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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 $\mathbb A^2$.)

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.

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 (see the top answer), but not proved, so I am hoping to find a proof here.

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.

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 $\mathbb A^2$.)

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.

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

It is possible that my definition of transversality needs to be strengthened. 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.

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.