors into linear factors in {{mvar|k}}), then {{mvar|G}} is the group {{math|{e, (12)(34), (13)(24), (14)(23)}}}.
• If {{math|m {{=}} 2}} (that is, if the resolvent cubic has one and, up to multiplicity, only one root in {{math|k}}), then, in order to determine {{mvar|G}}, one can determine whether or not {{math|P(x)}} is still irreducible after adjoining to the field {{mvar|k}} the roots of the resolvent cubic. If not, then {{mvar|G}} is a cyclic group of order 4; more precisely, it is one of the three cyclic subgroups of {{math|S₄}} generated by any of its six {{math|4}}-cycles. If it is still irreducible, then {{mvar|G}} is one of the three subgroups of {{math|S₄}} of order {{math|8}}, each of which is isomorphic to the dihedral group of order {{math|8}}.
• If {{math|m {{=}} 3}}, then {{mvar|G}} is the alternating group {{math|A₄}}.
• If {{math|m {{=}} 6}}, then {{mvar|G}} is the whole group {{math|S₄}}.

• Resolvent (Galois theory)

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Category:Algebra

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