$(ax)(ay) = a(xy) \in (a)$

Question

Lemma 26.1 Let $R$ be a commutative ring with unity element $e$. The set $(a) = \{ar : r \in R\}$ is an ideal of $R$.

Proof. First, we will show that $(a)$ is a subring of $R.$ Since $a = ae$ then $a ∈ (a)$ and so $(a) \ne \emptyset$. Let $ax \in (a)$ and $ay \in (a)$. Then $ax − ay = a(x − y) \in (a)$ since $x − y \in R$. Thus, $ax − ay \in (a).$ Also, $(ax)(ay) = a(xy) \in (a)$ since $xy ∈ R.$ $(a)$ is a subring of $R$.

Finally, for any $ar \in (a)$ and all $t ∈ R$ we have $t(ar) = t(ra) = (tr)a \in (a)$ and $(ar)t = a(rt) = (rt)a ∈ (a).$ Thus, $(a)$ is an ideal of $R.$

I didn't understand how $(ax)(ay) = a(xy) \in (a)$ in proof.

Please expain it.

Answer 1

The second $a$ appears to be dropped by accident, but the claim is still valid.

Since the ring is commutative, we can write $(ax)(ay) = a(axy)$. Since $a$, $x$, and $y$ all belong to $R$, so does $axy$. Thus, we've writetn $(ax)(ay)$ as "$a$ times a member of $R$", so $(ax)(ay) \in (a)$.

Answer 2

Note that you can safely drop the part of the proof (which contains the step you mention) where you show that if $u, v \in (a)$, then $uv \in (a)$. This is because you are showing next the stronger statement that if $u \in (a)$ and $v \in R$, then $u v \in (a)$.