Lemma for FTGT

Introduction

This post illustrates a key lemma that is used in proving the fundamental theorem of Galois theory (FTGT). Note that FTGT is not covered in this post. The focus of this post is on understanding and proving this lemma only. Here is the lemma from the book Galois Theory, 5th ed. by Stewart (2023):

The notation \( M^* \) denotes the group of all \( M \)-automorphisms of \( L \) with composition as the group operation. Note that Stewart writes \( \tau(M)^{*} = \tau M^* \tau^{-1} \) while stating the lemma but I have reversed the LHS and RHS to maintain consistency with the equations that appear in the discussion below.

To build intuition for this lemma, I'll first present an illustration, followed by a proof. The discussion below assumes familiarity with field extensions and field automorphisms, as several notations and results from these areas will be used implicitly without detailed justification. This post is meant to serve as a set of notes on the lemma, not a comprehensive tutorial.

Contents

• Introduction
• Illustration
  • Concrete Example
  • LHS ⊆ RHS
  • LHS ⊇ RHS
  • LHS = RHS
• Proof

Illustration

Concrete Example

Let \( L = \mathbb{Q}(\sqrt{2}, \sqrt{3}), \) \( K = \mathbb{Q}, \) and \( M = \mathbb{Q}(\sqrt{2}). \) Note that \begin{align*} L &= \{ a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} : a, b, c, d \in \mathbb{Q} \}, \\ M &= \{ k + l \sqrt{2} : k, l \in \mathbb{Q} \}. \end{align*} Now the group of \( K \)-automorphisms of \( L \) is \[ K^* = \{\phi_1, \phi_2, \phi_3, \phi_4 \} \] where each \( \phi_i \) is given by \begin{align*} \phi_1 &: a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \mapsto a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}, \\ \phi_2 &: a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \mapsto a - b \sqrt{2} + c \sqrt{3} - d \sqrt{6}, \\ \phi_3 &: a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \mapsto a + b \sqrt{2} - c \sqrt{3} - d \sqrt{6}, \\ \phi_4 &: a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \mapsto a - b \sqrt{2} - c \sqrt{3} + d \sqrt{6}. \end{align*} Then \( M^* = \{ \phi_1, \phi_3 \}. \) Let \( \tau = \phi_2. \) Then \begin{align*} \tau(M) &= \{ \tau(x) : x \in \mathbb{M} \} \\ &= \{ \tau(k + l \sqrt{2}) : k, l \in \mathbb{Q} \} \\ &= \{ k - l \sqrt{2} : k, l \in \mathbb{Q} \}. \end{align*} Note that in this case we ended up with \( \tau(M) = M \) but we will be careful not to utilise this fact. We will ensure that the steps below work without assuming \( \tau(M) = M. \) Next we find \begin{equation} \tau(M)^* = \{ \phi_1, \phi_3 \}. \label{eq-tau-m-ast} \end{equation} Now \begin{align*} \tau M^* \tau^{-1} &= \{ \tau \gamma \tau^{-1} : \gamma \in {M^*} \} \\ &= \{ \tau \phi_1 \tau^{-1}, \tau \phi_3 \tau^{-1} \}. \end{align*} Let us now find out how each element of \( \tau M^* \tau^{-1} \) transforms the elements of \( L. \) For all \( a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \in L, \) we get \begin{align*} (\tau \phi_1 \tau^{-1})(a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}) &= (\tau \phi_1)(a - b\sqrt{2} + c\sqrt{3} - d\sqrt{6}) \\ &= \tau (a - b\sqrt{2} + c\sqrt{3} - d\sqrt{6}) \\ &= a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}). \end{align*} Therefore \[ \tau \phi_1 \tau^{-1} = \phi_1. \] Similarly, \begin{align*} (\tau \phi_3 \tau^{-1})(a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}) &= (\tau \phi_3)(a - b\sqrt{2} + c\sqrt{3} - d\sqrt{6}) \\ &= \tau (a - b\sqrt{2} - c\sqrt{3} + d\sqrt{6}) \\ &= a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6}. \end{align*} Therefore \[ \tau \phi_3 \tau^{-1} = \phi_3. \] We have shown that \begin{equation} \tau M^* \tau^{-1} = \{ \phi_1, \phi_3 \}. \label{eq-tau-coset} \end{equation} From \eqref{eq-tau-m-ast} and \eqref{eq-tau-coset} we see that \[ \tau M^* \tau^{-1} = \tau(M)^*. \] Since we are working with a concrete example of \( \tau \) here, we know exactly how it behaves, so we succeeded in demonstrating the above equality. However, in a general proof, \( \tau \) is going to be an arbitrary \( K \)-automorphism of \( L, \) so we cannot know exactly how it behaves and as a result, we cannot obtain the above equation directly. Therefore, in a general proof, we are going to show that \( \tau M^* \tau^{-1} \subseteq \tau(M)^* \) and vice versa to obtain the same result.

LHS ⊆ RHS

Once again, let us see how each element of \( \tau M^* \tau^{-1} \) transforms the elements of \( \tau(M). \) Note that this time we are not going to examine how they transform arbitrary elements of \( L. \) We are only going to see how they transform the elements of \( \tau(M). \) For all \( k - l \sqrt{2} \in \tau(M), \) we get \begin{align*} (\tau \phi_1 \tau^{-1})(k - l \sqrt{2}) &= (\tau \phi_1)(k + l \sqrt{2}) \\ &= \tau(k + l \sqrt{2}) \\ &= k - l \sqrt{2}. \end{align*} Similarly, for all \( k - l \sqrt{2} \in \tau(M), \) we get \begin{align*} (\tau \phi_3 \tau^{-1})(k - l \sqrt{2}) &= (\tau \phi_3)(k + l \sqrt{2}) \\ &= \tau(k + l \sqrt{2}) \\ &= k - l \sqrt{2}. \end{align*} Note above that both \( \phi_1 \) and \( \phi_3 \) fix \( k + l \sqrt{2} \in M \) because \( \phi_1, \phi_2 \in M^*, \) the set of \( M \)-automorphisms of \( L. \) This detail will be used in the general proof.

Since both \( \tau \phi_1 \tau^{-1} \) and \( \tau \phi_3 \tau^{-1} \) fix the elements of \( \tau(M), \) they are both \( \tau(M) \)-automorphisms of \( L. \) Therefore \( \tau M^* \tau^{-1} \subseteq \tau(M)^{*}. \)

LHS ⊇ RHS

Consider the set \( \tau^{-1} \tau(M)^* \tau \) and examine how its elements transform the elements of \( M. \) For all \( k + l \sqrt{2} \in M, \) we get \begin{align*} (\tau^{-1} \phi_1 \tau)(k + l \sqrt{2}) &= (\tau^{-1} \phi_1)(k - \sqrt{2}) \\ &= \tau^{-1}(k - \sqrt{2}) \\ &= k + l \sqrt{2}. \end{align*} Similarly, for all \( k + l \sqrt{2} \in M, \) we get \begin{align*} (\tau^{-1} \phi_3 \tau)(k + l \sqrt{2}) &= (\tau^{-1} \phi_3)(k - \sqrt{2}) \\ &= \tau^{-1}(k - \sqrt{2}) \\ &= k + l \sqrt{2}. \end{align*} Here both \( \phi_1 \) and \( \phi_3 \) fix \( k - l \sqrt{2} \in \tau(M) \) because \( \phi_1, \phi_2 \in \tau(M)^*, \) the set of \( \tau(M) \)-automorphisms of \( L. \)

Since both \( \tau^{-1} \phi_1 \tau \) and \( \tau^{-1} \phi_3 \tau \) fix the elements of \( M, \) they are both \( M \)-automorphisms of \( L. \) Therefore \( \tau^{-1} \tau(M)^* \tau \subseteq M^* \) which implies \( \tau M^* \tau^{-1} \supseteq \tau(M)^*. \)

LHS = RHS

The previous two sections complete the illustration of the lemma with the chosen example. We have shown that \( \tau M^* \tau^{-1} \subseteq \tau(M)^{*} \) and \( \tau M^* \tau^{-1} \supseteq \tau(M)^*. \) Therefore \( \tau M^* \tau^{-1} = \tau(M)^*. \)

Proof

The ideas presented in the previous sections will now be extended to formulate a general proof. For clarity, the lemma is stated once again below before proceeding with the proof.

Lemma 12.1. Suppose that \( L/K \) is a field extension, \( M \) is an intermediate field, and \( \tau \) is a \( K \)-automorphism of \( L. \) Then \( \tau M^* \tau^{-1} = \tau(M)^{*}. \)

Proof. For all \( \gamma \in M^*, \) \( x' \in \tau(M), \) we use the notation \( x = \tau^{-1}(x') \in M \) and get \[ (\tau \gamma \tau^{-1})(x') = (\tau \gamma)(x) = \tau(x) = x'. \] In the second equality above, we have used the fact that \( \gamma \in M^* \) which implies that \( \gamma \) is an \( M \)-automorphism of \( L \) which allows us to conclude that \( \gamma(x) = x. \) Since every \( \tau \gamma \tau^{-1} \in \tau M^* \tau^{-1} \) fixes all elements \( x' \in \tau(M), \) each \( \tau \gamma \tau^{-1} \) must be a \( \tau(M) \)-automorphism of \( L. \) Thus \( \tau M^* \tau^{-1} \subseteq \tau(M)^*. \)

Similarly, for all \( \gamma' \in \tau(M)^*, \) \( x \in M, \) we use the notation \( x' = \tau(x) \in \tau(M) \) and get \[ (\tau^{-1} \gamma' \tau)(x) = (\tau^{-1} \gamma')(x') = \tau^{-1}(x') = x. \] In the second equality above, we have used the fact that \( \gamma' \in \tau(M)^* \) which implies that \( \gamma' \) is an \( \tau(M) \)-automorphism of \( L \) which allows us to conclude that \( \gamma'(x') = x'. \) Since every \( \tau^{-1} \gamma' \tau \in \tau^{-1} \tau(M)^* \tau \) fixes all elements \( x \in M, \) each \( \tau^{-1} \gamma' \tau \) must be an \( M \)-automorphism of \( L. \) Thus \( \tau^{-1} \tau(M)^* \tau \subseteq M^*. \) This implies \( \tau M^* \tau^{-1} \supseteq \tau(M)^*. \)

We have shown that \( \tau M^* \tau^{-1} \subseteq \tau(M)^* \) and \( \tau M^* \tau^{-1} \supseteq \tau(M)^*. \) Therefore \( \tau M^* \tau^{-1} = \tau(M)^*. \)