Let $R$ be a commutative ring. If $(x,y)$ is a regular sequence in $R$ then I can show that the image of $x$ in $R/yR$ is not a zero divisor and $x S=S$ where $S=ann_R(y)$. Hence if $R$ is Noetherian and $x \in J(R)$ [the Jacobson radical of $R$], then by Nakayama lemma , $S=0$ i.e. $y$ is not a zero divisor in $R$, so that $(y,x)$ is a regular sequence in $R$. My question is the following :
Let $R$ be a commutative Noetherian ring and $x \in R$ be a non-unit and not a zero divisor and such that $ y\in R$ and $(x,y)$ is Regular sequence in $R$ $\implies$ $(y,x)$ regular sequence in $R$. Then is it true that $x \in J(R)$ ?