Let $R$ be a commutative ring with unity and $I$ be a decomposable ideal of $R$.
Question 1
If $Q$ is a primary ideal containing $I$, then does there exist a minimal primary decomposition of $I$ which has $Q$ as its element?
Question 2
Let $P_1,...,P_n$ be the isolated prime ideals of $I$. Let $X_i$ be the collection of primary ideals whose radical is $P_i$ for each $i$. Then, is $|X_i|=|X_j|$?
Question 1. No, take any two different primary ideals with the same radical and intersect them. The intersection will again be primary.
For example, take $\mathfrak q_1 = (X^2,Y)$ and $\mathfrak q_2 = (X,Y^2)$ in $R = ℚ[X,Y]$. Then $\mathfrak m = (X,Y)$ is their radical (which is a maximal ideal, so both $\mathfrak q_1$ and $\mathfrak q_2$ are indeed primary); and $I = \mathfrak q_1 ∩ \mathfrak q_2$ is by definition decomposable. But as $I = (X^2, Y^2)$ is itself primary (again, its radical is $\mathcal m = (X,Y)$), the only minimal decomposition of it is trivial and hence does neither contain $\mathfrak q_1$ nor $\mathfrak q_2$.
Question 2. No, in $R = ℤ/12ℤ$, the (unique) minimal primary decomposition of zero is $0 = 4R ∩ 3R$ and its associated primes are $2R$ and $3R$. Both are obviously minimal as such, so they are both isolated. But $2R$ is the radical of both the primary ideals $2R$ and $4R$, whereas $3R$ is the only $3R$-primary ideal.
