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Let $R$ be a commutative ring and $\emptyset\neq A\subseteq R$ satisfy the conditions

  1. $a,b\in A\implies a+b\in A$;
  2. $r\in R$ and $a\in A\implies ar\in A$.

I was asked to prove that if $A$ is an ideal then for each $a\in A,$ there is an integer $n\neq 0$, which may depend on $a$, such that $na\in aR+Ra$. However the result looks not true due to the following reasons.

The so-called "counterexample" became apparent as a consequence of the answer to this post. Consider the ring of $\mathbb{Z}[x]$ and the ideal $R=(x)$. Then we consider $R$ as a ring in $\mathbb{Z}[x]$. $R$ satisfies conditions 1 and 2. But if the statement of the result were true then there would exist a nonzero integer $n$ such that $nx\in xR+Rx$ which is impossible since any nonzero member of $xR+Rx$ has degree greater than $1$. Hence the result to be proved is false.

I am not sure if this argument is correct or the result to be proved is simply correct. The reason I'm trying to disprove the result is that I tried proving it for roughly more than 12 hours and I couldn't. So is this argument correct? Or is the result simply true? If it is true could someone give me a hint to prove it? Thanks.

Janitha357
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1 Answers1

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Whether or not $R$ is commutative doesn't matter. What matters is if $R$ is unital. If $R$ is unital then for all $n \in \mathbf Z$

$$ aR + Ra \ni a0_R + (n \cdot 1_R)a = na $$

so any $n$ works.

On the other hand, if $R$ is not unital then this isn't always true as you've seen. If $R = x\mathbf{Z}[x]$ and $a = x$ then $aR + Ra = x^2\mathbf{Z}[x]$ which contains no integer multiples of $x$.

Trevor Gunn
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  • I have learnt that any ring $R$ is ismorphic to some subring of a unital ring. Does it have anything to do with this particular problem? I'm not sure if this question is relevant, but is $R$'s ability to be isomorphic to some subring of a unital ring significant here? – Janitha357 Jul 22 '17 at 16:10
  • @Janitha357 For $R = x \mathbf{Z}[x]$ we have $R \subseteq \widetilde{R} = \mathbf{Z}[x]$ where $\widetilde{R}$ is unital. Then we would have $na \in a\widetilde{R} + \widetilde{R}a$. But this doesn't give you any information about $aR + Ra$. – Trevor Gunn Jul 22 '17 at 16:13
  • Yes I understand it now. I greatly appreciate your help. – Janitha357 Jul 22 '17 at 16:17
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    @Janitha357 See this answer for the Dorroh unital extension and its advantages and disadvantages compared to other unital extensions. – Bill Dubuque Jul 22 '17 at 16:41