Let R be an infinite ring without an unity. Let I ⊆ R satisfying: for any a,b ∈ I, r ∈ R, we have a+b ∈ I, ar ∈ I, ra ∈ I. Then there are two possibilities about I:
(1) I must be an ideal, i.e., for any a ∈ I, we have -a ∈ I;
(2) there exists b ∈ I such that -b ∉ I.
Which one is right?

Thanks in advance.

  • $\begingroup$ Why do you believe that one of (1) and (2) has to be true in general? I mean, $I = R$ satisfies (1), but $I = 2\mathbb{Z}_{\geq 0} \subseteq 2 \mathbb{Z}$ satisfies (2). Also notice the third possibility $I = \emptyset$. $\endgroup$ – Matthias Klupsch Oct 16 '16 at 16:11
  • $\begingroup$ @MatthiasKlupsch I think (1) and (2) has to be true in general because "I" is amost an ideal except satisfying "I" is an additional subgroup,i.e.a-b∈ I,that is to say:-b ∈ I. $\endgroup$ – XSR Oct 17 '16 at 4:17
  • $\begingroup$ On the other hand,I think your example may not be proper, since if we take 4 ∈I= 2Z≥0,-6 ∈2Z,then 4(-6)=-24∉ I,which is contradicted with the conditions. All above let us assume that I ≠ ∅. $\endgroup$ – XSR Oct 17 '16 at 4:27
  • $\begingroup$ Yes, my example was wrong. I shall think about this a bit more. $\endgroup$ – Matthias Klupsch Oct 17 '16 at 7:33
  • $\begingroup$ Correct me again if I am wrong, but what happens when multiplication in $R$ is constant $0$? It seems to me that every subset which is additively closed then satisfies your conditions but not all of them need to contain additive inverses. $\endgroup$ – Matthias Klupsch Oct 17 '16 at 7:44

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