# Thread: Ideals - Intersection of Ideals

1. ## Ideals - Intersection of Ideals

If I and J are ideals of R, prove that their intersection $\displaystyle I \cap J$ is also an ideal of R

Hence show that the intersection of an arbitrary nonempty is again an ideal (do not assume that the collection is countable)

Would appreciate help with this exercise ... my main problem is with generalising to an uncountable collection

Peter

2. ## Re: Ideals - Intersection of Ideals

Originally Posted by Bernhard
If I and J are ideals of R, prove that their intersection $\displaystyle I \cap J$ is also an ideal of R

Hence show that the intersection of an arbitrary nonempty is again an ideal (do not assume that the collection is countable)

Would appreciate help with this exercise ... my main problem is with generalising to an uncountable collection

Peter
There should be nothing different in the uncountable case save for notation (which is possible to hide in a proof likes this). How did you prove the other cases?

3. ## Re: Ideals - Intersection of Ideals

I and J ideals of R $\displaystyle \longrightarrow I \cap J$ is an ideal of R

Proof

$\displaystyle x, y \in I \cap J$

$\displaystyle \longrightarrow$ $\displaystyle x, y \in I , x,y \in J$

$\displaystyle \longrightarrow$ $\displaystyle x - y \in I , x - y \in J$

$\displaystyle \longrightarrow$ $\displaystyle x - y \in I \cap J$ .... (1)

$\displaystyle x \in I \cap J$

$\displaystyle \longrightarrow$ $\displaystyle x \in I , x \in J$

$\displaystyle \longrightarrow$ $\displaystyle rx \in I , rx \in J$ for all $\displaystyle r \in R$

$\displaystyle \longrightarrow$ $\displaystyle rx \in I \cap J$ ..... (1)

(1) , (2) $\displaystyle \longrightarrow$ $\displaystyle I \cap J$ is ideal of R

Then since $\displaystyle I \cap J$ is an ideal, use the above proof on K and $\displaystyle I \cap J$ to show $\displaystyle I \cap J \cap K$ is an ideal and so on for countably many ideals.

4. ## Re: Ideals - Intersection of Ideals

Given an uncountable indexing $\displaystyle \Lambda$ of ideals, I modified your proof for the uncountable case. If you take $\displaystyle \Lambda$ to be a countable or finite indexing, you only really need this one proof.

Originally Posted by Bernhard
Proof

$\displaystyle x, y \in \displaystyle{\bigcap_{\alpha \in \Lambda}I_\alpha}$

$\displaystyle \longrightarrow$ $\displaystyle x, y \in I_\alpha$ for all $\displaystyle \alpha \in \Lambda$

$\displaystyle \longrightarrow$ $\displaystyle x - y \in I_\alpha$ for all $\displaystyle \alpha \in \Lambda$

$\displaystyle \longrightarrow$ $\displaystyle x - y \in \displaystyle{\bigcap_{\alpha \in \Lambda}I_\alpha}$ .... (1)

$\displaystyle x \in I \cap J$

$\displaystyle \longrightarrow$ $\displaystyle x \in I_\alpha$ for all $\displaystyle \alpha \in \Lambda$

$\displaystyle \longrightarrow$ $\displaystyle rx \in I_\alpha$ for all $\displaystyle \alpha \in \Lambda$ and for all $\displaystyle r \in R$

$\displaystyle \longrightarrow$ $\displaystyle rx \in \displaystyle{\bigcap_{\alpha \in \Lambda}I_\alpha}$ ..... (1)

(1) , (2) $\displaystyle \longrightarrow$ $\displaystyle \displaystyle{\bigcap_{\alpha \in \Lambda}I_\alpha}$ is ideal of R