Let $k \subset K$ be a separable field extension, $A = k[x_1, \ldots, x_n]$, $B=K[x_1, \ldots, x_n]$ and $I \leq A$ an ideal with minimal primary decomposition $$I = \mathfrak{q}_1 \cap \cdots \cap \mathfrak{q}_k.$$ Choose minimal primary decompositions $\mathfrak{q}_iB = \bigcap_j \mathfrak{q}_{i,j}$ in $B$. My question:
Is $$IB = \bigcap_{i,j} \mathfrak{q}_{i,j}$$ a minimal primary decomposition in $B$?
I know $\sqrt{\mathfrak{q}_{i,j}} \cap A = \sqrt{\mathfrak{q}_i}$, hence $\sqrt{\mathfrak{q}_{i,j}} \ne \sqrt{\mathfrak{q}_{i',j'}}$ for $(i,j) \ne(i',j')$ but how can I show, that $\cap \mathfrak{q}_{i,j} \not\subset \mathfrak{q}_{i',j'}$. Is it true, that $\mathfrak{q}_{i,j} \cap A = \mathfrak{q}_i?$