= Solution
Continue on the branch of <steady planar ideal-MHD field-line invariants>, with $k\ne0$ and positive $\rho,T$. Adiabatic entropy transport gives $\mathbf u\cdot\nabla s=0$. As all quantities are $z$-independent,
$$
\mathbf u\cdot\nabla=\frac{k}{\rho}\mathbf B\cdot\nabla,
\qquad\boxed{\mathbf B\cdot\nabla s=0}.
$$
The axial component of the <ideal magnetohydrodynamic momentum equation> has no pressure or gravitational term. Magnetic tension gives
$$
\rho\mathbf u_h\cdot\nabla u_z=\frac1{\mu_0}\mathbf B_h\cdot\nabla B_z.
$$
Since $k$ is constant along a <magnetic field line>, this becomes
$$
\boxed{\mathbf B\cdot\nabla V=0,\qquad V=u_z-\frac{B_z}{\mu_0k}}.
$$
Subtracting the two axial invariants yields the useful compatibility relation
$$
\boxed{B_z\left(\frac{\mu_0k^2}{\rho}-1\right)=\mu_0k(V-U)}.
$$
The squared ratio of horizontal flow speed to horizontal <Alfvén speed> is $M_h^2=\mu_0k^2/\rho$; vector equality with the horizontal <Alfvén velocity> additionally requires $k>0$. If $V-U\ne0$, the right side is a nonzero constant along the field line, so a regular finite $B_z$ cannot permit $M_h^2=1$ anywhere. This is the nondegenerate exclusion.
However, $B_z\not\equiv0$ does not imply $V-U\ne0$. The <Alfvénic degeneracy of steady planar MHD> is a genuine exception: take constant $\rho,p,s$, $\Phi=0$, and a constant magnetic field with nonzero horizontal and axial components, and set
$$
\mathbf u=\frac{\mathbf B}{\sqrt{\mu_0\rho}},\qquad
k=\sqrt{\rho/\mu_0},\qquad U=V=0.
$$
Every equation holds, but the horizontal velocity equals the horizontal <Alfvén velocity> everywhere despite $B_z\ne0$. Therefore \b[the printed pointwise exclusion needs a nondegeneracy hypothesis, such as $V\ne U$]. The general conclusion is the boxed compatibility relation; when $V=U$, it says $(M_h^2-1)B_z=0$ and allows the Alfvénic branch.
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