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

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 $$R$$?

• The $a_{\kappa}$ are given, the $r_{\kappa}$ range over all elements of $R$. – Daniel Fischer Dec 1 '19 at 13:59
• Why do you want $I$ to contain the multiplicative unit? A proper ideal doesn't contain it. – Bernard Dec 1 '19 at 14:06
• You are doing a mistake, notice that $r_i$'s are generalised elements of $R$ and $a_j$'s are fixed elements of ring, for $j=1,2,3 \cdots k$ ! Best of luck. – Alfha Dec 1 '19 at 14:08
• One example can clear your doubt, take ring $R$ = $(\mathbb{Z} , + , \circ)$ and take $a_i$'s = ${2,4}$ now think about the set $I$ ={ $2 \cdot r_1 + 4 \cdot r_2 : r_1 , r_2 \in R$ }. – Alfha Dec 1 '19 at 14:16
• @Alfha : In proper MathJax usage one would write $I = \{ 2\cdot r_1+4\cdot r_2 : r_1,r_2\in R\},$ with the "equals" sign and the $\{\text{curly braces} \}$ inside of MathJax. The three "equals" signs in your comment don't match the font size of the things that precede and follow them. $\qquad$ – Michael Hardy Dec 1 '19 at 14:31

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 $$r_1=\cdots=r_k=0.$$

• The OP probably was confused with multiplicative units and additive units... I guess. – WhatsUp Dec 1 '19 at 14:22
• @WhatsUp : It doesn't look that way to me. The OP clearly distinguishes between the two. – Michael Hardy Dec 1 '19 at 14:23
• To prove that $I$ is an ideal, it must be a subring. For $I$ to be a subring, it must be nonempty. To show that $I$ is nonempty, I thought maybe I could use the fact that $R$ has a unity. I didn't think that an ideal must necessarily contain the unity, but I appreciate you providing me an example. – combinat0ria1 Dec 1 '19 at 16:13
• @MichaelHardy I know that the zero element must be in $I$ since I'm trying to prove that it is an ideal (therefore it must be a subring), but how can we conclude that one of the $r_i$ is $0$? Edit: Since $r_i$ are just arbitrary elements of $R$, we can just say "Consider the case where $r_1 = ... = r_k = 0$," right? – combinat0ria1 Dec 1 '19 at 16:14
• @combinat0ria1 : Also note that in each ring with a unit element, there is ONLY ONE ideal that contains the unit element, and that ideal is $R$ itself. – Michael Hardy Dec 1 '19 at 19:10

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 $$(R, +, \circ)$$.