Let $R$ be a commutative ring with unity, prove that $I$ is an ideal of $R$.

Question

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 $R$?

Answer 1

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 $r_1=\cdots=r_k=0.$

Answer 2

Proof: Clearly $I$ is non empty subring of $(R, +, \circ)$ as addictive identity

$e \in I$ and $I$ is ring under the operations of $(R, + , \circ)$ ring. Now for any $r \in R$ we need to show $r \circ \mathbb{i}$ is in $I$.

Where $\mathbb{i} = a_1.r_1 + \cdots + a_k.r_k$ is any element in $I$. Consider $r \circ \mathbb{i} = a_1.r_1.r + \cdots + a_k.r_k.r$

[Note that: Ring is commutative]

Or

$r \circ \mathbb{i} = a_1.r_1' + \cdots + a_k.r_k'$

where $r_i.r =r_i'$ for some $r_i' \in R$

So $r \circ \mathbb{i} \in I$ [by the definition of $I$.]

And so $I$ is ideal of $(R, +, \circ)$.