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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 $R$?

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  • $\begingroup$ The $a_{\kappa}$ are given, the $r_{\kappa}$ range over all elements of $R$. $\endgroup$ – Daniel Fischer Dec 1 '19 at 13:59
  • $\begingroup$ Why do you want $I$ to contain the multiplicative unit? A proper ideal doesn't contain it. $\endgroup$ – Bernard Dec 1 '19 at 14:06
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    $\begingroup$ 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. $\endgroup$ – Alfha Dec 1 '19 at 14:08
  • $\begingroup$ One example can clear your doubt, take ring $R$ = $(\mathbb{Z} , + , \circ)$ and take $a_i$'s = ${2,4}$ now think about the set $I$ ={ $2 \cdot r_1 + 4 \cdot r_2 : r_1 , r_2 \in R$ }. $\endgroup$ – Alfha Dec 1 '19 at 14:16
  • $\begingroup$ @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. $\qquad$ $\endgroup$ – Michael Hardy Dec 1 '19 at 14:31
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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 $r_1=\cdots=r_k=0.$

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  • $\begingroup$ The OP probably was confused with multiplicative units and additive units... I guess. $\endgroup$ – WhatsUp Dec 1 '19 at 14:22
  • $\begingroup$ @WhatsUp : It doesn't look that way to me. The OP clearly distinguishes between the two. $\endgroup$ – Michael Hardy Dec 1 '19 at 14:23
  • $\begingroup$ 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. $\endgroup$ – combinat0ria1 Dec 1 '19 at 16:13
  • $\begingroup$ @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? $\endgroup$ – combinat0ria1 Dec 1 '19 at 16:14
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    $\begingroup$ @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. $\endgroup$ – Michael Hardy Dec 1 '19 at 19:10
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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 $(R, +, \circ)$.

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