Prove that $(\mathfrak{a} : \sum_i \mathfrak{b}_i ) = \bigcap_i (\mathfrak{a} : {\mathfrak{b}_i} ) $

Question

[1] Is the following result true?

$$\bigcap_i (\mathfrak{a} : \mathfrak{b}_i ) = \left(\mathfrak{a} : \bigcap_i \mathfrak{b}_i \right) $$

[2] Prove that $\left(\mathfrak{a} : \sum_i \mathfrak{b}_i\right) = \bigcap_i \left(\mathfrak{a} : \mathfrak{b}_i \right) $ where $\mathfrak{a}, \mathfrak{b}_i$'s are ideals and $(\mathfrak{a} : \mathfrak{b}_i ) $ denotes the ideal quotient ring.

Recall that ideal quotient ring $(\mathfrak{a} : \mathfrak{b} ) = \{x \in A : x \mathfrak{b} \subset \mathfrak{a} \} $.

My attempt:

[1]Proof: Let \begin{align*} & x \in \bigcap_i (\mathfrak{a} : \mathfrak{b}_i ) \\[2pt] & \Rightarrow x \in (\mathfrak{a} : \mathfrak{b}_i ) \; \forall i \\[2pt] & \Rightarrow x \mathfrak{b}_i \subset \mathfrak{a} \; \forall i \\[2pt] & \Rightarrow \; \bigcap_i (x \mathfrak{b}_i) \subset \mathfrak{a} \\[2pt] & \Rightarrow \; x \left( \bigcap_i \mathfrak{b}_i \right) \subset \mathfrak{a} \\[2pt] & \Rightarrow \; x \in \left(\mathfrak{a} : \bigcap_i \mathfrak{b}_i \right) \end{align*}

Conversely if

\begin{align*} & x \in \left(\mathfrak{a} : \bigcap_i \mathfrak{b}_i \right) \\[2pt] & \Rightarrow x \left( \bigcap_i \mathfrak{b}_i \right) \subset \mathfrak{a} \\[2pt] & \Rightarrow x ( \mathfrak{b}_i ) \subset \mathfrak{a} \; \forall i \\[2pt] & \Rightarrow x \subset (\mathfrak{a} : \mathfrak{b}_i ) \; \forall i \\[2pt] & \Rightarrow x \subset \bigcap_i (\mathfrak{a} : \mathfrak{b}_i ) \\[2pt] \end{align*}

No idea of how to proceed for [2].

Answer 1

I see that I have made a mistake in the following step.

\begin{align*} & x \in \left(\mathfrak{a} : \bigcap_i \mathfrak{b}_i \right) \\[2pt] & \Rightarrow x \left( \bigcap_i \mathfrak{b}_i \right) \subset \mathfrak{a} \nRightarrow x ( \mathfrak{b}_i ) \subset \mathfrak{a} \; \forall i \end{align*}

While the solution to problem (2) is:

\begin{align*} x \left(\sum_i \mathfrak{b}_i \right) & \subset \mathfrak{a} \\ \Rightarrow x \mathfrak{b}_i &\subset \mathfrak{a} \quad \forall i \\ \Rightarrow x &\in (\mathfrak{a} : \mathfrak{b}_i) \quad \forall i \\ \Rightarrow x &\in \bigcap_ i (\mathfrak{a} : \mathfrak{b}_i) \end{align*}

In step (2), we have used the fact that if sum of ideals contains any of them as suggested by @user26857 .