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:
 * The coefficients of the resolvent cubic can be obtained from the coefficients of $x^{4} + x^{3} – x^{2} – 7x/4 – 1/2$ using only sums, subtractions and multiplications.
 * Knowing the roots of the resolvent cubic of $R_{4}(y)$ is useful for finding the roots of $P(x)$ itself. Hence the name “resolvent cubic”.
 * The polynomial $P(x)$ has a multiple root if and only if its resolvent cubic has a multiple root.

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

In some cases, the concept of resolvent cubic is defined only when $C$ is a quartic in depressed form—that is, when $Q$.

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 $a_{3} = 0$.

First definition
Suppose that $P(x)$ is a depressed quartic—that is, that $2$. A possible definition of the resolvent cubic of $P(x)$ 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 $a_{3} = 0$. 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 $P(x)$. 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 $P(x)$ = 0.

If $x^{2} + a_{2}/2$ is a root of $R_{1}(y)$, then it is a consequence of the computations made above that the roots of $y_{0}$ 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 $R_{1}(y)$, but since the constant term of $P(x)$ is $y_{0} = 0$, $R_{1}(y)$ is a root of $–a_{1}^{2}$ if and only if $0$, and in this case the roots of $R_{1}(y)$ can be found using the quadratic formula.

Second definition
Another possible definition (still supposing that $a_{1} = 0$ 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:


 * $$\begin{align}P(x)=0&\Longleftrightarrow x^4=-a_2x^2-a_1x-a_0\\ &\Longleftrightarrow(x^2+y)^2=-a_2x^2-a_1x-a_0+2yx^2+y^2\end{align}$$

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 (again, supposing that $P(x)$ 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 $P(x)$ by expressing it as a product of two monic quadratic polynomials $P(x)$ and $P(x)$, 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 $x^{2} + αx + β$ (note that if $x^{2} – αx + γ$, 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, $α ≠ 0$ and $a_{1} ≠ 0$ 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 $β$ is a root of $γ$. So, again, knowing the roots of $α^{2}$ helps to determine the roots of $R_{3}(y)$.

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 $R_{3}(y)$ are $P(x)$, and $P(x)$, 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 $α_{1}, α_{2}, α_{3}$, $α_{4}$, and $α_{1}α_{2} + α_{3}α_{4}$.

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, $α_{1}α_{3} + α_{2}α_{4}$ has a multiple root if and only if $α_{1}α_{4} + α_{2}α_{3}$ has a multiple root. More precisely, $P(x)$ and $R_{4}(y)$ 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


 * $$R_5(y)=y^3-2a_2y^2+({a_2}^2+a_3a_1-4a_0)y+{a_1}^2-a_3a_2a_1+{a_3}^2a_0\text{.}$$

If, as above, the roots of $R_{4}(y)$ are $P(x)$, and $P(x)$, 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, $α_{1}, α_{2}, α_{3}$ is the monic polynomial whose roots are $α_{4}$, $R_{5}(y)$, and $(α_{1} + α_{2})(α_{3} + α_{4})$.

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 $(α_{1} + α_{3})(α_{2} + α_{4})$, $(α_{1} + α_{4})(α_{2} + α_{3})$ has a multiple root if and only if $R_{4}(y)$ has a multiple root. More precisely, $P(x)$ and $R_{5}(y)$ have the same discriminant. This is also a consequence of the fact that $P(x)$ = $R_{5}(y)$.

Note that if $R_{5}(y + a_{2})$ is a depressed polynomial, then


 * $$\begin{align}R_5(y)&=y^3-2a_2y^2+({a_2}^2-4a_0)y+{a_1}^2\\ &=-R_3(-y)\\ &=-R_1\left(-\frac y2\right)\text{.}\end{align}$$

Solving quartic equations
It was explained above how $-R_{4}(-y)$, $P(x)$, and $R_{1}(y)$ can be used to find the roots of $R_{2}(y)$ if this polynomial is depressed. In the general case, one simply has to find the roots of the depressed polynomial $R_{3}(y)$. For each root $P(x)$ of this polynomial, $P(x − a_{3}/4)$ is a root of $x_{0}$.

Factoring quartic polynomials
If a quartic polynomial $x_{0} − a_{3}/4$ is reducible in $P(x)$, 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 $k[x]$. 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 $k$ is a depressed polynomial. Then it was seen above that if the resolvent cubic $P(x)$ has a non-null root of the form $P(x)$, for some $R_{3}(y)$, then such a decomposition exists.

This can be used to prove that, in $α^{2}$, every quartic polynomial without real roots can be expressed as the product of two quadratic polynomials. Let $α ∈ k$ be such a polynomial. We can assume without loss of generality that $R[x]$ 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 $P(x)$ can and this polynomial is a reduced one. Then $P(x)$ = $P(x − a_{3}/4)$. There are two cases:
 * If $R_{3}(y)$ then $y^{3} + 2a_{2}y^{2} + (a_{2}^{2} − 4a_{0})y − a_{1}^{2}$ = $a_{1} ≠ 0$. Since $R_{3}(0)$ if $−a_{1}^{2} < 0$ is large enough, then, by the intermediate value theorem, $R_{3}(y) > 0$ has a root $y$ with $R_{3}(y)$. So, we can take $y_{0}$ = $y_{0} > 0$.
 * If $α$ = $√y_{0}$, then $a_{1}$ = $0$. The roots of this polynomial are $R_{3}(y)$ and the roots of the quadratic polynomial $y^{3} + 2a_{2}y^{2} + (a_{2}^{2} − 4a_{0})y$. If $0$, then the product of the two roots of this polynomial is smaller than $y^{2} + 2a_{2}y + a_{2}^{2} − 4a_{0}$ and therefore it has a root greater than $a_{2}^{2} − 4a_{0} < 0$ (which happens to be $0$) and we can take $0$ as the square root of that root. Otherwise, $−a_{2} + 2√a_{0}$ 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 $α$ is a real closed field, then every quartic polynomial without roots in $a_{2}^{2} − 4a_{0} ≥ 0$ can be expressed as the product of two quadratic polynomials in $k$. Indeed, this statement can be expressed in first-order logic and any such statement that holds for $k$ 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 $k[x]$ is reducible and, if it is, how to express it as a product of polynomials of smaller degree. Again, we will suppose that $R$ is monic and depressed. Then $P(x) ∈ Q[x]$ is reducible if and only if at least one of the following conditions holds: Indeed:
 * The polynomial $P(x)$ has a rational root (this can be determined using the rational root theorem).
 * The resolvent cubic $P(x)$ has a root of the form $P(x)$, for some non-null rational number $R_{3}(y)$ (again, this can be determined using the rational root theorem).
 * The number $α^{2}$ is the square of a rational number and $α$ = $a_{2}^{2} − 4a_{0}$.
 * If $a_{1}$ has a rational root $0$, then $P(x)$ is the product of $r$ by a cubic polynomial in $P(x)$, which can be determined by polynomial long division or by Ruffini's rule.
 * If there is a rational number $x − r$ such that $Q[x]$ is a root of $α ≠ 0$, it was shown above how to express $α^{2}$ as the product of two quadratic polynomials in $R_{3}(y)$.
 * Finally, if the third condition holds and if $P(x)$ is such that $Q[x]$=$δ ∈ Q$, then $δ^{2}$ = $a_{2}^{2} − 4a_{0}$.

Galois groups of irreducible quartic polynomials
The resolvent cubic of an irreducible quartic polynomial $P(x)$ can be used to determine its Galois group $(x^{2} + (a_{2} + δ)/2)(x^{2} + (a_{2} − δ)/2)$; 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 $G$ or $P(x)$; they have the same splitting field). Then the group $G$ is a subgroup of the symmetric group $R_{4}(y)$. More precisely:
 * If $R_{5}(y)$ (that is, if the resolvent cubic factors into linear factors in $k$), then $G$ is the group $S_{4}$.
 * If $m = 1$ (that is, if the resolvent cubic has one and, up to multiplicity, only one root in ${e, (12)(34), (13)(24), (14)(23)}$), then, in order to determine $G$, one can determine whether or not $m = 2$ 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 $k$ generated by any of its six $P(x)$-cycles. If it is still irreducible, then $G$ is one of the three subgroups of $S_{4}$ of order $4$, each of which is isomorphic to the dihedral group of order $S_{4}$.
 * If $8$, then $G$ is the alternating group $8$.
 * If $m = 3$, then $G$ is the whole group $A_{4}$.