# Prove or disprove

#### Krizalid

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

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Scholar
This is merely matter of working through the definitions.

For example, let $$\displaystyle r,r' \in I+J$$. This means that:

$$\displaystyle r = x + y, r' = x' + y', x,x' \in I, y,y' \in J$$.

So $$\displaystyle r - r' = (x + y) - (x' + y') = (x - x') + (y - y') \in I+J$$,

since $$\displaystyle I,J$$ are both ideals of $$\displaystyle A$$ (and thus additive subgroups).

This shows $$\displaystyle I+J$$ is an additive subgroup of $$\displaystyle (A,+)$$.

Now let $$\displaystyle a \in A$$ be any element. We have:

$$\displaystyle ar = a(x + y) = ax + ay \in I + J$$, because $$\displaystyle I,J$$ are both IDEALS.

The proof that $$\displaystyle ra \in I + J$$ 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 $$\displaystyle I \cap J$$. It should be clear that $$\displaystyle I \cap J$$ is an additive subgroup of $$\displaystyle A$$. I hope you can see how to prove that for any:

$$\displaystyle a \in A, r \in I \cap J$$ that $$\displaystyle ar,ra \in I \cap J$$.