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I watched a few videos about the solvability of a general quintic in radicals and I'm somewhat confused about a few concepts. My main confusion lies in the following definition;

$\textbf{Def}:$ Solvable group

A group $G$ is said to be solvable if it has a subnormal series $$ G_0 \triangleleft G_{1} \triangleleft \cdots \triangleleft G_{n-1} \triangleleft G_n$$ with $G_n = G$ and $G_0 = \left\lbrace e\right\rbrace$ such that each successive quotient group $G_i/G_{i-1}$ is abelian.

Why is the notion of "solvability" attached to this property?

More precisely, if I have a polynomial $p(x) \in F[x]$ with $F$ a field and Galois Group $\operatorname{Gal}(p)$, in an "explain it like I'm stupid"-sense, what is it about $\operatorname{Gal}(p)$ having these seemingly arbitrary properties that forces $p$ to have solutions in radicals?

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If the Galois group $ G $ of a finite Galois extension $ L/K $ where $ K $ has characteristic $ 0 $ is solvable, then letting $ n = [L:K] $, we may adjoin a primitive $ n $th root of unity $ \zeta_n $ to $ K $ to get the extension $ L(\zeta_n)/K(\zeta_n) $. This extension is also Galois, and its Galois group $ G' $ embeds into $ G $ by restricting automorphisms to $ L $. Since subgroups of solvable groups are solvable, it follows that $ G' $ is also solvable. Now, by first refining the composition series so that each successive quotient is cyclic, and then converting this into a chain of subfields by Galois correspondence, we get a chain

$$ K(\zeta_n) = L_0 < L_1 < L_2 \ldots < L_k = L(\zeta_n) $$

where each extension $ L_{i+1}/L_i $ is Galois with cyclic Galois group, and degree dividing $ n $. By Kummer theory, such extensions are obtained by adjoining the $ n $th root of some element in the ground field (this is why we adjoined $ \zeta_n $ before going through with this argument), thus each extension $ L_{i+1}/L_i $ is a simple radical extension, and the extension $ L(\zeta_n)/K(\zeta_n) $ is a radical extension. Since $ K(\zeta_n)/K $ is also a radical extension, it follows that $ L $ lies in some radical extension of $ K $. The converse direction is similar.

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  • $\begingroup$ Thanks a lot and congrats on 10k ;) $\endgroup$ – ÍgjøgnumMeg Apr 19 '17 at 8:01
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Starfall's answer is brilliant, as always! In addition to Starfall's answer, can I share a concrete example that I personally found helpful when I was learning this subject?

Consider the polynomial $X^n - 2$ over $\mathbb Q$. The splitting field is $\mathbb Q(\zeta_n, \sqrt[n]{2})$.

There is a sequence of normal field extensions: $$ \mathbb Q \subset \mathbb Q (\zeta_n) \subset \mathbb Q(\zeta_n , \sqrt[n]{2}).$$

Correspondingly, there is a sequence of group inclusions: $$ 1 \ \lhd \ {\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q(\zeta_n)) \ \lhd \ {\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q).$$

Now, notice that:

  • ${\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q(\zeta_n))$ is abelian, because it is a subgroup of the additive cyclic group $\mathbb Z_n$. Indeed, all automorphisms in ${\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q(\zeta_n))$ send $\sqrt[n]{2} \mapsto \zeta_n^a \sqrt[n]{2}$ for some $a \in \mathbb Z_n$.

  • The quotient ${\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q) / {\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q(\zeta_n))$ is isomorphic to ${\rm Gal}(\mathbb Q (\zeta_n) : \mathbb Q)$, by the Galois correspondence. This is also abelian, because it is isomorphic to $\mathbb Z_n^\times$, the multiplicative group of units modulo $n$: the automorphisms in ${\rm Gal}(\mathbb Q (\zeta_n) : \mathbb Q)$ are of the form $\zeta_n \mapsto \zeta_n^b$ for $b \in \mathbb Z_n^\times$.

Thus we have explicitly verified that ${\rm Gal}(\mathbb Q(\zeta_n, \sqrt[n]{2} : \mathbb Q)$ is a solvable group!

For more complicated examples, we may need to consider a longer sequence of field extensions. (We build our sequence by successively adjoining an $n$th root of an element of a field previously constructed, so the length of the sequnece will depend on how many times we need to adjoin $n$th roots until we've constructed the splitting field of the polynomial.) We may even wish to adjoin all $n$th roots of unity by hand, at the very beginning. But whatever example we come up with, the idea will always be fundamentally the same as in the example we just worked through: in the corresponding sequence of group inclusions, all quotients will be subgroups of the additive group $\mathbb Z_n$, except for the final one which will be isomorphic to ${\rm Gal}(\mathbb Q(\zeta_n):\mathbb Q)$, the multiplicative group of units modulo $n$.

Of course, this answer only addresses the "solvable by radicals" $\implies$ "solvable Galois group" implication. As Starfall explained, the reverse implication works too - and that properly addresses your original question. Nonetheless, I do find the forward implication more intuitive!

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