1. ## ring/ideal

Any input?

If A and B are ideals of R such that $\displaystyle A\cap B =0$ show that $\displaystyle ab=0=ba$.

2. By definition of ideal if $\displaystyle a\in A$ and $\displaystyle b\in B$ then $\displaystyle ab,ba\in A\cap B=0$ (I'm assuming your ring commutative, but I believe if A,B are two sided ideals the proof still works)

3. How does my reasoning sound?

By definition of ideal, $\displaystyle a$ multiplied by any element of $\displaystyle R$ will be in A and will also be in B. So, $\displaystyle b$ multiplied by any element $\displaystyle a$ in $\displaystyle R$ will be $\displaystyle \in B$ and same for $\displaystyle A$. Because the problem tells us that the union of A and B is 0, then because $\displaystyle ab\in A,B$, $\displaystyle ab=0$.

Hows this sound?

4. Originally Posted by sfspitfire23
How does my reasoning sound?

By definition of ideal, $\displaystyle a$ multiplied by any element of $\displaystyle R$ will be in A and will also be in B. So, $\displaystyle b$ multiplied by any element $\displaystyle a$ in $\displaystyle R$ will be $\displaystyle \in B$ and same for $\displaystyle A$. Because the problem tells us that the union of A and B is 0, then because $\displaystyle ab\in A,B$, $\displaystyle ab=0$.

Hows this sound?
Not quite. By definition of ideal if $\displaystyle a\in A$ and $\displaystyle b\in R$ then $\displaystyle ab,ba\in A$, if in addition $\displaystyle b \in B$, since $\displaystyle A\subset R$ we have $\displaystyle ab,ba\in B$ so all this implies that if $\displaystyle a\in A$ and $\displaystyle b\in B$ then $\displaystyle ab,ba \in A\cap B=0$ (not the union!)

5. Hm. I guess I get it all but this part here.

Originally Posted by Jose27
if in addition $\displaystyle b \in B$, since $\displaystyle A\subset R$ we have $\displaystyle ab,ba\in B$

6. Originally Posted by sfspitfire23
Hm. I guess I get it all but this part here.
It is the exact same reason why it is in A.