Refinement of two stratifications of an algebraic variety Let $(X,\mathscr{O}_{X})$ be a separated $\mathbb{C}$-scheme of finite type.
A locally finite decomposition $X=\coprod_{i\in I}X_{i}$ into locally closed subschemes is said to be a stratification of $X$, if each $X_{i_{0}}$ is smooth and for all $i_{0}\in I$ there exists a subset $I_{i_{0}}\subseteq I$ such that $\overline{X_{i}}=\coprod_{i\in I_{i_{0}}}X_{i}$.
Now given two such stratification $X=\coprod_{i\in I}X_{i}$ and $X=\coprod_{j\in J}Y_{j}$ of $X$, one might wonder whether there exists a stratification $X=\coprod_{k\in K}Z_{k}$ refining both, i.e. each $X_{i}$ and each $Y_{j}$ is a union of some subsets $Z_{k}$ respectively.
The most straightforward candidate seems to me to take the decomposition

$X=\coprod_{i\in I,j\in J}X_{i}\cap Y_{j}$

into locally closed subsets. Is there any hope that $X_{i}\cap Y_{j}$ can be equipped with the structure of a locally closed subscheme of $X$?
Here are my thoughts about this: If $\mathscr{I}$ and $\mathscr{J}$ is a quasi-coherent sheaf of ideals of $\mathscr{O}_{X}$
defining $X_{i}$ and $Y_{j}$, one could try to prove that $\mathscr{I}+\mathscr{J}\subseteq\mathscr{O}_{X}$ is a quasi-coherent sheaf of ideal defining $X_{i}\cap Y_{j}$. Now we can observe that

$supp(\mathscr{O}_{X}/\mathscr{I}+\mathscr{J})\subseteq supp(\mathscr{O}_{X}/\mathscr{I})\cap supp(\mathscr{O}_{X}/\mathscr{J})=X_{i}\cap Y_{j}$,

but I don't see if the converse inclusion is true in general.
Am I on the right track and if so, does anyone see how to continue?
 A: There's a bit of a can of worms here about stratifications: we want the intersection of any stratum-closures to be a union of stratum-closures. But if we talk about the scheme-theoretic intersection we can get in to some trouble: if we require the intersection to be the reduced union (instead of just asking that the underlying sets agree), then the orbits of $SO(n)$ on $GL(n)/B$ don't form a stratification, which is undesirable. (I originally learned this from Alan Knutson, and he put this in to a comment on StacksBlog here.) So let me assume we're talking about agreement of underlying sets when we ask for such equalities in this post - this will make our lives a bit easier.

The "most straightforwards candidate" in your post is unfortunately not quite correct: we may have to do some extra subdividing. Consider the following two stratifications of $\Bbb A^3$:


*

*$D(x-yz)$, $V(x-yz)$

*$D(x)$, $V(x)$
It is not difficult to see that these satisfy the assumptions from the question. On the other hand, if we take $\coprod_{i,j} X_i\cap Y_j$, here's what we get:


*

*$D(x^2-xyz)$

*$D(x)\cap V(x-yz)$

*$D(x-yz)\cap V(x)$

*$V(x,yz)$
and the final piece $V(x,yz)$ is singular. We would need to subdivide this in to smaller pieces (say $V(x,yz)\setminus V(x,y,z)$ and $V(x,y,z)$) to get a stratification satisfying your definition.

On the other hand, given any two locally finite stratifications $\coprod Y_j$ and $\coprod Z_k$ of a separated finite type $\Bbb C$-scheme $X$, it is true that there is a locally finite stratification refining both of them.
We make some preliminary observations. First, if $X_1\subset X$ is an irreducible component and we have a locally closed subset $L$ which contains the generic point of $X_1$, then $L$ contains an open subset of $X_1$. This is because a locally closed subset can be written as an intersection of a closed subset and an open subset, and any closed subset containing the generic point must contain all of $X_1$. Next, our scheme $X$ is Noetherian, because it is of finite type over a Noetherian ring. So $X$ has finitely many irreducible components and any locally finite collection of sets is actually globally finite.
Now on to the problem. Write $X=X_1\cup\cdots\cup X_n$ for the decomposition of $X$ in to irreducible components. For any irreducible component $X_i$ of the collection of points $X_i^\circ$ which belong to only that irreducible component and no others is open. Find a $Y_j$ and $Z_k$ which contain the generic point of $X_i$. By our observation above, each of $Y_j$ and $Z_k$ contain an open subset of $X_i$, and by taking the intersection of these open sets and the smooth locus of $X_i^\circ$ we can find a smooth open subset $W\subset X$ which is contained in $X_i^\circ$, $Y_j$, and $Z_k$ and who's complement inside $X_i$ consists of some finite number of closed subschemes of strictly smaller dimension (here we use that any locally closed subset of a noetherian scheme is again locally noetherian for the finiteness and irreducibility of $X_i$ to say the dimension must drop).
Now observe that in order to give a stratification of $X$ which is a common refinement of $\coprod Y_j$ and $\coprod Z_k$, it is enough to give a stratification of $X\setminus W$ which is a refinement of the stratifications $\coprod (Y_j\cap W^c)$ and $\coprod (Z_k\cap W^c)$. This suggests an inductive approach: consider the collection of maximal-dimensional irreducible components of $X$ and apply the above procedure for all of them. Then at each stage when we go from $X$ to $X\setminus W$, we eliminate one top-dimensional irreducible component at the cost of $X\setminus W$ having potentially finitely many more irreducible components of strictly lower dimension. So after applying this procedure a finite number of times, we get that it's enough to prove the proposition for a scheme of finite type over $\Bbb C$ of strictly smaller dimension. Eventually we get down to a finite discrete union of points, and we can just take the disjoint union of all of these points. So the claim is true.
