Prove $A=\{\,ra+na\mid r\in R, n\in\Bbb{Z}\,\}$ is an ideal.

Question

Let $R$ be a commutative ring, and let $a\in R$. Prove that $A=\{\,ra+na\mid r\in R, n\in\Bbb{Z}\,\}$ is an ideal belonging to $R$.

Remember that we cannot assume that the ring is unital.

I got stumped on this question, and could not proceed.

My approach: it is easy to prove that for all $p,q\in A$, $p-q\in A$. However, what is difficult to prove without assuming $R$ is unital is the fact that $A$ is closed under multiplication with elements of $R$.

Ay help would be appreciated.

Answer 1

Note that $s\cdot(ra+na) = (sr+ns)a+0a$.

Answer 2

Let $b = (ra+na)\in A$ and $c\in R$. We have $$ bc = a(cn+cr) = a(c(n+r) + 0c). $$ Since $c(n+r)\in R$ and $0\in \mathbb{Z}$ we conclude that $bc\in A$ and thus the closeness under multiplication holds.