# Help showing that set of all finite sums of $rx + lx$ is an ideal of commutative ring $R$.

This question is related to another one I asked here. Suppose I wanted to consider the set $(x)$, which consists of all finite sums of elements of the form $rx + lx$, where $r \in R$, a commutative ring (not necessarily with unity), and $l \in \mathbb{Z}$.

If I wanted to show that $(x)$ is an ideal in the ring $R$, I would need to show that $rx + lx$ is an additive subgroup of $R$, and also that $(x)R \subseteq (x)$ and $R(x) \subseteq (x)$.

I'm a little confused about how to actually go about showing this though.

Here's what I've done:

$$\text{To show that}\, (x)\, \text{is an additive subgroup of}\, R, \,$$

1. $0_{R} \in R, \, 0 \in \mathbb{Z}, \, \text{so when}\, r = 0_{R},\, l=0,\, rx + lx = 0_{R}x + 0x = 0.\, \text{So,}\, 0 \in (x).$

2. $\text{Let}\, r_{1},r_{2} \in R \,\text{and}\, l_{1}, l_{2} \in \mathbb{Z}. \,$ Then, $r_{1}x+l_{1}x-(r_{2}x+l_{2}x) = r_{1}x+l_{1}x - r_{2}x-l_{2}x = (r_{1}-r_{2})x + (l_{1}-l_{2})x \in (x), \, \text{since}\, r_{1}-r_{2} \in R$ because $R$ is a ring, and thus is closed under additive inversion; and since $l_{1}-l_{2} \in \mathbb{Z}$, because $\mathbb{Z}$ is certainly closed under additive inversion.

$$\text{To show that}\,R(x) \subseteq (x),$$

Consider $rx + lx \in (x)$. Then, consider any element $s \in R$.

$s(rx+lx)=srx + slx = srx + lsx$ (because $R$ is commutative).

However, and now's the part where I'm in real trouble, while $sr \in R$, $ls$ is not necessarily in $\mathbb{Z}$, so I'm obviously doing something wrong.

Could somebody please help me finish the proof that this set $(x)$ is an ideal of the commutative ring $R$? Thank you.

$srx+lsx =(sr+ls)x+0.x$ and $sr+ls=r'\in R$. So $srx+lsx=r'x+0.x$.
• and same thing for $(rx+lx)s$? – ALannister Mar 19 '17 at 1:40
• Yes, $(rx+lx)s=(rs+ls)x+0.x$ since the ring is commutative. – Tsemo Aristide Mar 19 '17 at 1:41