If $R$ is a ring with identity $1\neq 0$, and Ideals $I=(a,b),J=(b,c), a,b,c\in R$ is it true that $$ IJ =(a,b)(b,c)=(ab, ac, b^2, bc) $$ I know it holds for principal ideals (so $(a)(b)=(ab)$) but I'm not sure if it holds in general. I tried to prove it by writing $I=(a)+(b), J=(b)+(c)$ and then $$ IJ=((a)+(b))((b)+(c))=(ab)+(ac)+(b^2)+(bc)=(ab,ac,b^2,bc) $$ but I'm not sure if that's correct.
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Write $\,A = (a)\,$ etc. Then applying the ideal distributive law $$\begin{align} (a,b)(b.c) &= (A+B)(B+C)\\ &= (A+B)B + (A+B)C\\ &= AB+BB+AC+BC\\ &= (ab)+(bb)+(ac)+(bc)\\ &= (ab,bb,ac,bc)\end{align}\qquad$$
Remark $ $ The same equality is true if you read $\,(a,b)\,$ as a gcd because - like ideals - gcds also satisfy the associative, commutative and distributive laws (the same notation is used in order to exploit this commonality, e.g. we can give proofs that work for both gcds and ideals if we use only these common laws).
Bill Dubuque
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Thank you! Also, the remark about the gcd is really interesting. I thought it was just abusive notation... – 14159 Oct 16 '20 at 11:52
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@14159 See also here for the same question about gcds. – Bill Dubuque Oct 16 '20 at 11:55
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@14159 See here for more on the motivation for the common notation for gcds and ideals. – Bill Dubuque Feb 28 '21 at 09:36