https://susam.net/lemma-for-ftgt.html

Lemma for FTGT

By Susam Pal on 09 Mar 2025

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):

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 \( \tau M^* \tau^{-1} = \tau(M)^{*}. \)

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 we will first
show that \( \tau M^* \tau^{-1} \subseteq \tau(M)^* \) and then we
will show that \( \tau M^* \tau^{-1} \supseteq \tau(M)^* \) in order
to prove the above equation.

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 \( \tau M^* \
tau^{-1} = \tau(M)^*. \)

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