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 8: Solvability

In the previous section we saw that the Galois Group reduces into a normal subgroup when we factorize the resolvent polynomial. The opposite of this statement is also true.

Suppose we have an irreducible resolvent polynomial $g(x)$ with coefficents from field $K$. Let $G$ be its Galois Group. If we know a normal subgroup of $G$, say $H$ with size $1/p$, then there exists an element $r$ such that $r^p \in K$, the Galois Group of $g(x)$ (the irreducible factor of $g(x)$ containing $t$ as a root) over $K(r)$ is $H$.

Before we prove this statement, we explore the consequences of having a Normal Subgroup of size $1/p$ times the Group.

Quotient Group

For any normal subgroup $H$ of $G$, the cosets form a group defined by:

\[(Ha) (Hb) = (Ha) (bH) = H(abH) = H(Hab) = Hab\]

This is represented by the Quotient Group $G/H$. The size of the Quotient Group $=$ number of cosets $= \text{size}(G)/\text{size}(H) = p$.

Each coset corresponds to permutations that map $t$ to some other roots of some specific factor of $g(x)$. Thus, the quotient group can be seen as permutations that map one factor of the $g$ to another factor of $g$.

Quotient Group is Cyclic

For $\sigma \in G/H$, we can form the subgroup:

\[\{e, \sigma, \sigma^2,\ldots,\sigma^{k-1}\}\]

where $k$ is the smallest number such that $\sigma^k = e$. This subgroup is called the cyclic subgroup generated by $\sigma$ and has size $k$. Quotient Group is of size $p$. So, $k$ divides $p$ implying $k=1$ or $k=p$.

$k=1$ is not possible, since the subgroup has atleast two elements $e$ and $\sigma$.

$k=p$ implies that the entire group is the subgroup. This means that the entire group ends up becoming cyclic, irrespective of the choice of $\sigma$ (as long as, it is not the identity element).

For our purpose, let us represent the cosets as $H$, $\sigma H$, $\sigma^2 H$, $\ldots$, $\sigma^{p-1} H$, where $\sigma \in G, \sigma \notin H$. It does not matter which $\sigma$ we choose. We can still represent the quotient group using the abovce formulation. As a consequence, any permutation in $G$ can be represented as $\sigma^i h$, where $h \in H$.

Proof of Solvability

Here is a sketch of the proof of the converse statement, inspired from Lagrange Resolvents. Given $H$, we can apply the permutations in $H$ on $t$ to obtain $t_1$, $t_2$, $\ldots$, $t_m$ with $mp = n = \text{deg}(g)$. We can obtain the polynomial $h(x) = (x-t_1)(x-t_2)\ldots(x-t_m)$. The coefficients of this polynomial cannot be in $K$, since the irreducible polynomial containing $t$ is of degree $n = mp$. The coefficients of this polynomial are candidates in $K(r)$. Let us choose one such candidate and call it $\theta_1$.

Consider any permutation $\sigma \in G, \sigma \notin H$. Let $\sigma(\theta_1) = \theta_2$. If $\theta_2 = \theta_1$ for all $\sigma \in G$, then $\theta_1 \in K$, which is not possible. Hence, there is some $\sigma$ for which $\sigma(\theta_1) \neq \theta_2$. For this $\sigma$, we can define the Quotient Group containing $p$ cosets. We can pick the representative permutations: ${e, \sigma, \sigma^2, \ldots, \sigma^{p-1}}$. Applying these on $\theta_1$, we get: $\sigma^{i}(\theta_1) = \theta_{i+1}$.

We know that $\sigma^p \in H$ and hence $\sigma^p(\theta_1) = \theta_1$. But if $\sigma^i(\theta_1) = \theta_1$ for some $1 < i < p$, then we have $ai = pb + 1$ for some positive integers $a,b$.

\[\begin{align*} \theta_1 &= \sigma^i(\theta_1) = \sigma^{2i}(\theta_1) = \ldots = \sigma^{ai}(\theta_1)\\ &= \sigma^{1 + pb}(\theta_1)=\sigma^{1 + p(b-1)}(\theta_1)=\ldots=\sigma^{1}(\theta_1)\\ &= \sigma(\theta_1) \end{align*}\]

However, we have already shown this is not possible. Thus, all the $\theta_i\text{s}$ are distinct.

Consider:

\[r_1 = \theta_1 + \alpha\theta_2 + \alpha^2\theta_3 + \ldots + \alpha^{p-2}\theta_{p-1}+ \alpha^{p-1}\theta_p\]

We could apply $\sigma$ on both sides to get:

\[\begin{align*} \sigma(r_1)&=\theta_2 + \alpha\theta_3 + \alpha^2\theta_4 + \ldots + \alpha^{p-2}\theta_p + \alpha^{p-1}\theta_1\\ &=r_1/\alpha \end{align*}\]

Any permutation in $G$ can be represented as $\sigma^ih$. $h$ leaves $\theta_i\text{s}$ unchanged.

\[\sigma^i(h(r_1)) = \sigma^i(r_1)=r_1/\alpha^i\]

If we apply the same permutation on $r_1^p$, we get:

\[\sigma^i(h(r_1^p)) = (r_1/\alpha^i)^p = r_1^p\]

Thus, we have shown that all permutations in $G$ fix $r_1^p$. Thus $r_1^p \in K$.

Note: The above proof works except when $r_1 = 0$. To make sure that it does not happen, consider:

\[r_1 = \theta_1 + \alpha\theta_2 + \alpha^2\theta_3 + \ldots + \alpha^{p-1}\theta_p\] \[r_2 = \theta_1 + \alpha^2\theta_2 + \alpha^4\theta_3 + \ldots + \alpha^{2(p-1)}\theta_p\] \[r_3 = \theta_1 + \alpha^3\theta_2 + \alpha^6\theta_3 + \ldots + \alpha^{3(p-1)}\theta_p\] \[\vdots\] \[r_{p-1} = \theta_1 + \alpha^{p-1}\theta_2 + \alpha^{2(p-1)}\theta_3 + \ldots + \alpha^{(p-1)(p-1)}\theta_p\]

Atleast one of these has to be non-zero. If not, we can sum it all up to get:

\[0 = p\theta_1 - \theta_1 - \theta_2 - \theta_3 \ldots - \theta_p\] \[p\theta_1 = \theta_1 + \theta_2 + \theta_3 +\ldots + \theta_p\]

We could apply $\sigma$ on both sides to get: $\theta_1 = \theta_2 = \ldots = \theta_p$, which is not true. Thus, if $r_1 = 0$, we can use any one of $r_2,r_3,\ldots, r_{p-1}$ in the place of $r_1$. All of them cannot be zero.

We have shown that there is a $r$ such that $r^p \in K$. If we were to extend $K$ into $K(r)$ and attempt factorizing $g(x)$, its Galois Group would be either $G$ or a normal subgroup of $G$ of size $1/p$.

If the Galois Group is $G$, then $\sigma(r)=r$ for all $\sigma \in G$. But we clearly have a $\sigma$ that violates this condition: the one used in the proof above.

Let us call the new Galois Group as $N$. Let us choose some $n \in N$. Then $n = \sigma^i h$ with $n(r) = r$. This means

\[\sigma^i(h(r)) = r\]

But

\[\sigma^i(h(r))=r/\alpha^i\]

This can happen if and only if $i=0$, which implies $n = \sigma^ih = h \in H$. This shows that $N \subseteq H$. But, $N$ and $H$ are of the same size. Thus $N = H$.

This shows that the Galois Group of $g(x)$ (after factorization) over $K(r)$ is $H$.

Conclusion

We have shown that the process of solving a polynomial by radicals can be understood entirely through the structure of its Galois group. Each time the base field $K$ is extended by adjoining an element $r$ with $r^p \in K$, the corresponding resolvent polynomial factors in a highly structured way, and the Galois group shrinks to a normal subgroup of size $1/p$ of the original Galois Group. Conversely, the existence of a normal subgroup guarantees the existence of a field extension obtained by adjoining a $p^{th}$ root. Following this trail of Galois Groups from that of the original polynomial to a group of only identity element, we can derive a way to solve the polynomial using basic algebraic operations and radicals.