# Why can any primary decomposition be reduced to a minimal one

By definition we have

A primary decomposition of an ideal $\mathfrak{a}$ is a decomposition $\mathfrak{p}=\cap_{i=1}^r\mathfrak{q}_i$ into primary ideal $\mathfrak{q}_i$. The decomposition is called minimal if the corresponding prime ideals $\mathfrak{p}_i=rad(\mathfrak{q}_i)$ are pairwise different and $\cap_{i\neq j}\mathfrak{q}_i\not\subset \mathfrak{q}_j$ for alle $j=1,...,r$.

I'm having trouble understanding why the following lemma implies that any primary decomposition of ideal can be reduced to a minimal one?

Let $\mathfrak{q}_1,...,\mathfrak{q}_r$ be $\mathfrak{p}$-primary ideals for some prime ideal $\mathfrak{p}$. Then the intersection $\mathfrak{q}=\cap_{i=1}^r\mathfrak{q}_i$ is $\mathfrak{p}$-primary.

## 1 Answer

Because it allows you to replace the family of $q_i$'s having a common radical $p$ with the intersection of the family. This guarantees the primes are pairwise different.

The other part of minimality is just common sense: if $\cap_{i\neq j}q_i\subset q_j$, then you can just omit $q_j$ from the factorization altogether. You don't need the lemma above to do this portion.