$\DeclareMathOperator\codim{codim}$Let $X=V(I)$, $Y=V(J)$ be two affine varieties. I'd like to know possibile strategies to understand when their schematic intersection, i.e. $X\cap Y=V(I+J)$, is reduced.
I am aware that reduceness is equivalent to Serre's conditions $R\_0$+$S\_1$ (at least under Noetherian hypothesis), but I don't really know how to use them efficiently, at least in my specific case.
More precisely, the case I am interested in is the following:


Let $V=\mathbb A^{2n}$ be an even dimensional vector space, let $X$ be an half dimensional vector subspace and let $Y=\bigcup\_i Y\_i \cup \bigcup\_j Z\_j$, where each $Y\_i$, $Z\_j$ is again a half dimensional vector subspace of $V$ (the reason for using different notations for them will become clear).


In general, $X\cap Y$ is not reduced: for example it's easy to see that, if $X=V(y-x)$ is the diagonal and $Y=V(xy)$ is the cross inside $\mathbb A^2$, then $X\cap Y=V(x^2, y-x)$ is not reduced…so one certainly needs to put more assumptions on $X,Y$.


I'd like to know if the following two assumptions are actually enough to ensure reduceness for $X\cap Y$:


(*i*) The $X\cap Y\_i$'s are all distinct subspaces, of codimension 1 in $X$.


(*ii*) For each $Z\_j$, there exists at least one $Y\_i$ such that $X\cap Z\_j \subsetneq X\cap Y\_i$ (in particular, $\codim\_X X\cap Z\_j\geq 2$).


My intuition is that such an intersection should be at least generically reduced, namely reduced outside the closed locus $\bigcup\_j X\cap Z\_j$, and I hope these conditions are enough to ensure that nothing bad happens even inside such a locus, but I don't know how to prove that (basically I don't know how to exclude embedded primes).


P.S. If this turns out to be actually false, anyway I'd be interested to know which sort of other ‘reasonable’ conditions on the $Y\_i$'s and $Z\_j$'s could ensure reducedness of such an intersection.



How to show that the intersection of two certain affine varieties is reduced?
It seems to me that if there are no $Z$'s, then this works. Indeed, both $X$ and the $Y\_i$ are linear spaces, so the intersection $X\cap Y$ is just a union of pairwise different linear spaces, so it is reduced. I think one could write down this with equations.


However, adding the $Z\_i$'s is problematic. I think the following example works:
Let $Y=Z(xz,xt,yz,yt)\subseteq \mathbb A^4$. Then $Y=Y\_1\cup Z\_1$, where $Y\_1=Z(z,t)$ and $Z\_1=Z(x,y)$. Further let $X=Z(y-z,t)$. Then $X\cap Y\_1=Z(y,z,t)$, a hyperplane in $X$ and $X\cap Z\_1=Z(x,y,z,t)$, which is contained, but not equal to $X\cap Y\_1$. Finally, $X\cap Y=Z(xy,y^2,y-z,t)$ is not reduced.


Here is also a theoretical reason for this example. (This had been the motivation behind my previous attempts at giving an example. In fact, this is essentially the first example I gave, but it didn't work, because I tried to do it without the Z's). Anyway the point is, that $Y$ is a simple example of a union of linear spaces that is not $S\_2$. A hyperplane section of something that's not $S\_2$ is not $S\_1$, so cannot be reduced. The only trick is to figure out how to choose $X$, so its intersection with $Y$ is actually a hyperplane cut of $Y$.


Finally, a little nit picking: $\mathbb A^n$ is **not** a vectorspace, nor are the $X$, the $Y\_i$'s and the $Z\_i$. They are affine spaces. If you want to say that they all share a common point, then say that.


