Commutative Algebra/Fractions, annihilator, quotient ideals

The quotient of two ideals

Definition 19.1:

Let $R$ be a ring and let $I,J\leq R$ be ideals. Then the quotient ideal $(I:J)$ (also written $I/J$) is defined to be

$(I:J):=\{r\in R|rJ\subseteq I\}$.

We note some properties:

Theorem 19.2 (properties of the quotient ideal):

Let $R$ be a ring and $I,J,K\leq R$ ideals.

1. $I\cdot J\subseteq K\Leftrightarrow I\subseteq (K:J)$
2. $J\subseteq I\Leftrightarrow (I:J)=R$
3. $(I:J+K)=(I:J)\cap (I:K)$
4. $(I\cap J:K)=(I:K)\cap (J:K)$ and, more generally, $\cap _{i\in I}(I_{i}:K)=(\cap _{i\in I}I_{i}:K)$

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.

${\begin{aligned}r\in (I:J+K)&\Leftrightarrow r(J+K)\subseteq I\\&\Leftrightarrow rJ\subseteq I\wedge rK\subseteq I\\&\Leftrightarrow r\in (I:J)\cap (I:K),\end{aligned}}$

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.

${\begin{aligned}r\in (\cap _{i\in I}I_{i}:K)&\Leftrightarrow rK\in \cap _{i\in I}I_{i}\\&\Leftrightarrow \forall i\in I:rK\in I_{i}\\&\Leftrightarrow \forall i\in I:r\in (I_{i}:K)\\&\Leftrightarrow r\in \cap _{i\in I}(I_{i}:K)\end{aligned}}$

$\Box$

Definition 19.3:

In the case $J=\langle x\rangle$ for an $x\in R$, we write

$(I:J)=:(I:x)$

for $I\leq R$.

Exercises

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