Let R be a ring and A,B,C ideals such that A+B=R and BC $\subset$ A, then C $\subset$ A.
My attempt:
Let x $\in$ C, and as C is an ideal of R, then x=a+b for some b $\in$ B and a $\in$ A. And we can have a'=bx=b(a+b).
I don't know how to prove the problem after that.
Any help will be much appreciated.
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Mubarak Alsaeedi
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See the linked dupe for the various forms of Euclid's Lemma, including this ideal form. – Bill Dubuque May 25 '21 at 20:29
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If R has an identity then you can write 1 as the sum of A and B ... $1=a+b$ for $a\in A, b\in B$. Then multiply by c from the right.
Christian Lomp
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@Lomp Thanks, that was helpful. Could we do something if R is a ring with no 1? – Mubarak Alsaeedi May 25 '21 at 13:16
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Well, at least a condition like $RC=C$ is needed. Then $C=RC=(A+B)C\subseteq AC+BC\subseteq AR+A\subseteq A$. – Berci May 25 '21 at 13:33
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@MubarakAlsaeedi no, not in general; for a counterexample, take $R=B=C=2\mathbb{Z}$ and $A=4\mathbb{Z}$ – Atticus Stonestrom May 25 '21 at 13:34