Part 3: Lagrange Resolvent

In this section, we unify the solution methods used for solving the quadratic, cubic and quartic equations.

Let $\alpha$ be a primitive $n^{th}$ root of unity. For $n \leq 4$, we have:

For $n=2$, $\alpha=-1$.
For $n=3$, $\alpha=\omega, \omega^2$ and is a root of the quadratic equation $x^2 + x + 1 = 0$
For $n=4$, $\alpha=i, -i$ and is a root of the quadratic equation $x^2 + 1 = 0$

Instead of solving the $n^{th}$ degree polynomial $f(x)$ with roots $r_1$, $r_2$, $\ldots$, $r_n$, let us try solving for a polynomial $g(x)$ whose roots are $t = r_1 + \alpha r_2 + \alpha^2 r_3 + \ldots + \alpha^{n-2} r_{n-1} + \alpha^{n-1} r_n$ formed by the $n!$ permutations of the $n$ roots $r_1$, $r_2$, $\ldots$, $r_n$. Hence $g(x)$ is a $n!$ degree polynomial. $t$ is called the Lagrange Resolvent. From the previous section, we observe that this method seems to work for degree $2$ and degree $3$ equations. Degree $4$ seems to use a different resolvent.

Resolvent Polynomial

$g(x)$ is called the resolvent polynomial. For the polynomials that we have tried solving so far, the resolvent polynomial always had coefficients that can be expressed using known values (coefficients of $f$). Here is why this happens. Let the roots of $g$ be $t_1$, $t_2$, $\ldots$, $t_{n!}$. The coefficients of $g$ are elementary symmetric polynomials in $t_i$. Any symmetric polynomial in $t_i$ is also symmetric in $r_i\text{s}$, since permuting $r_i\text{s}$ only permutes $t_i\text{s}$. Thus, the coefficients of $g$ are symmetric polynomials in $r_i\text{s}$ and are hence known quantities.

For the polynomials that we have tried solving so far, the resolvent polynomial has also been solvable. Here is how this works for degree $2$ and degree $3$ polynomials.

Solving quadratic with Lagrange Resolvent

For the quadratic polynomial $f(x) = (x-r_1)(x-r_2)$, the resolvents are

\[t_1 = r_1 - r_2\] \[t_2 = r_2 - r_1\]

The resolvent polynomial $g(x) = (x-t_1)(x-t_2)$ is also a quadratic. However, in this case, $t_1 = -t_2$.

\[g(x) = (x-t_1)(x + t_1) = x^2 - t_1^2\]

This is a simpler quadratic that lacks the linear term and can be solved with only a square root. Solving this, gives two possible values of $t_1$, one corresponding to $r_1- r_2$ and the other corresponding to $r_2- r_1$. We can choose any one of these without loss of generality. Let us say our choice corresponds to $t_1 = r_1 - r_2$. Since we know $r_1 + r_2$ and $r_1 - r_2$, we can solve this linear system to get $r_1$ and $r_2$.

Solving cubic with Lagrange Resolvent

For the cubic polynomial $f(x) = (x-r_1)(x-r_2)(x-r_3)$, the resolvents are

\[t_1 = r_1 + \omega r_2 + \omega^2 r_3\] \[t_2 = r_3 + \omega r_1 + \omega^2 r_2\] \[t_3 = r_2 + \omega r_3 + \omega^2 r_1\] \[t_4 = r_1 + \omega r_3 + \omega^2 r_2\] \[t_5 = r_2 + \omega r_1 + \omega^2 r_3\] \[t_6 = r_3 + \omega r_2 + \omega^2 r_1\]

The resolvent polynomial $g(x) = \prod_{i=1}^{6} (x-t_i)$ is a $6^{th}$ degree polynomial. However, in this case, $t_2 = \omega t_1$, $t_3 = \omega^2 t_1$ and similarly $t_5 = \omega t_4$, $t_6 = \omega^2 t_4$.

\[g(x) = (x-t_1)(x-\omega t_1)(x-\omega^2 t_1)(x-t_4)(x-\omega t_4)(x-\omega^2 t_4)\] \[g(x) = (x^3-t_1^3)(x^3-t_4^3)\]

where:

\[t_1 = r_1 + \omega r_2 + \omega^2 r_3\] \[t_4 = r_1 + \omega r_3 + \omega^2 r_2\]

This is actually a quadratic in $x^3$ and can be solved using the above method. We observe that any odd permutation of the roots $r_1$, $r_2$, $r_3$ maps $t_1^3$ to $t_4^3$ (and vice versa). Any even permutation of the roots $r_1$, $r_2$, $r_3$ maps $t_1^3$ to $t_4^3$ itself. Hence, the two roots of the quadratic equation $g(x) = (x^3-t_1^3)(x^3-t_4^3)$ can be assigned to $t_1^3, t_4^3$ in any order.

From this, we can recover $t_1$ using a cube root. There are $3$ possible cube roots, which are equivalent to cyclic permutations (or the 3 even permutations) of the roots.

Once we know $t_1$, we can recover the corresponding $t_4$ since $t_1t_4 = (r_1 + \omega r_2 + \omega^2 r_3)(r_1 + \omega r_3 + \omega^2 r_2)$ is a symmetric polynomial in the roots and is hence a known quantity. From

\[\begin{align*} t_1 &= r_1 + \omega r_2 + \omega^2 r_3\\ t_4 &=r_1 + \omega r_3 + \omega^2 r_2\\ -a &= r_1 + r_2 + r_3 \end{align*}\]

we can solve for the individual roots by solving the system of linear equations.

Solving quartic with Lagrange Resolvent

In the previous section, we showed that the quartic can be solved using the resolvent:

\[t_1 = r_1r_2 + r_3r_4\] \[t_2 = r_1r_3 + r_2r_4\] \[t_3 = r_1r_4 + r_2r_3\]

The resulting resolvent polynomial is a cubic. Solving this, followed by solving a few quadratic equations solves the quartic polynomial.

Alternatively, we can also solve the quartic using the Lagrange Resolvent $t = r_1 + ir_2 - r_3 - ir_4$. Similar to the cubic case, the resolvent polynomial in this case can be written as:

\[g(x) = (x^4-t_1^4)(x^4-t_2^4)(x^4-t_3^4)(x^4-t_4^4)(x^4-t_5^4)(x^4-t_6^4)\]

This is a $24^{th}$ degree equation, but a $6^{th}$ degree equation in $x^4$, where:

\[t_1 = (r_1 - r_2) + i(r_3 - r_4)\] \[t_2 = (r_1 - r_2) + i(r_4 - r_3)\] \[t_3 = (r_1 - r_3) + i(r_4 - r_2)\] \[t_4 = (r_1 - r_3) + i(r_2 - r_4)\] \[t_5 = (r_1 - r_4) + i(r_2 - r_3)\] \[t_6 = (r_1 - r_4) + i(r_3 - r_2)\]

This is a $6^{th}$ degree polynomial that we do not know how to solve. Instead of solving the resolvent polynomial directly, we go on to solve other auxilary polynomials. We observe that:

\[t_1t_2 = (r_1 - r_2)^2 + (r_3 - r_4)^2 = r_1^2 + r_2^2 + r_3^2 + r_4^2 - 2(r_1r_2 + r_3r_4)\] \[t_3t_4 = (r_1 - r_3)^2 + (r_4 - r_2)^2 = r_1^2 + r_2^2 + r_3^2 + r_4^2 - 2(r_1r_3 + r_2r_4)\] \[t_5t_6 = (r_1 - r_4)^2 + (r_2 - r_3)^2 = r_1^2 + r_2^2 + r_3^2 + r_4^2 - 2(r_1r_4 + r_2r_3)\]

$t_1t_2, t_3t_4, t_5t_6$ are roots of a cubic polynomial with known coefficients and can be solved. This is similar to the cubic polynomial with roots $(r_1r_2+r_3r_4, r_1r_3+r_2r_4, r_1r_4+r_2r_3)$ used in solving the quartic in the previous section.

We also observe that:

\[\begin{align*} t_1^2 - t_2^2 &= 4i(r_1 - r_2)(r_3 - r_4)\\ &= 4i(r_1r_3+r_2r_4)-4i(r_1r_4+r_2r_3)\\ &= 2i(t_5t_6-t_3t_4) \end{align*}\]

This means that once we know $t_1t_2$, $t_3t_4$, $t_5t_6$, by solving the auxilary cubic equation, we also know $t_1^2 - t_2^2$, $t_3^2 - t_4^2$, $t_5^2 - t_6^2$. Thus, by solving the auxilary cubic equation, we can factorize $g(x)$ further into $6$ factors:

\[g(x) =\] \[\big( (x^2-t_1^2)(x^2+t_2^2) \big) \hspace{5mm} \big( (x^2-t_3^2)(x^2+t_4^2) \big) \hspace{5mm} \big( (x^2-t_5^2)(x^2+t_6^2) \big) \hspace{5mm}\] \[\big( (x^2+t_1^2)(x^2-t_2^2) \big) \hspace{5mm} \big( (x^2+t_3^2)(x^2-t_4^2) \big) \hspace{5mm} \big( (x^2+t_5^2)(x^2-t_6^2) \big) \hspace{5mm}\]

Each factor is a polynomial with known coefficients. Each of them is a quadratic in $x^2$. We can solve each of these quadratics to get $t_1^2$, $t_2^2$, $t_3^2$, $t_4^2$, $t_5^2$, and $t_6^2$. This helps us factorize $g(x)$ fully into its $12$ factors of the form $(x^2 - t_i^2)$.

We can recover $t_i\text{s}$ by taking the square roots. There are two possible roots in each case. However, once we fix $t_1$, we can find the corresponding $t_2$ since we already know $t_1t_2$. Similarly, once we fix $t_3$, we can find the corresponding $t_4$. Similarly for $t_5$ and $t_6$. The three choices of signs of $t_1, t_3, t_5$ have to be aligned such that:

\[t_1 + t_3 + t_5 = t_2 + t_4 + t_6\]

But we still need to distinguish between $(t_1, t_2)$ and $(t_2, t_1)$ and similarly for the other two pairs. We can do this by observing that:

\[(t_1 + it_2)(t_3 + it_4)(t_5 + it_6)\] \[=(1+i)^3(r_1+r_2-r_3-r_4)(r_1+r_3-r_4-r_2)(r_1+r_4-r_2-r_3)\]

is symmetric in the roots $r_1$, $r_2$, $r_3$, $r_4$ and is a known quantity. With these restriction, all $t_i\text{s}$ get fixed. From these, the roots can be solved using a system of linear equations.

Roots of unity with Lagrange Resolvent

We can solve for any primitive root of unity using Lagrange Resolvent, going beyond the quartic equation.

We know that the primitive $n^{th}$ root of unity satisfies

\[x^{n-1} + x^{n-2} + \ldots + x^2 + x +1=0\]

We also know that if $n$ is not a prime with $n=ab$ where $1<a<n$ and $1<b<n$, the $n^{th}$ primitive root of unity is a product of the $a^{th}$ primitive root of unity and $b^{th}$ primitive root of unity. Hence, to solve for primitve roots of unity, it is enough to consider cases where $n = p$ is a prime.

Let the roots of this polynomial be $\alpha$, $\alpha^2$, $\alpha^3$, $\ldots$, $\alpha^{p-1}$. This can also be arranged in a slightly different order: $\alpha$, $\alpha^g$, $\alpha^{g^2}$, $\ldots$, $\alpha^{g^{p-2}}$ for some $g<p$. This is always possible due to a result in number theory. There always exists a generator $g$ such that $g^i$ mod($p$) generates all of $1,2,3,\ldots, p-1$ (mod $p$). The resolvent corresponding to this order is:

\[t = \alpha + \beta\alpha^g + \beta^2\alpha^{g^2} + \ldots + \beta^{p-2}\alpha^{g^{p-2}}\]

where $\beta$ is a primitive $(p-1)^{th}$ root of unity. We assume $\beta$ to be a known quantity. The resolvent polynomial of degree $(p-1)!$ would have a factor of the form $(x^{p-1} - t^{p-1})$. Generally, $t^{p-1}$ is not a known value (as shown in the case of the general quartic case). But in the case of the above polynomial, this factor is expressible in terms of known values.

When $t^{p-1}=(\alpha + \beta\alpha^g + \beta^2\alpha^{g^2} + \ldots + \beta^{p-2}\alpha^{g^{p-2}})^{p-1}$ is expanded in terms of $\alpha$ and $\beta$, it can be expressed as a polynomial in $\alpha$ whose coefficients are polynomials in $\beta$:

\[t^{p-1} = P_0(\beta) + P_1(\beta)\alpha + P_2(\beta)\alpha^g + P_3(\beta)\alpha^{g^2}+ \ldots + P_{p-1}(\beta)\alpha^{g^{p-2}}\]

When $\alpha$ is replaced by $\alpha^g$ in the above, $t$ becomes $t/\beta$. The left hand side remains unchanged since $(t/\beta)^{p-1} = t^{p-1}$. On the right hand side, this implies $P_1(\beta) = P_2(\beta) = \ldots =P_{p-1}(\beta)$. This gives:

\[t^{p-1} = P_0(\beta) + P_1(\beta)(\alpha + \alpha^g + \alpha^{g^2}+ \ldots + \alpha^{g^{p-2}}) = P_0(\beta) - P_1(\beta)\]

Hence, $t^{p-1}$ is a known quantity.

Once we know $t$ we can solve for the individual roots. Here is a sketch of the method. Consider the quantities:

\[u_1 = \alpha + \beta\alpha^g + \beta^2\alpha^{g^2} + \ldots + \beta^{p-2}\alpha^{g^{p-2}}\] \[u_2 = \alpha + \beta^2\alpha^g + \beta^4\alpha^{g^2} + \ldots + \beta^{2(p-2)}\alpha^{g^{p-2}}\] \[\vdots\] \[u_k = \alpha + \beta^k\alpha^g + \beta^{2k}\alpha^{g^2} + \ldots + \beta^{k(p-2)}\alpha^{g^{p-2}}\] \[\vdots\] \[u_{p-1} = \alpha + \beta^{p-1}\alpha^g + \beta^{2({p-1})}\alpha^{g^2} + \ldots + \beta^{(p-1)(p-2)}\alpha^{g^{p-2}}\]

We see that $u_1 = t$ is a known quantity. $u_{p-1} = \alpha + \alpha^g + \alpha^{g^2} + \ldots + \alpha^{g^{p-2}} = -1$ is also a known quantity. In general, any $u_iu_1^{p-1-i}$ can be shown to be a known quantity by the exact same method as in showing $t^{p-1}$ to be a known quantity. Once we find all $u_i\text{s}$, we can solve for $\alpha$ from the system of linear equations.

Lagrange Theorem on Resolvents

So far, we have observed that, once we have solved the resolvent polynomial $g(x)$, we can find the roots of the polynomial $f(x)$.

Theorem: Suppose $t$ is an expression in terms of the roots and known quantities, and $u$ is another expression in roots and known quantities. Let us evaluate all possible values of $t$ and $u$ under the various root permutations. If it so happens that the root permutations that change the value of $u$ also change the value of $t$, then $u$ can be expressed using $t$ and other known values.

Proof: Let us assume there are $k$ different values that $u$ can take under the root permutations: $u_1$, $u_2$, $u_3$, $\ldots$, $u_k$. Let the corresponding values of $t$ be $t_1$, $t_2$, $t_3$, $\ldots$, $t_k$. Consider the following:

\[u_1 + u_2 + \ldots + u_k\] \[t_1u_1 + t_2u_2 + \ldots + t_ku_k\] \[t_1^2u_1 + t_2^2u_2 + \ldots + t_k^2u_k\] \[t_1^3u_1 + t_2^3u_2 + \ldots + t_k^3u_k\] \[\vdots\] \[t_1^{k-1}u_1 + t_2^{k-1}u_2 + \ldots + t_k^{k-1}u_k\]

All these expressions are symmetric in roots and are known quantities. Hence, $u_1$ can be expressed in terms of $t_i^j$:

\[u_1 = \frac{D_1}{D} = \frac{D_1D}{D^2}\]

where D is the Vandermonde Determinant, $D^2 = \prod (t_i - t_j)^2$. $D^2$ is symmetric in $t_i$ and hence symmetric in roots and is a known quantity. $D_1$ is same as D but $t_1^j$ is replaced with the value of the $j^{th}$ expression in the above system of equations. Hence, the numerator is a polynomial in $t_1$ whose coefficients are symmetric polynomials in $t_2$, $t_3$, $\ldots$, $t_k$, which in turn can be expressed as polynomials in $t_1$. Hence $u_1$ can be expressed as polynomial in $t_1$ (with coefficients being known quantities).

If we can find a $t$ such that it takes $n!$ different values under root permutations, we have got our resolvent. Indeed, such a resolvent can always be formed using a linear combination of roots. For such a linear combination to not change under some permutation requires the coefficients to satisfy elaborate constraints. Finding all such constraints for our given polynomial and then choosing coefficients that violate these constraints provides us the resolvent. We have not only proved the existence of a resolvent, but also a resolvent that is linear in the roots.

Here is a summary of what we know so far:

We can create a new variable $t$ called the resolvent which is expressed entirely using the roots of the polynomial and known quantities (like coefficients of the polynomial and roots of unity).
$t$ can be expressed as a root of a polynomial with known coefficients. We have been able to solve this polynomial so far in the cases that we have explored. Though, this is not always guaranteed.
Once we find $t$, we can obtain all the roots using known quantities.
We can always find a $t$ that is linear in roots. Lagrange Resolvent is one such resolvent. For polynomials of degree $2$ and $3$, the resolvent polynomial can be directly solved. For polynomials of degree $4$, the resolvent polynomial can not be directly solved. However, by solving an auxilary cubic polynomial, the resolvent polynomial can be factorized into simpler polynomials, that can then be solved. The resolvent method can also be used for solving for primitive roots of unity.

Continue Reading: Part 4: Fields

Rediscovering Galois Theory: Part-3

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