I have began understanding Galois theory and I had a question regarding the relationships of normal subgroups to field extensions.

So given an irreducible polynomial over the rationals $$a_1 + a_2x +\cdots+ a_nx^{n-1} + x^n $$ the initial group of symmetries of the roots $r_1,\ldots,r_n$ is the set of permutations of roots that uphold

$$ \begin{pmatrix} r_1 + r_2 +\cdots+ r_n = -a_n \\ r_1r_2 +r_1r_3 +\cdots+ r_{n-1}r_n = a_{n-1} \\ \vdots \\ r_1r_2\cdots r_n = (-1)^n a_1 \end{pmatrix} $$ which of course is the symmetric group $S_n$.

Now if we take the field $\mathbb{Q}$ and extend it by one of the roots

$$ \Bbb{Q} \rightarrow \Bbb{Q}(r_1)$$

The equations above obviously reduce to a smaller collection

$$ \begin{pmatrix} r_2 + \cdots+ r_n = -a_n -r_1 \\ r_2r_3 + \cdots + r_{n-1}r_n = -a_{n-1} - r_1{a_n}+r_1^2\\ \vdots \\ r_2 r_3 \cdots r_n = (-1)^n \frac{a_1}{r_1} \end{pmatrix} $$

Which is preserved by the group $S_{n-1}$ of permutations of the roots.

So this sequence of permutation groups converges to the identity once enough field extensions have occurred.

$$ G \rightarrow G' \rightarrow G'' \rightarrow\cdots \rightarrow e $$

$$ \Bbb{Q} \rightarrow \Bbb{Q}(r_1) \rightarrow \Bbb{Q}(r_1)(r_2) \rightarrow\cdots\rightarrow \Bbb{Q}(\text{all roots}) $$

And I understand that $$G' \subset G, \qquad G'' \subset G',\qquad\ldots $$ In other words each of those groups is a subgroup of its predecessor. But, it is not immediately clear to me why they need to be normal subgroups.

Could someone explain that?

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    $\begingroup$ The Galois group of an irreducible polynomial of degree $n$ need not be the entire group $S_n$ but just a subgroup. What makes you suspect that the chain of subgroups needs to be normal subgroups? $\endgroup$ – Taylor Jul 7 '15 at 14:33
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    $\begingroup$ Well, if you have a splitting field of multiple irreducible polynomials, then the Galois group is a subgroup of the direct product of the Galois groups of the individual irreducibles, each of which will be normal in the whole group. $\endgroup$ – Taylor Jul 7 '15 at 14:50
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    $\begingroup$ Note that for a field extension $F/K$ and intermediate field $L$, $G(L/K)$ need not be a subgroup of $G(F/K)$. $G(F/L)$ is a subgroup, and if it is normal in $G(F/K)$, then $G(F/K)/G(F/L)\cong G(L/K)$. $\endgroup$ – Taylor Jul 7 '15 at 15:04
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    $\begingroup$ I can give a brief explanation: consider the mapping $\pi : G(F/K)\rightarrow G(L/K)$ defined by restriction, i.e., $\alpha \in G(F/K)$ is mapped to $\alpha|_L$. Show that $\pi$ is well defined ($\alpha(x)\in L$ for all $x\in L$) if and only if $L/K$ is Galois (hence if $G(F/L)$ is normal). In this case, $\pi$ is an epimorphism with kernel $G(F/L)$. $\endgroup$ – Taylor Jul 7 '15 at 16:36
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    $\begingroup$ It means restricting the domain to $L$. So $\alpha |_L : L\rightarrow F$ whereas $\alpha: F \rightarrow F$ and both map $x$ to $\alpha(x)$. $\endgroup$ – Taylor Jul 7 '15 at 16:42

What you're asking about is false in general.

Let's set $F_0=\mathbb{Q}$ and $F_k=\mathbb{Q}(r_1,\ldots,r_k)$, so that $F_n=\mathbb{Q}(\text{all roots})$.

Now, $F_n/L$ is Galois for any field $\mathbb{Q}\subseteq L\subseteq F_n$ because $F_n$ is a splitting field. In particular, $F_n/F_k$ is a Galois extension for any $0\leq k\leq n$. Let's use proper notation to keep everything clear, and set $G_k=\mathrm{Gal}(F_n/F_k)$. Then we do have $$G_0\supseteq G_1\supseteq\cdots\supseteq G_n=\text{trivial group}$$ but there is no reason why we would have $G_k\trianglerighteq G_m$ for any $k<m$ except for $k=0$ and $m=n$ (or rather, $m=n$ and any $k$ for which $F_k=F_n$). Certainly, the only case when we'd have $G_0\trianglerighteq G_m$ — that is, the only case when we'd have $\mathrm{Gal}(F_n/\mathbb{Q})\trianglerighteq\mathrm{Gal}(F_n/F_m)$ — is when $F_m/\mathbb{Q}$ is a Galois extension, which can't happen unless $F_m=F_n$, i.e., unless $F_m$ is the splitting field of the irreducible polynomial we started with.

For example, consider the polynomial $x^3-2$ with roots $$r_1=\sqrt[3]{2},\qquad r_2=\zeta_3\sqrt[3]{2},\qquad r_3=\zeta_3^2\sqrt[3]{2}$$ Then we have $G_0=\mathrm{Gal}(\mathbb{Q}(\zeta_3,\sqrt[3]{2})/\mathbb{Q})\cong S_3$, but $G_1=\mathrm{Gal}(\mathbb{Q}(\zeta_3,\sqrt[3]{2})/\mathbb{Q}(\sqrt[3]{2}))$ is a subgroup of order $2$ in $S_3$, which will not be a normal subgroup.

  • $\begingroup$ Then I think I'm having trouble understanding, en.wikipedia.org/wiki/…. Since there it is given that field extensions correspond to factor groups derived from the composition series of initial Group of Permutations, which I thought was a sequence of normal subgroups $\endgroup$ – frogeyedpeas Jul 7 '15 at 14:45
  • $\begingroup$ Chronoles, I'm trying to understand very specifically what structure the subgroups must have. Since the proof of the unsolvability of the Quintic was written as $A_5$ is a simple normal, non abelian subgroup of $S_5$ therefore it doesn't correspond to a field extension and thus some 5th degree equations are unsolvable. If $A_5$ was not normal but still not abelian does that imply that 5th degree equations are unsolvable? In other words, is the fact it is normal, just extraneous information that was unncessary to mention, OR if it was necessary to mention, why? $\endgroup$ – frogeyedpeas Jul 7 '15 at 15:06
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    $\begingroup$ That is a poorly worded Wikipedia entry; it's extremely vague what field extensions they're talking about, mainly because they aren't using any names for the relevant objects. So let's set $E/F$ to be some finite Galois extension, with $G=\mathrm{Gal}(E/F)$. Then since any finite group has a composition series, we can find a composition series for $G$, $$1 = H_0\triangleleft H_1\triangleleft \cdots \triangleleft H_n = G$$ which corresponds to a sequence of fields $$E=L_0\supset L_1\supset\cdots\supset L_n=F$$ where $H_k=\mathrm{Gal}(E/L_k)$. $\endgroup$ – Zev Chonoles Jul 7 '15 at 15:09
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    $\begingroup$ Then yes, it is true that $H_{k+1}/H_k\cong\mathrm{Gal}(L_k/L_{k+1})$. However, there is no reason why it would be true that the fields $L_k$ are formed by adjoining the roots of a polynomial in the way you asked about in your post. $\endgroup$ – Zev Chonoles Jul 7 '15 at 15:09
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    $\begingroup$ Okay so $H_{k+1}/H_K $ is isomorphic to $Gal(L_k/L_{k+1})$ .From here I have the following. I read somewhere that if for all $k= 0 ... n-1$ $H_{k+1}/H_k$ is cyclic then the Galois Group, $Gal(E/F)$ is solvable. Why must this be the case? $\endgroup$ – frogeyedpeas Jul 18 '15 at 3:50

The link between (normal) subgroups and field subextensions of a Galois extension $L$, with Galois group $G$ is there is a bijection between subgroups of $G$ and subextensions of $L$, defined by $\;H\mapsto L^H$ (the fixed points of $L$ under the action of $H$).

Furthermore $L^H$ is Galois if and only if $H$ is a normal subgroup of $G$; in which case the Galois group of $L^H$ over $\mathbf Q$ is the quotient group $G/H$.

  • $\begingroup$ I don't follow what the notation $L^H$ means, whereas H appears to be a group and $L$ is a field? $\endgroup$ – frogeyedpeas Jul 7 '15 at 15:01
  • $\begingroup$ @frogeyedpeas: Bernard defines it in the post, and it is standard notation for Galois theory - you've never seen it? Given a Galois extension $L/F$, and a subgroup $H\subseteq\mathrm{Gal}(L/F)$, the definition is $$L^H=\{\alpha\in L:\sigma(\alpha)=\alpha\text{ for all }\sigma\in H\}$$ $\endgroup$ – Zev Chonoles Jul 7 '15 at 15:04
  • $\begingroup$ Let me add (this was implicit in my post) that $L^H$ is a subfield of $L$ that contains $\mathbf Q$. $\endgroup$ – Bernard Jul 7 '15 at 15:19

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