# Why restriction to $B(\alpha)$ is a homomorphism from $Gal(E/B)$ to a group with kernel $Gal(E/B(\alpha))$?

I'm reading Galois Theory for Beginners by John Stillwell. It's a good introduction, giving the essence of the idea with minimum algebra complexity.

However, I'm a bit lost at his Theorem 2 (the details are at the end).

The proof intend to find a homomorphism of $\texttt{Gal}(E/B)$, with kernel $\texttt{Gal}(E/B(\alpha))$, into an abelian group. John says:

The obvious map with kernel $\texttt{Gal}(E/B(\alpha))$ is restriction to $B(\alpha)$, $\lvert_{B(\alpha)}$, since by definition $$\sigma \in \texttt{Gal}(E/B(\alpha)) \Leftrightarrow \sigma \lvert_{B(\alpha)} \texttt{ is the identity map}.$$

Why restriction to $B(\alpha)$ is a homomorphism from $\texttt{Gal}(E/B)$ to a group with kernel $\texttt{Gal}(E/B(\alpha))$? What does this "restriction" mean?

Trying to understand it, I'm taking $B = \mathbb Q$, $\alpha = \zeta$, $B(\alpha) =\mathbb Q(\zeta)$, $E = \mathbb Q (\zeta, \sqrt{2})$, where $\zeta^5 = 1$.

So $$\texttt{Gal}(E/B(\alpha)) = \texttt{Gal}(\mathbb Q(\zeta, \sqrt{2})/\mathbb Q(\zeta)),$$ in this example it is

$\begin{array}{c|cc} \mathbb Q(\zeta, \sqrt{2}) & \sigma_1 & \sigma_2 \\ \hline \sqrt{2} & \sqrt{2} & -\sqrt{2} \\ -\sqrt{2} & -\sqrt{2} & \sqrt{2} \\ \zeta & \zeta & \zeta \\ \zeta^2 & \zeta^2 & \zeta^2 \\ \zeta^3 & \zeta^3 & \zeta^3 \\ \zeta^4 & \zeta^4 & \zeta^4, \end{array}$

so it's isomorphism to $$S_2 = \{(1), (1,2)\}.$$

Then $$\texttt{Gal}(E/B) = \texttt{Gal} (\mathbb Q(\zeta, \sqrt{2})/\mathbb Q),$$ in this example it is

$\begin{array}{c|cccccccc} \mathbb Q(\zeta, \sqrt{2}) & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 & \sigma_5 & \sigma_6 & \sigma_7 & \sigma_8 \\ \hline \sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2} & -\sqrt{2} & -\sqrt{2} & -\sqrt{2} & -\sqrt{2}\\ -\sqrt{2} & -\sqrt{2} & -\sqrt{2} & -\sqrt{2} & -\sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2} & \sqrt{2}\\ \zeta & \zeta & \zeta^2 & \zeta^3 & \zeta^4 & \zeta & \zeta^2 & \zeta^3 & \zeta^4 \\ \zeta^2 & \zeta^2 & \zeta^4 & \zeta & \zeta^3 & \zeta^2 & \zeta^4 & \zeta & \zeta^3 \\ \zeta^3 & \zeta^3 & \zeta & \zeta^4 & \zeta^2 & \zeta^3 & \zeta & \zeta^4 & \zeta^2 \\ \zeta^4 & \zeta^4 & \zeta^3 & \zeta^2 & \zeta & \zeta^4 & \zeta^3 & \zeta^2 & \zeta, \end{array}$

so it's isomorphism to $$\{(1), (3,4,5,6), (3,5,6,4), (3,6)(4,5), (1,2), (1,2)(3,4,5,6), (1,2)(3,5,6,4), (1,2)(3,6)(4,5) \}.$$

My guess is, so that the "restriction to $B(\alpha)$" actually means a subset of $\texttt{Gal} (E/B)$ that only change elements in $B(\alpha)$, so itcan be defined as $$\Sigma := \{ \tau \in \texttt{Gal}(E/B) \lvert \tau(\beta) = \beta, \forall \beta \in E \setminus B(\alpha) \},$$ in this example

$\begin{array}{c|c} \mathbb Q(\zeta, \sqrt{2}) & A = f(\texttt{Gal}(\mathbb Q(\zeta, \sqrt{2})/\mathbb Q)) \\ \hline (1) & (1) \\ (1,2) & (1) \\ (3,4,6,5) & (3,4,6,5) \\ (1,2)(3,4,6,5) & (3,4,6,5) \\ (3,5,6,4) & (3,5,6,4) \\ (1,2)(3,5,6,4) & (3,5,6,4) \\ (3,6)(4,5) & (3,6)(4,5) \\ (1,2)(3,6)(4,5) & (3,6)(4,5) \end{array}$

Is my understanding correct?

-- Theorem 2 details --

Any radical extension $F(\alpha_1, \dots, \alpha_k)$ is the union of an ascending tower of fields $F=F_0 \subseteq F_1 \subseteq \cdots \subseteq F_k = F(\alpha_1, \dots, \alpha_k)$ where each $F_i = F_{i-1}(\alpha_i)$, $\alpha_i$ is the $p_i$-th rot of an element in $F_i{i-1}$, $p_i$ is prime, and $F_i$ contains no $p_i$-th roots of unity not in $F_i{i-1}$ unless $\alpha_i$ is itself a $p_i$-th root of unity.

Theorem 2. If $E\supseteq B(\alpha) \supseteq B$ are fields with $\alpha^p \in B$ for some prime $p$, and if $B(\alpha)$ contains no $p$th roots of unity not in $B$ unless a itself is a $p$th root of unity, then $\texttt{Gal}(E/B(\alpha))$ is a normal subgroup of $\texttt{Gal}(E/B)$ and $\texttt{Gal}(E/B)/\texttt{Gal}(E/B(\alpha))$ is abelian.

Proof: By the homomorphism theorem for groups, it suffices to find a homomorphism of $\texttt{Gal}(E/B)$, with kernel $\texttt{Gal}(E/B(\alpha))$, into an abelian group (i.e., onto a subgroup of an abelian group, which of course is also abelian). The obvious map with kernel $\texttt{Gal}(E/B(\alpha))$ is restriction to $B(\alpha)$, $\lvert_{B(\alpha)}$, since by definition $$\sigma \in \texttt{Gal}(E/B(\alpha)) \Leftrightarrow \sigma \lvert_{B(\alpha)} \texttt{ is the identity map}.$$

The homomorphism property, $$\sigma' \sigma \lvert_{B(\alpha)}=\sigma'\lvert_{B(\alpha)} \sigma\lvert_{B(\alpha)}, \forall \sigma', \sigma \in \texttt{Gal}(E/B),$$ is automatic provided $\sigma \lvert_{B(\alpha)}(b) \in B(\alpha)$ for each $b \in B(\alpha)$, i.e. provided $B(\alpha)$ is closed under each $\sigma \in \texttt{Gal}(E/B)$.

Since $\sigma$ fixes $B$, $\sigma \lvert_{B(\alpha)}$ is completely determined by the value $\sigma(\alpha)$. If $\alpha$ is a $p$th root of unity then $$(\sigma(\alpha))^p = \sigma(\alpha^p) = \sigma(\zeta^p) = \sigma(1) = 1,$$ hence $\sigma(\alpha) = \zeta^i= \alpha^i \in B(\alpha)$, since each $p$th root of unity is some $\zeta^i$. If $\alpha$ is not a root of unity then $$(\sigma(\alpha))^p = \sigma(\alpha^p) = \alpha^p \texttt{ since } \alpha^p \in B,$$ hence $\sigma(\alpha) = \zeta^ia$ for some $p$th root of unity $\zeta$; and $\zeta \in B$ by hypothesis, so a\texttt{Gal}in $\sigma(\alpha) \in B(\alpha)$. Thus $B(\alpha)$ is closed as required.

This also implies that $I_{B(\alpha)}$ maps $\texttt{Gal}(E/B)$ into $\texttt{Gal}(B(\alpha)/B)$, so it now remains to check that $\texttt{Gal}(B(\alpha)/B)$ is abelian. If $\alpha$ is a root of unity then, as we have just seen, each $\sigma \lvert_{B(\alpha)} \in \texttt{Gal}(B(\alpha)/B)$ is of the form $\sigma_i$, where $\sigma_i(\alpha) = \alpha^i$, hence $$\sigma_i\sigma_j(\alpha) = \sigma_i(\alpha^j) = \alpha^{ij} = \sigma_j\sigma_i(\alpha).$$ Likewise, if $\alpha$ is not a root of unity then each $\sigma \lvert_{B(\alpha)} \in \texttt{Gal}(B(\alpha)/B)$ is of the form $\sigma_i$ where $\sigma_i(\alpha) = \zeta^i\alpha$, hence $$\sigma_i\sigma_j(\alpha) = \sigma_i(\zeta^j\alpha) = \zeta^{i+j}\alpha = \sigma_j\sigma_i(\alpha)$$ since $\zeta \in B$ and therefore $\zeta$ is fixed. Hence in either case $\texttt{Gal}(B(\alpha)/B)$ is abelian.

• Your understanding is not correct. Restriction means what the word really suggests: restrict the domain of the function. When $f\colon X \rightarrow Y$ and $Z \subset X$ then the restriction $f|_Z$ means the function $Z \rightarrow Y$ where $z \mapsto f(z)$ for all $z \in Z$. Theorem: If $E/B$ is a Galois extension and there is a field $K$ between $E$ and $B$ such that $K/B$ is Galois, then restriction ${\rm Gal}(E/B) \rightarrow {\rm Gal}(K/B)$ is a surjective group homomorphism with kernel ${\rm Gal}(E/K)$. – KCd Mar 15 '15 at 10:50
• $\mathbb Q \subset \mathbb Q(\sqrt[p]{2}) \subset \mathbb Q(\sqrt[p]{2}, \zeta_p)$ contradicts the theorem, doesn't it? – MooS Mar 15 '15 at 10:56
• @MooS John is using a special definition of radical extension actually. Such the contradiction does not happen. My bad. Have just included the definition in. – athos Mar 15 '15 at 11:17
• @KCd I don't get it. Let $f$ be the homomorphism, so $f: \texttt{Gal} E/B \rightarrow A$, where group $A$ is abelian; and $\texttt{Ker}(f) = \texttt{Gal} E / B(\alpha)$, in other words, $f(\beta) = 1_A, \forall \beta \in \texttt{Gal} E / B(\alpha)$. So this $f$ is defined on the whole $\texttt{Gal} E/B$, how could this be limited to a smaller domain? – athos Mar 15 '15 at 13:20
• In the example when we are restricting elements of $Gal(\Bbb{Q}(\zeta,\sqrt2)/\Bbb{Q})$ to those of $Gal(\Bbb{Q}(\sqrt2)/\Bbb{Q})$ the homomorphism (in your notation), maps a permutation in the described subgroup $S_6$ to a permutation in $S_2$. So $\sigma_1,\sigma_2,\sigma_3,\sigma_4$ are all mapped to $(1)$, but $\sigma_j, 5\le j\le8$ are all mapped to $(12)$ - all according to how they permute $\pm\sqrt2$. The kernel of the restriction homomorphism is thus $\{\sigma_1,\sigma_2,\sigma_3,\sigma_4\}$, which is exactly $Gal(\Bbb{Q}(\zeta,\sqrt2)/\Bbb{Q}(\sqrt2)$ as claimed. – Jyrki Lahtonen Mar 15 '15 at 15:12