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 \( \tau(M) = M \).  Nex