2 Edited to fix typos and correct grammar throughout

# When is the product of two ideals equal to their intersection?

Consider a ring $A$ and an affine scheme $X=SpecA$ . Given two ideals $I,J$ I$and$J$and their asociated associated subschemes$V(I), V(J)$, V(I)$ and $V(J)$, we know that the intersection $I\cap J$ corespond corresponds to the union : $V(I\cap J)=V(I)\cup V(J)$. But a product $I.J$ give gives a new subscheme $V(I.J)$ which has same support as the union but is maybe can be bigger in an infinitesimal sense. For example if $I=J$ you get a scheme $V(I^2)$ which is equal to "double" $V(I)$.

Vague Question : What is geometric interpretation of $V(I.J)$ in general?

Precise question : When is $I\cap J=I.J$? Every body know Everybody knows the case $I+J=A$ but this is absolutely not necessary. For example if $A$ is UFD and $f,g$ are primes between themself relatively prime then for coresponding ideals $(f).(g)=(f)\cap(g)$ but in general $(f)+(g)\neq A$ ; simplest case is (e.g. $f=X, g=Y \in k[X, Y]$)

Thank you very much.

1

# When is product of two ideals equal their intersection?

Consider ring $A$ and affine scheme $X=SpecA$ . Given two ideals $I,J$ and their asociated subschemes $V(I), V(J)$ , intersection $I\cap J$ corespond to union: $V(I\cap J)=V(I)\cup V(J)$. But product $I.J$ give new subscheme $V(I.J)$ which has same support as union but is maybe bigger in infinitesimal sense. For example if $I=J$ you get scheme $V(I^2)$ equal to "double" $V(I)$.

Vague Question : What is geometric interpretation of $V(I.J)$ in general?

Precise question : When is $I\cap J=I.J$? Every body know case $I+J=A$ but this is absolutely not necessary. For example if $A$ is UFD and $f,g$ are primes between themself then for coresponding ideals $(f).(g)=(f)\cap(g)$ but in general $(f)+(g)\neq A$; simplest case is $f=X, g=Y \in k[X, Y]$

Thank you very much.