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 6: Groups

In the last section, we observed that the allowed root permutations of $(r_1,r_2,\ldots,r_n)$ map one root of $g$ to another root. And we also observed that, for any any two roots $t$ and $t’$, we can find the corresponding root permutation that maps $t \to t’$.

Let $\sigma_1$ be the root permutation derived from $t \to t’$. Then, $\sigma_1(t)=t’$. Let $\sigma_2$ be another allowed root permutation and let $\sigma_2(t’) = t’’$. We know that $t’’$ is another root of $g$. We observe that $\sigma_2(\sigma_1(t)) = t’’$. We can find a $\sigma_3$ that maps $t \to t’’$. Thus, $\sigma_2(\sigma_1(t)) = \sigma_3(t)$ This shows that the allowed root permutations are closed under composition.

A set of permutations that contains the identity permutation and are closed under compositions is called a “Group”. The group of allowed root permutations is called the Galois Group: $\text{Gal}(K(r_1,r_2,\ldots,r_n)/K)$. The size of the Galois Group = degree of $g(x)$. All possible $n!$ permutations form the permutation group $S_n$. Galois Group is a subgroup of $S_n$.

Automorphisms

Given any permutation $\sigma \in \text{Gal}(K(r_1,r_2,\ldots,r_n)/K)$ that maps $t \to t’$, we can apply it on the entire field, since any element of the field can be expressed as a polynomial in $(r_1,r_2,\ldots,r_n)$ with coefficients from $K$. By defining $\sigma(k) = k$ for all $k \in K$ and $\sigma(r_i) = \phi_i(t’)$. For any element $u \in K(t)$, we can define:

\[\sigma(u) = \sigma(\psi(r_1,r_2,\ldots,r_n)) = \psi(\sigma(r_1),\sigma(r_2),\ldots,\sigma(r_n))\]

This is equivalent to saying:

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

For this definition to be well-defined, we need that, whenever we have two different representations of $u$, $\sigma(u)$ defined using either of those representations is consistent with the other. If:

\[u = \psi_1(r_1,r_2,\ldots,r_n)=\psi_2(r_1,r_2,\ldots,r_n)\]

Then,

\[\psi_1(r_1,r_2,\ldots,r_n) - \psi_2(r_1,r_2,\ldots,r_n) = 0\] \[\psi_1(\phi_1(t),\phi_2(t),\ldots,\phi_n(t)) - \psi_2(\phi_1(t),\phi_2(t),\ldots,\phi_n(t)) = 0\]

This can be seen as a polynomial with coefficients in $K$ with a root $t$. It immediately follows that $t’$ is also a root of the same polynomial:

\[\psi_1(\phi_1(t'),\phi_2(t'),\ldots,\phi_n(t')) - \psi_2(\phi_1(t'),\phi_2(t'),\ldots,\phi_n(t')) = 0\]

which is same as saying:

\[\sigma(\psi_1(\phi_1(t),\phi_2(t),\ldots,\phi_n(t))) = \sigma(\psi_2(\phi_1(t),\phi_2(t),\ldots,\phi_n(t)))\]

It is easy to see that the mapping $\sigma$ preserves sums and products:

\[\sigma(x + y) = \sigma(x) + \sigma(y)\] \[\sigma(xy) = \sigma(x)\sigma(y)\]

where $x, y$ are polynomials of the form $\psi(r_1,r_2,\ldots,r_n)$. Hence $\sigma$ defined this way is an automorphism from $K(r_1,r_2,\ldots,r_n)$ to itself.

Examples of Galois Groups

For the general polynomial of $n^{th}$ degree, the Galois Group contains all the $n!$ permutations.

For some special polynomials like the ones for the $p$-th root of unity, the roots have more algebraic structure in them. The roots are of the form $\alpha$, $\alpha^2$, $\alpha^3$, $\ldots$, $\alpha^{p-1}$. Any allowed root permutation is a automorphism that preserves algebraic structure. Thus defining $\sigma(\alpha) = \alpha^i$ defines the root permutation for all the other roots, since for any other root $\alpha^j$, $\sigma(\alpha^j) = \sigma(\alpha)^j = \alpha^{ij}$. In this way, there are only $p$ possible root permutations: $\sigma(\alpha) = \alpha^i$ where $i = 1,2,3\ldots,p$.

Fixed Points

From the previous section, we know that:

  • If any polynomial expression in roots is equal to some $k \in K$, then, it must be invariant to all the root permutations in the Galois Group.
  • If any polynomial expression in roots in invariant to all the root permutations in Galois Group, then it must be in $K$.

The first statement is equivalent to saying $\sigma(k) = k$ for all $k \in K$. This is true by defintion. The second statement is equivalent to saying if for some $x \in K(r_1,r_2,\ldots,r_n)$, $\sigma(x) = x$ for all $\sigma \in $ Galois Group, then $x \in K$.

What we know so far:

  1. The allowed root permutations form a Group called the Galois Group.
  2. Any $\sigma$ in the Galois Group can be extended to an automorphism on $K(r_1,r_2,\ldots,r_n)$ such that it leaves $K$ fixed.
  3. The permutations in the Galois Group collectively fix only $K$ and nothing more. If some $x$ is fixed under all $\sigma \in$ Galois Group, then $x \in K$.

Continue Reading: Part 7: Factorization of the Resolvent Polynomial