Commutative Algebra/Fractions, annihilator, quotient ideals

The quotient of two ideals
We note some properties:

The points 1. and 2. make calling those ideals "quotient" plausible, 3. and 4. less so (although the ideal still gets smaller when adding something to the denominator or shrinking the numerator).

Proof:

1. $$I \cdot J \subseteq K \Leftrightarrow \forall i \in I: iJ \subseteq K \Leftrightarrow I \subseteq (K:J)$$

2. $$J \subseteq I \Leftrightarrow \forall r \in R: rJ \subseteq J \subseteq I$$

3.


 * $$\begin{align}

r \in (I:J+K) & \Leftrightarrow r(J + K) \subseteq I \\ & \Leftrightarrow rJ \subseteq I \wedge rK \subseteq I \\ & \Leftrightarrow r \in (I:J) \cap (I:K), \end{align}$$ where the middle equivalence follows since $$r(J + K)$$ is the smallest ideal containing $$rJ$$ and $$rK$$, and thus is contained in every ideal where the latter two are contained.

4.


 * $$\begin{align}

r \in (\cap_{i \in I} I_i : K) & \Leftrightarrow rK \in \cap_{i \in I} I_i \\ & \Leftrightarrow \forall i \in I: rK \in I_i \\ & \Leftrightarrow \forall i \in I: r \in (I_i : K) \\ & \Leftrightarrow r \in \cap_{i \in I} (I_i:K) \end{align}$$

Exercises

 * Exercise 19.1.1: Prove that for a ring $$R$$ and any ideal $$I \le R$$, $$(R:I) = R$$ and $$(I:R) = I$$.