MHB Prove/Disprove: $I+J$ and $I\cap J$ are Ideals of $A$

[Krizalid1]

Let $I$ and $J$ be ideals of a ring $A.$ Are $I+J$ and $I\cap J$ ideals of $A$?

 

[Fernando Revilla]

More general. Let $\{I_j:j\in J\}$ be a family of ideals of a ring $A$. Are $\displaystyle\sum_{j\in J}I_j$ and $\displaystyle\bigcap_{j\in J}I_j$ ideals of $A$?

 

[Deveno]

This is merely matter of working through the definitions.

For example, let [math]r,r' \in I+J[/math]. This means that:

[math]r = x + y, r' = x' + y', x,x' \in I, y,y' \in J[/math].

So [math]r - r' = (x + y) - (x' + y') = (x - x') + (y - y') \in I+J[/math],

since [math]I,J[/math] are both ideals of [math]A[/math] (and thus additive subgroups).

This shows [math]I+J[/math] is an additive subgroup of [math](A,+)[/math].

Now let [math]a \in A[/math] be any element. We have:

[math]ar = a(x + y) = ax + ay \in I + J[/math], because [math]I,J[/math] are both IDEALS.

The proof that [math]ra \in I + J[/math] is similar, and left to the reader.

This proof clearly generalizes to any family of ideals indexed by a FINITE set. The infinite case has some complications better off discussed elsewhere.

A similar approach works for [math]I \cap J[/math]. It should be clear that [math]I \cap J[/math] is an additive subgroup of [math]A[/math]. I hope you can see how to prove that for any:

[math] a \in A, r \in I \cap J[/math] that [math] ar,ra \in I \cap J[/math].