Show that $\sqrt{I}:J=\bigcap_{P\in M}P$

Question

If $I,J$ are ideals in a ring $R$, the colon ideal is $$I:J=\{a\in R\mid ab\in I\text{ for all } b\in J\}.$$

(a) Set $M=\{P\in\mathrm{Spec}(R)\mid I\subset P\text{ and } J\not\subset P\}$. Show that \begin{equation} \sqrt{I}:J=\bigcap_{P\in M}P. \end{equation} (b) Let $K$ be a field and $X,Y\subset K^n$ such that $Y$ is affine variety. Show that \begin{equation} I(X):I(Y)=I(X\setminus Y). \end{equation}

My attempt.

(a) We have \begin{equation} \sqrt{I}=\bigcap_{P\in\mathrm{Spec}(R),\ I\subset P}P. \end{equation} Since $aJ\subset\bigcap_{P\in\mathrm{Spec}(R),\ I\subset P}P$, then $a\in\bigcap_{P\in\mathrm{Spec}(R),\ I\subset P}P$ or $J\subset\bigcap_{P\in\mathrm{Spec}(R),\ I\subset P}P$.

If there are some $P\supset I$ such that $P\supset J$, then would be $I\subset J$. I don't know how to get a contradiction.

For (b), I wish to use (a).

Answer 1

Hint: We always have $$(\cap_i \mathfrak a_i : \mathfrak b)=\cap_i(\mathfrak a_i:\mathfrak b).$$ Now if $\mathfrak b\subset \mathfrak a_i$, then $(\mathfrak a_i:\mathfrak b)=(1)$ and we can discard it from the intersection. If $\mathfrak p$ is prime and $\mathfrak b\not\subset \mathfrak p$, then $(\mathfrak p:\mathfrak b)=\mathfrak p$.