Exploration 9: Solvability

December 21, 2023.

Questions:

When exactly does a formula for the roots of a polynomial exist?

Recall that we discussed quadratic extensions and cyclotomic extensions. What are their degrees and Galois groups?

For quadratic extensions, it is very straightforward. Since all quadratic extensions are simple extensions by a root of a degree $2$ polynomial, every quadratic extension is degree $2$ , with Galois group $C_2$ .

What about cyclotomic extensions? Recall that the $n$ th cyclotomic extension is created by adjoining a primitive $n$ th root of unity. Since the minimal polynomial of such a root is the $n$ th cyclotomic polynomial, whose roots are the $\varphi(n)$ primitive $n$ th roots of unity, the degree of a cyclotomic extension is $\varphi(n)$ with a cyclic Galois group $\ZZ/n\ZZ^\times$ of order $\varphi(n)$ .

Now let’s move on to cubics.

Solving cubic polynomials

Cubic polynomials have the form $x^3+a_2x^2+a_1x+a_0$ . That’s a lot, but we can eliminate the $a_2x^2$ term with the substitution $x\mapsto x-\frac{a_2}{3}$ (the depressed cubic substitution):

$\begin{aligned} &~(x-\frac{a_2}{3})^3+a_2(x-\frac{a_2}{3})^2+a_1(x-\frac{a_2}{3})+a_0\\ =&~(x^3-\frac{3a_2x^2}{3}+\frac{3a_2^2x}{9}-\frac{a_2^3}{27})+a_2(x^2-\frac{2xa_2}{3}+\frac{a_2^2}{9})+a_1(x-\frac{a_2}{3})+a_0\\ =&~x^3-a_2x^2+\frac{a_2^2x}{3}-\frac{a_2^3}{27}+a_2x^2-\frac{2a_2^2x}{3}+\frac{a_2^3}{9}+a_1x-\frac{a_1a_2}{3}+a_0\\ =&~x^3+(a_1-\frac{a_2^2}{3})x+\frac{2a_2^3}{27}-\frac{a_1a_2}{3}+a_0\\ =&~x^3+\frac{3a_1-a_2^2}{3}x+\frac{2a_2^3-9a_1a_2+27a_0}{27}\\ \end{aligned}$

Thus we can always express cubic polynomials as $x^3+a_1x+a_0$ . In the splitting field, we have $x^3+a_1x+a_0=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$ Expanding the RHS symmetric polynomial gives us

$\begin{aligned} \alpha_1+\alpha_2+\alpha_3&=0\\ \alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha_3&=a_1\\ \alpha_1\alpha_2\alpha_3&=-a_0 \end{aligned}$

The first equation, $\alpha_1+\alpha_2+\alpha_3=0$ , shows that $\alpha_3$ is generated by $\alpha_1,\alpha_2$ . Thus the splitting field $K/F$ can be written $F(\alpha_1,\alpha_2)$ if $\alpha_1$ does not generate $\alpha_2$ , and $F(\alpha_1)$ otherwise.

So there’s a tower of fields:

Consider $f=(x-\alpha_1)h$ over $F(\alpha_1)$ . Either $h$ splits (so $\alpha_1$ generates $\alpha_2$ ) and the top two fields are equal, or $h$ doesn’t split and is irreducible (so $\alpha_2$ is degree $2$ , since $h$ is quadratic).

This means the degree of the splitting field of a cubic is either $[K:F]=3$ or $[K:F]=6$ .

Splitting fields of a polynomial over $F$ are Galois extensions over $F$ . Since the degree of this splitting field is either $3$ or $6$ , and the Galois group must consist of permutations of the three roots, $G(K/F)$ must be a subgroup of the symmetric group $S_3$ , so it must be either the dihedral group $S_3$ or the cyclic group $C_3$ .

In conclusion: while the splitting fields of quadratics are always degree $2$ with Galois group $C_2$ , splitting fields of cubics are either degree $3$ or $6$ , with Galois group either $S_3$ or $C_3$ .

Solving quartic polynomials

There’s some dark magic involving nested square roots and factoring into quadratics, so I’ve ignored this bit. Just know that for any quartic polynomial $f$ , you can define a resolvent cubic $g$ and a discriminant $D$ , which together determine the Galois group.

It turns out that the Galois group $G$ for splitting fields of quartic polynomials are also subgroups of $S_4$ for the same reason (they permute the four roots). $S_4$ is order $24$ , so there are a lot more than just two subgroups like in the cubic case.

The full breakdown is as follows:

If $g$ is irreducible, then $G\iso A_4$ when $D$ is a rational square, and $G\iso S_4$ otherwise.
If $g$ splits partially into a linear times quadratic, then $G\iso D_4$ when $f$ is irreducible over $F(\sqrt{D})$ , and $G\iso C_4$ otherwise.
If $g$ splits completely, then $G\iso K_4$ .

Again, I won’t get into quartics too much, since the important one is quintics. Just know that we can solve quartics similarly to how we solved cubics.

Solving quintic polynomials

Solving quintic (degree $5$ ) equations was the original goal of Galois.

Like with the other cases, the Galois group of the splitting field is a subgroup of $S_5$ . It turns out $S_5$ is not solvable in the sense of $S_4$ and $S_3$ .

In this section, we describe solvability and what it means.

A root $\alpha\in\CC$ is solvable over $F$ if you can use radicals (like square roots, $n$ th roots, nested roots) in $F$ to write $\alpha$ .

Motivation: Galois showed that the roots $\alpha\in\CC$ of certain irreducible $f\in\QQ[x]$ of degree $5$ are not solvable.

For instance, take the root $\sqrt[5]{10}$ of $x^5-10\in\QQ[x]$ . It is solvable over $\QQ$ since we expressed it as a radical over $\QQ$ . In other words, $\QQ\subseteq\QQ(\sqrt[5]{10})=\QQ(\alpha)$ .

Another example: take the root $\sqrt[5]{1+\sqrt[3]{2}}$ of $(x^5-1)^3-2\in\QQ[x]$ . It is solvable over $\QQ$ since we could express it in radicals over $\QQ$ . In other words, $\QQ\subseteq\QQ(\sqrt[3]{2})\subseteq\QQ(\sqrt[5]{1+\sqrt[3]{2}})$ .

In general, to prove $\alpha$ is solvable over $F$ , we prove there’s a chain of fields $F\subseteq F_1\subseteq F_2\subseteq F_3\subseteq\ldots\subseteq K$ such that (conditions $A$ ):

$\alpha\in K$ (root is in the final extension)
$\forall i\ldotp\exists\alpha_i\in F_i\ldotp F_i=F_{i-1}(\alpha_i)$ (each extension is built by adjoining to previous extension)
$\forall i\ldotp\exists m>0\ldotp\alpha_i^m\in F_{i-1}$ (adjoined element is a $m$ th root of an element in the previous extension)

We shall prove that to prove the above, it is enough to prove the following:

$\alpha$ is solvable over $F$ if there’s a chain of subfields $F\subseteq F_1\subseteq F_2\subseteq F_3\subseteq\ldots\subseteq K$ such that (conditions $B$ ):

$\alpha\in K$ (root is in the final extension)
$\forall i\ldotp F_i/F_{i-1}$ is Galois of prime degree

Theorem: Every Galois extension

K/F

of prime degree

p

can be broken down into a cyclotomic and a Kummer extension.

Since $K$ is Galois and thus normal, $K$ is a splitting field for $x^p-a$ for some $a\in F$ , which implies $K$ contains the $p$ th roots of unity.
Let $L=F(\zeta_p)$ so that $L/F$ is a cyclotomic extension that contains the $p$ th roots of unity, so that $L$ is a subfield of $K$ that extends $F$ , i.e. $K\supseteq L\supseteq F$ .
Then since $K/F$ is Galois, $K/L$ is also Galois. Since $L$ contains the $p$ th roots of unity, and $K/L$ is Galois, by Kummer’s theorem, $K/L$ is a Kummer extension of degree $p$ .

Since both cyclotomic and Kummer extensions are obtained by adjoining some root of an element in their base field, this means every Galois extension is equivalent to extensions that satisfy conditions $A$ . Thus, given a chain that satisfies conditions $B$ , we can construct a chain that satisfies conditions $A$ .

Because of the Fundamental Theorem of Galois Theory, towers of field extensions $F\subseteq F_1\subseteq F_2\subseteq F_3\subseteq\ldots\subseteq K$ correspond to chains of subgroups. In particular, towers of Galois extensions correspond to chains of normal subgroups:

Theorem: Every intermediate Galois extension

L_1/L_2

corresponds to a normal subgroup relation

H_1\lhd H_2

in the original Galois group

G(K/F)

( $\to$ $\to$ )
- According to the fundamental theorem of Galois theory, $L_1/L_2$ indeed corresponds to a subgroup relation $H_1\le H_2$ between unique subgroups $H_1$ and $H_2$ . To prove that this is a normal subgrouping, we use the fact that $L_1/L_2$ is Galois, and therefore normal.
- Since $H_1=G(K/L_1)$ and $H_2=G(K/L_2)$ under the correspondence, we can see that the elements of $H_1$ are $L_1$ -automorphisms and the elements of $H_2$ are $L_2$ -automorphisms. In particular, the subgrouping $H_1\le H_2$ implies that all $L_1$ -automorphisms are $L_2$ -automorphisms.
- Recall that in normal extensions like $L_1/L_2$ , $L_2$ -automorphisms are invariant under conjugation. Thus, the elements of $H_1$ , being $L_2$ -automorphisms, are invariant under conjugation, implying that $H_1\lhd H_2$ .
( $\from$ $\leftarrow$ )
- Say you have $H_1\lhd H_2$ . Again, $H_1=G(K/L_1)$ contains $L_1$ -automorphisms $\tau$ and $H_2=G(K/L_2)$ contains $L_2$ -automorphisms $\sigma$ under the correspondence. The normal subgrouping implies that $\sigma^{-1}\tau\sigma$ is also a $L_1$ -automorphism in $H_1$ .
- Let $\alpha\in L_1$ be the root of some polynomial $f$ irreducible over $L_2$ . Since $G(K/L_2)=H_2$ permutes the roots of $f$ , $\sigma(\alpha)$ represents an arbitrary conjugate of $\alpha$ . To show $L_1/L_2$ is normal, we just need to show that $\sigma(\alpha)$ is in $L_1$ , implying that $f$ splits in $L_1$ , and thus $L_1/L_2$ is normal.
- But since $\sigma^{-1}\tau\sigma$ fixes $L_1$ , it must fix $\alpha\in L_1$ . So we have $\sigma^{-1}\tau(\sigma(\alpha))=\alpha$ , implying $\tau(\sigma(\alpha))=\sigma(\alpha)$ , implying that $\tau$ fixes $\sigma(\alpha)$ . But $\tau$ is a $L_1$ -automorphism, meaning elements it fixes are in $L_1$ .

This implies:

Theorem: Every Galois extension

K/F

whose Galois group is abelian can be broken down into cyclotomic and Kummer extensions.

If $G(K/F)$ is abelian, every subgroup $H$ is normal, and by the above theorem, must correspond to an intermediate Galois field extension $K/L$ .
If $G(K/F)$ is abelian, then it decomposes into a direct product of cyclic groups. By the Chinese Remainder Theorem we can decompose those cyclic groups into a direct product of cyclic groups of prime order. Write this decomposition in any order: $G(K/F)\iso C_{p_1}\times C_{p_2}\times\ldots\times C_{p_k}$ Then we can construct a chain of normal subgroups (where each quotient is of prime order) by incrementally popping off the rightmost factor: $\begin{aligned} \{1\}~~\lhd~~ C_{p_1}~~\lhd~~ C_{p_1}\times C_{p_2}~~\lhd~~\ldots~~\lhd~~ C_{p_1}\times C_{p_2}\times\ldots\times C_{p_k} \end{aligned}$
These subgroups all correspond to a prime degree intermediate Galois field extension, and we can break each of those down into a cyclotomic and Kummer extension.

This gives us conditions $C$ :

$\alpha$ is solvable over $F$ if there’s a chain of subfields $F\subseteq F_1\subseteq F_2\subseteq F_3\subseteq\ldots\subseteq K$ such that (conditions $C$ ):

$\alpha\in K$ (root is in the final extension)
$\forall i\ldotp F_i/F_{i-1}$ is Galois with an abelian Galois group

Finally, we can convert this into the realm of group theory using the fact that intermediate Galois extensions correspond to normal Galois subgroups. This gives us conditions $D$ :

$\alpha$ is solvable over $F$ if there’s a chain of normal subgroups $\{1\}\lhd\ldots\lhd G(K/L_2)\lhd G(K/L_1)\lhd G(K/F)$ such that (conditions $D$ ):

$\alpha\in K$ (root is in the final extension)
$\forall i\ldotp |G(K/L_i)|/|G(K/L_{i+1})|$ is abelian

Using this final version of conditions for solvability, we can prove some basic theorems about solvability.

Theorem: Subgroups of solvable groups are solvable.

Let $H$ be said subgroup and let $G_i$ be an element in the chain that proves the group solvable. Then in each $G_i$ of the chain, just take the intersection $H_i=H\cap G_i$ . This just gets rid of some elements in each $G_i$ , which doesn’t change the condition “its elements commute with everything” in the definition of normal subgroup and abelian quotient. So each subgroup relation in the chain is still normal, and each quotient is still abelian.

Theorem: Non-abelian simple groups are not solvable.

If $G$ is simple, then the only proper normal subgroup is $\{1\}$ , meaning our chain can only look like $\{1\}\lhd G$ . But $G/\{1\}$ is non-abelian, and thus the only possible chain does not satisfy conditions $D$ , thus not solvable.

In this section, we prove that degree $5$ polynomials are not solvable.

Note that all degree $5$ polynomials have $5$ roots, and thus the Galois group is at maximum $S_5$ , the symmetric group for five elements. This means all it takes to show not all degree $5$ polynomials are solvable is to prove that $S_5$ is not solvable.

Theorem:

S_5

is not solvable.

We know that $A_n$ is a subgroup of $S_n$ by definition. That means if $S_5$ is solvable, then $A_5$ would be solvable. However, $A_5$ is simple, and also not abelian since $(1~2~3)(2~3~4)=(1~2)(3~4)$ but $(2~3~4)(1~2~3)=(1~3)(2~4)$ , and is therefore is not solvable. Thus, $S_5$ cannot be solvable.