= Solution
Write $\mathbf u=\partial_t\boldsymbol\xi$. In the uniform equilibrium, the <linearized ideal magnetohydrodynamic equations> reduce to
$$
\partial_t\delta\rho=-\rho\nabla\cdot\partial_t\boldsymbol\xi,\qquad
\partial_t\delta p=-\gamma p\nabla\cdot\partial_t\boldsymbol\xi,\qquad
\partial_t\delta\mathbf B=\nabla\times(\partial_t\boldsymbol\xi\times\mathbf B).
$$
Integrating from the undisplaced reference state, and using the uniform background <magnetic field>, gives
$$
\boxed{\delta\rho=-\rho\nabla\cdot\boldsymbol\xi,\quad
\delta p=-\gamma p\nabla\cdot\boldsymbol\xi,\quad
\delta\mathbf B=(\mathbf B\cdot\nabla)\boldsymbol\xi-\mathbf B\nabla\cdot\boldsymbol\xi.}
$$
These are the perturbations induced by the <fluid displacement>; independent time-independent changes to the reference state are excluded. The linear <Lorentz force density> is $(\nabla\times\delta\mathbf B)\times\mathbf B/\mu_0$. Its <magnetic pressure> and <magnetic tension> parts give
$$
\boxed{\rho\partial_t^2\boldsymbol\xi=-\nabla\left(\delta p+\frac{\mathbf B\cdot\delta\mathbf B}{\mu_0}\right)+\frac{(\mathbf B\cdot\nabla)\delta\mathbf B}{\mu_0}.}
$$
For a <Fourier mode>, put $s=\mathbf k\cdot\boldsymbol\xi$, $q=\mathbf B\cdot\boldsymbol\xi$ and $K=\mathbf k\cdot\mathbf B$. Then $\delta p=-i\gamma p s$ and $\delta\mathbf B=i(K\boldsymbol\xi-\mathbf B s)$. Substitution gives
$$
\boxed{\rho\omega^2\boldsymbol\xi=\mathbf k\left[\left(\gamma p+\frac{B^2}{\mu_0}\right)s-\frac{Kq}{\mu_0}\right]+\frac{K}{\mu_0}(K\boldsymbol\xi-\mathbf B s).}
$$
If $s=q=0$, the <fluid displacement> is perpendicular to both the <wave vector> and the <magnetic field>. Only <magnetic tension> restores it, and
$$
\boxed{\omega^2=(\mathbf k\cdot\mathbf v_a)^2,\qquad \mathbf v_a=\frac{\mathbf B}{\sqrt{\mu_0\rho}}.}
$$
This is the <Alfvén wave>. The <Alfvén velocity> is the field-directed propagation vector: $\omega=\pm\mathbf k\cdot\mathbf v_a$. The signed <phase velocity> normal to the wavefront is $\pm v_a\cos\theta\,\widehat{\mathbf k}$; the <group velocity> is $\pm\mathbf v_a$. The distinction matters for oblique propagation. For generic directions the <Alfvén wave> has one transverse polarization, with $\delta\rho=\delta p=0$.
To obtain the other <magnetohydrodynamic waves>, take the scalar products with $\mathbf k$ and $\mathbf B$:
$$
\rho\omega^2s=k^2\left[\left(\gamma p+\frac{B^2}{\mu_0}\right)s-\frac{Kq}{\mu_0}\right],\qquad
\rho\omega^2q=\gamma p K s.
$$
The <determinant> condition for a nonzero pair $(s,q)$ is
$$
\omega^4-k^2(c_s^2+v_a^2)\omega^2+k^4c_s^2v_a^2\cos^2\theta=0,\qquad c_s^2=\frac{\gamma p}{\rho}.
$$
Thus the <fast magnetosonic wave> and <slow magnetosonic wave> have
$$
\boxed{v_{\mathrm f,\mathrm{sl}}^2=\frac{c_s^2+v_a^2}{2}\pm\sqrt{\frac{(c_s^2+v_a^2)^2}{4}-c_s^2v_a^2\cos^2\theta}.}
$$
Their <fluid displacements> lie in the plane of $\mathbf k$ and $\mathbf B$ and are generally compressive. In the <fast magnetosonic wave>, gas and <magnetic pressure> provide the stronger restoring combination; in the <slow magnetosonic wave>, their perturbations oppose one another. The <phase speeds> depend on direction. They satisfy $v_{\mathrm{sl}}\leq\min(c_s,v_a)$ and $v_{\mathrm f}\geq\max(c_s,v_a)$. For parallel propagation the two speeds are $c_s$ and $v_a$, with a degeneracy between a transverse branch and the <Alfvén wave>. For perpendicular propagation $v_{\mathrm f}^2=c_s^2+v_a^2$ and $v_{\mathrm{sl}}=0$; the latter is a nonpropagating limiting disturbance. These special directions require interpreting the polarizations by continuity rather than assuming three distinct nonzero frequencies.
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