# If $A$ and $B$ are ideals of a ring $R$. Then $A+B$ is an ideal of $R$ generated by $A \cup B$? [closed]

I have proved that $$A+B$$ is an ideal of $$R$$. But I'm not able to prove that it is generated by $$A \cup B$$.

## closed as off-topic by Javi, YuiTo Cheng, Arnaud D., Paul Frost, callculusMay 6 at 17:42

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Guide:

Ideal $$J$$ is by definition the ideal generated by $$A\cup B$$ iff

• $$J$$ is an ideal.
• $$A\cup B\subseteq J$$.
• If $$I$$ is an ideal with $$A\cup B\subseteq I$$ then $$J\subseteq I$$.

Now try to prove $$J=A+B$$ has these properties in the special case where $$A,B$$ are ideals .

(You said that you already proved yourself that it has the first property)

• Thanks for the help! And yes I've proved the first property but now dont know how to prove the other two.🙁 – Mariam May 5 at 12:45
• Do you agree that $A\subseteq A+B$ and $B\subseteq A+B$? That gives the second property. – drhab May 5 at 12:47
• Yes!! Okay and what about the third one? – Mariam May 5 at 12:51
• It remains to prove that $A,B\subseteq I$ implies that $A+B\subseteq I$ for any ideal $I$. Well, elements of $A+B$ have the shape $a+b$ with $a\in A\subseteq I$ and $b\in B\subseteq I$. Then it follows that $a+b\in I$ since ideal $I$ is closed under addition. – drhab May 5 at 12:55
• It's okay.I got it.Thanks a lot for clearing my doubts! – Mariam May 5 at 13:08