Ideal Quotient is Ideal
Jump to navigation
Jump to search
Theorem
Let $R$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b$ be ideals of $R$.
Then the ideal quotient $\ideal {\mathfrak a : \mathfrak b}$ is indeed an ideal.
Proof
We shall check $(1)$-$(3)$ of Test for Ideal.
(1)
We have:
- $0 \mathfrak b = \set 0 \subseteq \mathfrak a$
That is:
- $0 \in \ideal {\mathfrak a : \mathfrak b}$
$\Box$
(2)
Let $x,y \in \ideal {\mathfrak a : \mathfrak b}$.
Then:
- $x \mathfrak b \subseteq a$
and:
- $\paren {-y} \mathfrak b \subseteq a$
Therefore:
- $\paren {x - y} \mathfrak b \subseteq x \mathfrak b + \paren {-y} \mathfrak b \subseteq a$
That is:
- $x - y \in \ideal {\mathfrak a : \mathfrak b}$
$\Box$
(3)
Let $x \in \ideal {\mathfrak a : \mathfrak b}$.
Let $ r \in R$.
Then:
- $\paren{r x} \mathfrak b \subseteq r \paren {x \mathfrak b} \subseteq r \mathfrak a \subseteq \mathfrak a$
That is:
- $r x \in \ideal {\mathfrak a : \mathfrak b}$
$\blacksquare$