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?



Flatness and intersections of Cohen-Macaulay subvarieties
No, that is not true. Let $A$ be $\mathbb{P}^3$ with homogeneous coordinates $[X\_0,X\_1,X\_2,X\_3]$. Let $[T\_0,T\_1]$ be homogeneous coordinates on the base $\mathbb{P}^1$. Let $Y$ be the common zero scheme of $X\_2^2-X\_0X\_1$ and $X\_3$. Let $Z$ be the common zero scheme of $T\_0^2X\_3^2-X\_0(T\_1X\_0-T\_0X\_1)$ and $X\_2$. Both of these are irreducible, closed subschemes of $\mathbb{P}^3\times \mathbb{P}^1$ that are flat over $\mathbb{P}^1$. But the intersection is the common zero locus of $X\_2$, $X\_3$, $X\_0X\_1$ and $X\_0(T\_1X\_0-T\_0X\_1)$. This is not flat over $\mathbb{P}^1$. 


