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).

Thanks a lot for any comment about that!

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 $Y_i,Z_j$'s could ensure reduceness of such an intersection.

$\DeclareMathOperator\codim{codim}$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.