= Solution
Take the inviscid limit of the <Navier-Stokes equation>, neglect gravity, and use <ideal magnetohydrodynamics>. The governing equations are
$$
\nabla\cdot\mathbf v=0,\qquad
\rho(\partial_t\mathbf v+\mathbf v\cdot\nabla\mathbf v)=-\nabla P+\frac{(\nabla\times\mathbf B)\times\mathbf B}{\mu_0},\qquad
\partial_t\mathbf B=\nabla\times(\mathbf v\times\mathbf B),\quad\nabla\cdot\mathbf B=0.
$$
The <continuity equation> also gives $D_t\rho=0$. Write the first-order magnetic perturbation as $\mathbf b=\mathbf B'$ and linearize about the uniform stationary background:
$$
\rho_0\partial_t\mathbf v=-\nabla\left(P'+\frac{B_0b_z}{\mu_0}\right)+\frac{B_0}{\mu_0}\partial_z\mathbf b,
\qquad\partial_t\mathbf b=B_0\partial_z\mathbf v,
\qquad\nabla\cdot\mathbf v=\nabla\cdot\mathbf b=0.
$$
For a nonzero-<wavenumber> <plane wave> proportional to $e^{i(kz-\omega t)}$, the two divergence constraints imply $v_z=b_z=0$. The longitudinal momentum equation then gives $P'=0$ for that mode, and $\partial_t\rho'=0$ gives $\rho'=0$ for nonzero $\omega$. Thus the propagating <Alfvén wave> is transverse. Its two independent transverse directions have the same <dispersion relation>.
The <Linear Alfvén wave equations> reduce to
$$
-i\omega\rho_0\mathbf v=\frac{ikB_0}{\mu_0}\mathbf b,\qquad-i\omega\mathbf b=ikB_0\mathbf v.
$$
Eliminating either <amplitude> gives
$$
\boxed{\omega^2=k^2v_A^2,\qquad v_A=\frac{B_0}{\sqrt{\mu_0\rho_0}},\qquad\mathbf b=-\frac{kB_0}{\omega}\mathbf v.}
$$
For $B_0>0$, the branch traveling along $+z$ has $\omega=kv_A$ and $\mathbf b=-\sqrt{\mu_0\rho_0}\mathbf v$. The opposite branch reverses that <amplitude> sign. Finite <viscosity> would additionally damp the motion; it is neglected in obtaining the real ideal <dispersion relation>.
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