]> Commutative Algebra - Ideal Quotients

Ideal Quotients

Let $R$ be a ring. Let $I,J◃R$. The ideal quotient of $I$ by $J$ is

$\left(I:J\right)=\left\{x\in R\mid Jx\subset I\right\}$

We call

$\left(0:J\right)=\left(\left\{0\right\}:J\right)=\left\{x\in R\mid Jx=\left\{0\right\}\right\}$

the annihilator of $J$, also denoted by $\mathrm{Ann}\left(J\right)$.

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

Example: Take $R=ℤ$. Let $m,n\in {ℤ}^{+}$. 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

$\left(mℤ:n\right)=\left\{z\in ℤ\mid \mathrm{zn}\in mℤ\right\}=qℤ$

where $q={p}_{1}^{{\gamma }_{1}}...{p}_{k}^{{\gamma }_{k}}$ and

${\gamma }_{i}=\mathrm{max}\left\{{\alpha }_{i}-{\beta }_{i},0\right\}={\alpha }_{i}-\mathrm{min}\left\{{\alpha }_{i},{\beta }_{i}\right\}$

In other words, $\left(mℤ:n\right)=qℤ$ where $q=m/\mathrm{gcd}\left(m,n\right)$.

The following are easily verified.

1. $I\subset \left(I:J\right)$

2. $\left(I:J\right)J\subset I$

3. $\left(\left(I:J\right):K\right)=\left(I:JK\right)=\left(\left(I:K\right):J\right)$

4. If ${I}_{l}◃R$ for $l\in X$ then $\left({\cap }_{l\in X}{I}_{l}:J\right)={\cap }_{l\in X}\left({I}_{l}:J\right)$

5. If ${J}_{l}◃R$ for $l\in X$ then $\left(I:{\sum }_{l\in X}{J}_{l}\right)={\cap }_{l\in X}\left(I:{J}_{l}\right)$