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Let $R$ be a commutative ring with identity and $a\in R$ has the property that for $b\in R$ if $\langle a\rangle\cap\langle b\rangle=\{0\}$, then $ann_R(a)+ann_R(b)=R$, where $ann_R(x) :=\{r\in R\mid rx=0 \}$ for $x\in R$. Is there any characterization for such an element?

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1 Answer 1

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This is just an extended comment:

If $Ra$ is essential in a direct summand of $R,$ then $a$ has this property. To see this, fix some idempotent $e\in R$ such that $Ra$ is essential in $Re.$ Then if $b\in R$ satisfies $\langle a\rangle \cap \langle b\rangle=\{0\},$ we have $b\in R(1-e).$ Then $ann_R(a)+ann_R(b)\supseteq R(1-e)+Re=R.$

As a consequence, any element in an arbitrary direct product of chain rings and domains has this property.

In general, the equation $ann_R(a)+ann_R(b)=R$ is equivalent to the existence of some $s\in ann_R(a)$ such that $1-s\in ann_R(b).$ In other words, we just need some $s\in R$ such that $sa=0$ but $sb=b.$ This is a local condition, where $s$ might depend on $b.$

Let $F$ be a field, and consider the ring $$ R=F[a,b_1,b_2,\ldots\, :\, ab_i=0,\, b_ib_j=b_i \text{ for }i<j]. $$ An easy argument, using degrees of monomials, shows that $R$ has only the trivial idempotents. The set of those $b\in R$ with $\langle a\rangle \cap \langle b\rangle=\{0\}$ are the polynomials in $b_1,b_2,\ldots$ with zero constant term. Given any such polynomial $b$, then for $n$ large enough we have $b_na=0$ and $b_nb=b,$ as wanted. Thus, this element $a$ has the property you desire, in only a local manner.

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