There's a commutative algebra fact that I would very much like to be true but could, for all I know, be completely false. One version that would be sufficient is:

Say $A$ is a smooth projective variety and $Y$ and $Z$ are closed, irreducible, Cohen-Macaulay subvarieties of $A\times\mathbb{P}^1$ which are flat over $\mathbb{P}^1$. If all the fibers (over $\mathbb{P}^1$) of $Y\cap Z$ have the same dimension as each other, then $Y\cap Z$ is flat over $\mathbb{P}^1$.

If it helps, $A$ is secretly a Grassmannian, $Y$ is the product of $\mathbb{P}^1$ with some subvariety of $A$, all the fibers of $Z$ are isomorphic to each other, and all the fibers of $Y\cap Z$ except the one over $\infty$ are isomorphic to each other. (So I'm assuming the fiber over $\infty$ has the same dimension as the rest, but not that it's isomorphic to the rest.) Crucially, note that I'm NOT assuming that $\mathrm{codim}Y+\mathrm{codim}Z=\mathrm{codim}(Y\cap Z)$.

Does anyone know if this is true, and if so, a simple proof or reference?