Resolvent cubic

Suppose that the coefficients of belong to a field whose characteristic is different from . Whenever roots of are mentioned, they belong to some extension of such that factors into linear factors in . If is the field of rational numbers, then can be the field of complex numbers or the field of algebraic numbers. In some cases, the concept of resolvent cubic is defined only when is a quartic in depressed form—that is, when . Note that the fourth and fifth definitions below also make sense and that the relationship between these resolvent cubics and are still valid if the characteristic of is equal to . First definition Suppose that is a depressed quartic—that is, that . A possible definition of the resolvent cubic of is: :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 . 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, , to . 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  = 0. If is a root of , then it is a consequence of the computations made above that the roots of 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 , but since the constant term of is , is a root of if and only if , and in this case the roots of can be found using the quadratic formula. Second definition Another possible definition (again, supposing that is a depressed quartic) 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 by expressing it as a product of two monic quadratic polynomials and , 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 (note that if , 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, and 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 is a root of . So, again, knowing the roots of helps to determine the roots of . Note that :R_3(y)=R_1\left(\frac y2\right)\text{.} Fourth definition Still another possible definition is :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 are , and , 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 , , and . 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, has a multiple root if and only if has a multiple root. More precisely, and have the same discriminant. If 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 :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 are , and , 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, is the monic polynomial whose roots are , , and . 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 , has a multiple root if and only if has a multiple root. More precisely, and have the same discriminant. This is also a consequence of the fact that  = . Note that if 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 , , and can be used to find the roots of if this polynomial is depressed. In the general case, one simply has to find the roots of the depressed polynomial . For each root  of this polynomial, is a root of . Factoring quartic polynomials If a quartic polynomial is reducible in , 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 has a root in . In order to determine whether or not can be expressed as the product of two quadratic polynomials, let us assume, for simplicity, that is a depressed polynomial. Then it was seen above that if the resolvent cubic has a non-null root of the form , for some , then such a decomposition exists. This can be used to prove that, in , every quartic polynomial without real roots can be expressed as the product of two quadratic polynomials. Let be such a polynomial. We can assume without loss of generality that is monic. We can also assume without loss of generality that it is a reduced polynomial, because can be expressed as the product of two quadratic polynomials if and only if can and this polynomial is a reduced one. Then  = . There are two cases: • If then  = . Since if is large enough, then, by the intermediate value theorem, has a root with . So, we can take  = . • If  = , then  = . The roots of this polynomial are  and the roots of the quadratic polynomial . If , then the product of the two roots of this polynomial is smaller than  and therefore it has a root greater than  (which happens to be ) and we can take as the square root of that root. Otherwise, and then, ::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 is a real closed field, then every quartic polynomial without roots in can be expressed as the product of two quadratic polynomials in . Indeed, this statement can be expressed in first-order logic and any such statement that holds for 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 is reducible and, if it is, how to express it as a product of polynomials of smaller degree. Again, we will suppose that  is monic and depressed. Then  is reducible if and only if at least one of the following conditions holds: • The polynomial has a rational root (this can be determined using the rational root theorem). • The resolvent cubic  has a root of the form , for some non-null rational number  (again, this can be determined using the rational root theorem). • The number is the square of a rational number and  = . Indeed: • If has a rational root , then is the product of by a cubic polynomial in , which can be determined by polynomial long division or by Ruffini's rule. • If there is a rational number  such that is a root of , it was shown above how to express  as the product of two quadratic polynomials in . • Finally, if the third condition holds and if is such that =, then  = . Galois groups of irreducible quartic polynomials The resolvent cubic of an irreducible quartic polynomial can be used to determine its Galois group ; that is, the Galois group of the splitting field of . Let  be the degree over of the splitting field of the resolvent cubic (it can be either or ; they have the same splitting field). Then the group  is a subgroup of the symmetric group . More precisely: • If (that is, if the resolvent cubic factors into linear factors in ), then  is the group {{math|{e, (12)(34), (13)(24), (14)(23)}}}. • If (that is, if the resolvent cubic has one and, up to multiplicity, only one root in ), then, in order to determine , one can determine whether or not is still irreducible after adjoining to the field the roots of the resolvent cubic. If not, then is a cyclic group of order 4; more precisely, it is one of the three cyclic subgroups of  generated by any of its six -cycles. If it is still irreducible, then is one of the three subgroups of  of order , each of which is isomorphic to the dihedral group of order . • If , then is the alternating group . • If , then is the whole group . ==See also==