Let $X \subseteq \mathbb{C}^n$ be an irreducible algebraic set that forms a cone, and let $I=I(X) \subseteq \mathbb{C}[x\_1,...,x\_n]$. Let $m < n$ and $k\leq m$ be positive integers. Is it true that for a generic collection of elements $v\_1,..., v\_k \in X$ and $v\_{k+1},..., v\_m \in \mathbb{C}^n$, it holds that $I + I(\text{span}\{v\_1,...,v\_m\})$ is radical?



Is the sum of a radical ideal and the ideal of a generic linear space intersecting that ideal radical?
In general no. I'm only familiar with this question with $k=0$. Here are two basic counter-examples (both with $k=0$)


(1) Take $X$ singular and dimension $n-m$, so we are intersecting $X$ with a generic codimension $n-m$-plane. Then the intersection is a point of multiplicity equal to the multiplicity of $0$ as a singular point of $X$.


(2) Let $X$ be non-Cohen Macaulay with depth $\delta < \dim X$. Then intersecting $X$ with a codimension $\delta$ plane gives a variety with an embedded point.


If we impose that $X$ is Cohen-Macaulay, that $k=0$ and that $n-m < \dim X$, then the answer is "yes". Recall Serre's Theorem that "reduced" is equivalent to R0 and S1. $X$ Cohen-Macaulay means that $X$ is $S\_{\dim X}$, and $X$ reduced means that $X$ is $R\_0$. Taking $q$ generic hyperplane sections (through the singularity of $X$) turns this into $S\_{\dim X - q}$ and $R\_0$, so still reduced. I can't quickly find good references for this; I'm willing to try harder if someone else doesn't beat me too it.


The problem with $k>0$ seems more interesting to me, and I haven't seen it before.


