If $(a)+(b)$ is a principal ideal then $(a)\cap(b)$ is also a principal ideal.

Asked Jun 28 '21 at 03:16

Active Jun 28 '21 at 03:16

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Suppose $R$ be a commutative ring with $1$ and $a,b \in R$. I have to show if $(a)+(b)$ is a principal ideal then $(a)\cap(b)$ is also a principal ideal.

Suppose $(a)+(b)=(d)$ for some $d \in R$. Then let $a=pd,b=qd$ and $(a+b)=rd$ for some $p,q,r \in R$. I want to show that $(a)\cap(b)=(pqd)$. Clearly $(pqd)\subset (a)\cap(b)$. How do I prove the converse ?

asked Jun 28 '21 at 03:16

user371231

2,481

I don't think you're using the definition of an ideal addition. $I+J={pi+qj: i\in I, j\in J, p,q\in R}$. – daruma Jun 28 '21 at 03:32
Why would you want to consider $(a+b)=rd$? – user10354138 Jun 28 '21 at 03:38
yes, thanks a lot. – user371231 Jun 28 '21 at 04:08

If $(a)+(b)$ is a principal ideal then $(a)\cap(b)$ is also a principal ideal.

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