Is there a way to extract a primary decomposition of $(0)\subset R/I$ given a decomposition of $I\subset R$?

Question

Suppose we have a (minimal) primary decomposition of an ideal $I$ in a polynomial ring $R$ over a field; for simplicity assume $I=I_1\cap I_2$ where $I_j$ is $P_j$-primary. Is there a way to extract from this a primary decomposition of $(0)$ in $R/I$? If so, say $(0)=J_1\cap\dots\cap J_k$ where $J_i$ is $Q_i$-primary.

Can I just take the equality $I=I_1\cap I_2$ and replace all generators by their residues mod $I$? If so, how to justify this? How are $Q_i$ and $P_j$ related?

Answer 1

Here are three results that together solve your problem:

1. For $I\subset R$ an ideal, there's a bijective correspondence between ideals of $R$ containing $I$ and ideals in $R/I$, given by sending $J\supset I \mapsto J/I$. (Correspondence theorem)

2. If $I\subset J$ are ideals in $R$, then $(R/I)/(I/J)\cong R/J$, and the maps are the obvious maps. (Third isomorphism theorem)

3. An ideal $I\subset R$ is primary iff the only zero divisors of $R/I$ are nilpotent. (Definition of primary ideal)

So the images of the ideals $I_1$ and $I_2$ in $R/I$ are still primary, since $R/I_i \cong (R/I)/(I_i/I)$ and thus the condition on zero divisors being nilpotent is still satisfied. Further, any primary decomposition of $(0)\subset R/I$ gives rise bijectively to a primary decomposition of $I\subset R$ and vice-versa via 1. In short, your final line about "[replacing] all generators by their residue mod $I$" is correct.