Commutative Algebra

Let $R$ be a ring. Let $I,J◃R$. The ideal quotient of $I$ by $J$ is $$(I:J)=\{x\in R\mid Jx\subset I\}$$ We call $$(0:J)=(\{0\}:J)=\{x\in R\mid Jx=\{0\}\}$$ the annihilator of $J$, also denoted by $\mathrm{Ann}(J)$.

If $y\in R$ we write $(I:y)=(I:Ry)$ and $\mathrm{Ann}(y)=\mathrm{Ann}(Ay)$. Using this notation, we see that the set of zero divisors of $R$ is precisely the set ${\cup }_{x\ne 0}\mathrm{Ann}(x)$.

Example: Take $R=\mathbb{Z}$. Let $m,n\in {\mathbb{Z}}^{+}$. Write $m=p_1^{{\alpha }_1}...p_k^{{\alpha }_k},n=p_1^{{\beta }_1}...p_k^{{\beta }_k}$ for some primes $p_1,...,p_k$ and nonnegative exponents ${\alpha }_1,...{\alpha }_k,{\beta }_1,...,{\beta }_k$. Then $$(m\mathbb{Z}:n)=\{z\in \mathbb{Z}\mid \mathrm{zn}\in m\mathbb{Z}\}=q\mathbb{Z}$$ where $q=p_1^{{\gamma }_1}...p_k^{{\gamma }_k}$ and $${\gamma }_i=\max \{{\alpha }_i-{\beta }_i,0\}={\alpha }_i-\min \{{\alpha }_i,{\beta }_i\}$$ In other words, $(m\mathbb{Z}:n)=q\mathbb{Z}$ where $q=m/\gcd (m,n)$.

The following are easily verified.

1. $I\subset (I:J)$
2. $(I:J)J\subset I$
3. $((I:J):K)=(I:JK)=((I:K):J)$
4. If $I_l◃R$ for $l\in X$ then $({\cap }_{l\in X}{I}_l:J)={\cap }_{l\in X}(I_l:J)$
5. If $J_l◃R$ for $l\in X$ then $(I:{\sum }_{l\in X}{J}_l)={\cap }_{l\in X}(I:J_l)$