Table of Contents

  1. Prerequisites
  2. Solving Polynomials upto degree 4
  3. Lagrange Resolvent
  4. Fields
  5. Root Permutations
  6. Groups
  7. Factorization of the Resolvent Polynomial
  8. Solvability

Part 5: Root Permutations

Let us revisit the solution to the quartic using the Lagrange Resolvent by following the irreducible polynomial containing $t_1$ as a root.

Initially, the resolvent polynomial is the full $24$ degree polynomial. The only expression in roots that can be evaluated (as a known quantity) at this point are the ones that remain unchanged by all the $24$ permutations of the roots.

In the next step, we end up knowing the values of $r_1r_2 + r_3r_4$, $r_1r_3 + r_2r_4$, $r_1r_4 + r_2r_3$. These are not symmetric polynomials in roots. The root permutations that preserve them are:

  1. $(r_1, r_2, r_3, r_4)$: The identity permutation

  2. $(r_2, r_1, r_4, r_3)$: Flip $(r_1, r_2)$ and Flip $(r_3, r_4)$

  3. $(r_3, r_4, r_1, r_2)$: Flip $(r_1, r_3)$ and Flip $(r_2, r_4)$

  4. $(r_4, r_3, r_2, r_1)$: Flip $(r_1, r_4)$ and Flip $(r_2, r_3)$

At this point, the resolvent polynomial is $(x^2-t_1^2)(x^2+t_2^2)$. These permutations take $t_1$ to the other roots of the resolvent polynomial: $t_1$, $-t_1$, $it_2$, $-it_2$. The only expressions in roots that we can evaluate at this stage happens to be polynomials in roots that remain unchanged by these permutations. For example, we can evaluate $(r_1 + r_2)(r_3 + r_4)$.

We notice the following pattern:

  1. Polynomial expressions in roots that can be evaluated as known quantities are characterized by a subset of root permutations.

  2. These same permutations map one root of the resolvent polynomial to the other roots of the resolvent polynomial.

Allowed Root Permutations

$K(t) = K(r_1,r_2,\ldots,r_n)$. $t$ is a root of $g(x)$. Once we solve for $t$, we can recover the roots $r_1$, $r_2$, $\ldots$, $r_n$, using a polynomial in $t$:

\[r_1 = \phi_1(t), r_2 = \phi_2(t), \ldots r_n = \phi_2(t)\]

However, we cannot distinguish one root of $g(x)$ from another. Thus, if $t’$ is another root of $g(x)$, we could also get a permutation of the roots $r_1$, $r_2$, $\ldots$, $r_n$ as

\[\phi_1(t'), \phi_2(t'), \ldots \phi_2(t')\]

We can show that this is a permutation of $r_1$, $r_2$, $\ldots$, $r_n$.

$f(\phi_i(x))$ has $t$ as a root. Then, the irreducible polynomial of $t$: $g(x)$ divides $f(\phi_i(x))$.

So, $f(\phi_i(t’)) = 0$ establishing that $\phi_1(t’)$, $\phi_2(t’)$, $\ldots$, $\phi_2(t’)$ are all roots of $f(x)$.

Further, if $\phi_i(t’) = \phi_j(t’)$, then the polynomial $\phi_i(x) - \phi_j(x)$ has a root $t’$. Then $g(x)$ $|$ $\phi_i(x) - \phi_j(x)$ and $t$ is also a root of $\phi_i(x) - \phi_j(x)$ implying that $\phi_i(t) = \phi_j(t)$, which is not possible.

Thus, $\phi_1(t’)$, $\phi_2(t’)$, $\ldots$, $\phi_2(t’)$ is a rearrangement of $r_1$, $r_2$, $\ldots$, $r_n$. If we were to somehow know $t$, and then find the roots $r_i$ using $t$, we will not be able to distinguish between these permutations. Let us call these permutations as “Allowed Permutations”. Let us call the specific permutation

\[(r_1,r_2,\ldots,r_n) = (\phi_1(t),\phi_2(t),\ldots,\phi_n(t)) \to (\phi_1(t'),\phi_2(t'),\ldots,\phi_n(t'))\]

as “Allowed Permutation” derived from $t \to t’$.

Allowed Root Permutation maps one root of $g$ to another root of $g$

Consider a root permutation $\sigma$ that maps

\[(r_1,r_2,\ldots,r_n) = (\phi_1(t),\phi_2(t),\ldots,\phi_n(t)) \to (\phi_1(t'),\phi_2(t'),\ldots,\phi_n(t'))\]

Let $t_1$ be any root of $g$. Since $t_1$ can be expressed as a polynomial in $(r_1,r_2,\ldots,r_n)$, we have:

\[\begin{align*} t_1 &= \psi(r_1, r_2, \ldots, r_n)\\ &= \psi(\phi_1(t),\phi_2(t),\ldots,\phi_n(t)) \end{align*}\]

When the permutation $\sigma$ is applied on $t_1$, we get:

\[\psi(\phi_1(t'),\phi_2(t'),\ldots,\phi_n(t'))\]

Since $g(t_1) = 0$, we get that

\[g(\psi(\phi_1(t),\phi_2(t),\ldots,\phi_n(t))) = 0\]

This is a polynomial which shares a root with the irreducible polynomial $g$. Then $g$ divides it and all roots of $g$ are also roots of this polynomial. Thus:

\[g(\psi(\phi_1(t'),\phi_2(t'),\ldots,\phi_n(t'))) = 0\]

This shows that $\psi(\phi_1(t’),\phi_2(t’),\ldots,\phi_n(t’))$ is also a root of $g$.

This shows that for any root $t_1$ of $g$, the root permutation

\[(r_1,r_2,\ldots,r_n) = (\phi_1(t),\phi_2(t),\ldots,\phi_n(t)) \to (\phi_1(t'),\phi_2(t'),\ldots,\phi_n(t'))\]

maps $t_1 \to t_2$ where $t_2$ is another root of $g$.

In the special case when $t_1 = t$, we have:

\[t = \psi(\phi_1(t),\phi_2(t),\ldots,\phi_n(t))\]

We observe that this is a polynomial which shares a root with the irreducible polynomial $g$. Then $g$ divides it and all roots of $g$ are also roots of this polynomial. Thus:

\[t' = \psi(\phi_1(t'),\phi_2(t'),\ldots,\phi_n(t'))\]

Thus, when $t_1 = t$, we have $t_2 = t’$.

We have shown that given any two roots $t, t’$ of $g$, we can define a root permutation that maps $t \to t’$. This same root permutations maps every root of $g$ to other roots of $g$.

Polynomial Expressions in Roots that can be evaluated

Consider any polynomial expression in roots: $\psi(r_1, r_2,\ldots,r_n)$. We know that

\[\psi(r_1, r_2,\ldots,r_n) = \psi(\phi_1(t), \phi_2(t), \ldots, \phi_n(t)) \in K(t)\]

since it is a polynomial in $t$. Suppose that we can evaluate this expression, and assign it a known value. Then, $\psi(r_1, r_2,\ldots,r_n) = c \in K$. This implies that $t$ is a root of $\psi(\phi_1(x), \phi_2(x), \ldots, \phi_n(x)) - c =0$. Again, by the same logic as before, any $t’$ is also a root of $\psi(\phi_1(x), \phi_2(x), \ldots, \phi_n(x)) - c = 0$. This shows that if we can evaluate any polynomial expression in roots, then it is invariant under all the allowed root permutations.

The converse of this statement is also true. If we have an expression that is invariant under all the allowed root permutations, then it can be assigned a value from $K$. Suppose $\psi$ is invariant to all the allowed permutations. Then,

\[\psi(\phi_1(t), \phi_2(t), \ldots, \phi_n(t)) = \psi(\phi_1(t), \phi_2(t), \ldots, \phi_n(t))\] \[\psi(\phi_1(t), \phi_2(t), \ldots, \phi_n(t)) = \psi(\phi_1(t'), \phi_2(t'), \ldots, \phi_n(t'))\] \[\psi(\phi_1(t), \phi_2(t), \ldots, \phi_n(t)) = \psi(\phi_1(t''), \phi_2(t''), \ldots, \phi_n(t''))\] \[\vdots\] \[\psi(\phi_1(t), \phi_2(t), \ldots, \phi_n(t)) = \psi(\phi_1(t^{(k)}), \phi_2(t^{(k)}), \ldots, \phi_n(t^{(k)}))\]

Adding them all up, we get:

\[\psi(\phi_1(t), \phi_2(t), \ldots, \phi_n(t)) = \frac{1}{k+1} \sum_{i=0}^{k} \psi(\phi_1(t^{(i)}), \phi_2(t^{(i)}), \ldots, \phi_n(t^{(i)}))\]

The right side of this expression is symmetric in the roots of $g(x)$. Hence, it is a known quantity.

Thus the only expressions that can be evaluated as a known quantity are those that are invariant under the allowed root permutations.

So far we have found that:

  1. If we were to somehow solve $g(x)$ and get $t$, it could be any one of the roots of $g(x)$. If we then were to recover the roots of $f(x)$ using $(\phi_1(t’),\phi_2(t’),\ldots,\phi_n(t’))$, it would be a permutation of $r_i\text{s}$.
  2. When the roots $t$ of $g(x)$ are expressed as a polynomial in roots $(r_1,r_2,\ldots,r_n)$, these same permutations map one root of $g$ to another root of $g$. For any two roots of $t$, and $t’$ of $g$, we can define a root permutation that maps $t \to t’$ and any root of $g$ to some other root of $g$.
  3. A polynomial in $(r_1,r_2,\ldots,r_n)$ can be evaluated to a known
    quantity in $K$ if and only if it is invariant to these same permutations.

Continue Reading: Part 6: Groups