Ideal generated by a subset of ring.

Question

The definition of Ideals generated by a subset :

Let $S$ be any subset of ring $R$ then an ideal $I$ of $R$ is said to be generated by $S$ if :

(1) $S \subseteq I$.

(2) for any ideal $J$ of $R$ containig $S$,we have:$~~~~I \subseteq J$.

The thing I can't understand is that if $I$ is an Ideal generated a subset $S$ and $J$ is also generated of $S$,then what are the kind of elements that $J$ has which $I$ doesn't have .Also i'm struck with the meaning of smallest Ideal generated by $S$.Please help....

Answer 1

Here's an example of an ideal containing $S$ but not being generated by $S$.

Let $R$ be the ring of integers, and let $S = \{ 12 \}$.

Then the ideal generated by $S$ is the set of all integers that are divisible by 12. This set is usually written as $(12)$ or sometimes $\langle 12 \rangle$.

All of the other ideals that contain $S$ are:

• $(6)$, the ideal of all integers divisible by $6$
• $(4)$, the ideal of all integers divisible by $4$
• $(3)$, the ideal of all integers divisible by $3$
• $(2)$, the ideal of all integers divisible by $2$

and if you also consider improper ideals,

• $(1)$, the ideal of all integers.

You'll observe, for example, that $(12) \subseteq (3)$: every number divisible by $12$ is also divisible by $3$. But these ideals are different: while $(3)$ does contain every integer that is a linear combination of $S$ (i.e. that are multiples of $12$), $(3)$ also contains other integers, such as $3$.

Answer 2

The second condition means you are defining the smallest ideal generated by the set $S$. In this case, if both $I$ and $J$ are generated by $S$ (satisfying both conditions) they are not distinct; to see this note that by the condition 2 we must have $I\subset J$ and $J\subset I$, and so we have $I=J$.

Answer 3

A good exercise is to prove that $I$ is $$ \bigcap_{S \subseteq J} J $$ the intersection of all ideals containing S. Check that this is an ideal - and your condition (2) is obvious. Note that (2) does not say that $S$ generates $J$ (in which case $J=I$), it is only required that $S \subseteq J$