Given $A=K[x_1,\dots,x_n]$ a polynomial ring on a field $K$, let $p(x)\in A$ be an element, and $M_1,\dots,M_s$ some maximal ideals.
Is it true that $$\cap(M_i,p) = (\cap M_i,p)?$$
I obtained that it's true if $K$ is algebraically closed, since you can work well with their varieties, but I don't know how to dis/prove it in general.
Some addictional facts:
I proved that, in any case,
$$\cap(M_i,p) = \sqrt{(\cap M_i,p)}$$ So it's sufficient to prove that the ideal is radical.
Another thing I discovered is that this fact is equivalent to
For every radical $0$-dimensional ideal $Q$, if $Q\subseteq J$, then $J$ is radical
Ok, I just found the solution.
It's obvious that $$\cap(M_i,p) \supseteq (\cap M_i,p)$$
but if $p\not\in M_1,\dots,M_r$, $p\in M_{r+1},\dots,M_s$ then $$\exists m_i\in M_i, \exists a_i : m_i+a_ip =1 \quad \forall i\le r$$ so $$q\in \cap(M_i,p) \iff q\in M_{r+1}\cap\dots\cap M_s$$ $$q=q\prod(m_i+a_ip)=q\prod m_i + p(\dots)\in (\cap M_i,p)$$ resulting in $$\cap(M_i,p) \subseteq (\cap M_i,p)$$