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Adam
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When a sum of the ideals is radical

Let $X,Y$ be algebraic subsets of $\mathbb A^n.$ I would like to show that if $X$ and $Y$ intersect "transversely" then $I(X)+I(Y)$ is radical (so $I(X\cap Y)=I(X)+I(Y)$). How to prove it?

"transversely" can mean that $T_p\, X\cup T_p\, Y=T_p\, \mathbb A^n$ for all $p\in X\cap Y$ and $X$ and $Y$ are smooth at all intersection points, but I hope this condition can be relaxed somewhat. (Though I realize limitations. For example, for $X=V(u^2-v^2)$ and $Y=V(v)$, $I(X)+I(Y)=(u^2,v)$ is not radical, even though the tangent spaces at $(0,0)$ span $\mathbb A^2$.)

This statement was conjectured in this math stackexchange answer (see the top answer), but not proved, so I am hoping to find a proof here.

Adam
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