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:

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  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 a3 = 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 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

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 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.

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 x2 + αx + β and x2  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  0 (note that if a1  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 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

R_3(y)=R_1\left(\frac y2\right)\text{.}

Fourth definition

Graph of the polynomial function x4 + x3  x2  7x/4  1/2 (in green) together with the graph of its resolvent cubic R4(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 α1, α2, α3, and α4, 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 α1α2 + α3α4, α1α3 + α2α4, and α1α4 + α2α3.

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 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

\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 α1, α2, α3, and α4, 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, 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

(\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 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

\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 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).

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 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:

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)  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:

Indeed:

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 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]

See also

References

  1. 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
  2. 1 2 Brookfield, G. (2007), "Factoring quartic polynomials: A lost art" (PDF), Mathematics Magazine 80 (1): 67–70, JSTOR 27642994, Zbl 1227.97040
  3. 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
  4. 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
  5. Rotman, Joseph (1998), "Galois groups of quadratics, cubics, and quartics", Galois Theory (2nd ed.), Springer-Verlag, ISBN 0-387-98541-7, Zbl 0924.12001
  6. 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

External links

This article is issued from Wikipedia - version of the Thursday, April 28, 2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.