Let $R$ be a commutative ring with unity. If $a_1, a_2, \ldots,a_k ∈ R$, prove that $I = \{a_1r_1 + a_2r_2 + \cdots + a_kr_k \mid r_1, r_2, \ldots, r_k ∈R\}$ is an ideal of $R$.
I wanted to start by showing that $I$ is a subring of $R$, but I'm stuck trying to show that $I$ is nonempty. It's not clear to me from the problem statement that the zero element or the unity element are necessarily in $a_1, a_2, \ldots ,a_k$ or $r_1, r_2, \ldots ,r_k$.
What am I missing? Isn't it possible that the zero and the unity are not in either of these two subsets of $R$?
Best Answer
Why would you expect the unit element to be in $I$? Ideals do not generally contain the unit element. For example, the set $\{0, \pm 6, \pm12, \pm18,\ldots\}$ of all integer multiples of $6$ is an ideal in $\mathbb Z.$
The zero element is in $I$ because that is the case in which $r_1=\cdots=r_k=0.$