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 $T\_p\, X+ T\_p\, Y=T\_p\, \mathbb A^n$ 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 $T\_{(0,0)}\,\mathbb A^2$.)


This statement was conjectured in this [math stackexchange answer](https://math.stackexchange.com/questions/322872/conditions-for-sqrt-mathfraka-b-sqrt-mathfraka-sqrt-mathfrak) (see the top answer), but not proved, so I am hoping to find a proof here.



When a sum of the ideals is radical
(Adapting earlier comment into an answer.) Assume for simplicity that $X$ and $Y$ are irreducible, or in general take an irreducible component of each. Let $Z$ be an irreducible component of the intersection $X \cap Y$ and let $p \in Z$. Let $a = \dim X$, $b = \dim Y$.


Since $X$ and $Y$ are smooth at $p$, $\dim T\_p X = a$ and $\dim T\_p Y = b$. By transversality, $\dim T\_p X \cap T\_p Y = a+b-n$.


On the one hand, $T\_p Z \subseteq T\_p X \cap T\_p Y$. So $\dim T\_p Z \leq a+b-n$.


On the other hand, $\dim Z \geq a+b-n$, see for example <https://math.stackexchange.com/a/3419739/343280>.


It follows that $\dim T\_p Z = \dim Z$. So $Z$ is reduced at the point $p$.


(I make no claim of originality in this argument, but I can't remember where I might have read it. If anyone knows where this might be found - somewhere in Fulton's Intersection Theory maybe? - I'll add a citation to this answer.)


