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Prove or disprove

Krizalid

Active member
Feb 9, 2012
118
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
Jan 29, 2012
661
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
Feb 15, 2012
1,967
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 \(\displaystyle ar,ra \in I \cap J\).