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[Q-1] Can we find subrings in $\mathbb{Z}$ that are not ideals of $\mathbb{Z}$?

Edit: There was another question, that I was trying to answer:

[Q-2] Find a subring of $\mathbb{Z}\oplus\mathbb{Z}$ that is not an ideal.

While solving [Q-2], It came to my mind, whether we can find any subring in $\mathbb{Z}$ that is not an ideal of $\mathbb{Z}$. Now, the first thing to do was find out subrings of $\mathbb{Z}$ and some of the examples are the set $n\mathbb{Z}$ under the normal operation. But can we find other subrings of $\mathbb{Z}$ itself?

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    $\begingroup$ What is an "ideal ring"? $\endgroup$
    – russoo
    Sep 27, 2016 at 10:18
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    $\begingroup$ I think he means subrings that are not ideals. $\endgroup$
    – Janik
    Sep 27, 2016 at 10:49
  • $\begingroup$ In the integers, the ideals are precisely the subgroups of $(\mathbb Z,+)$. Since a subring is automatically such a subgroup, all of them are ideals. $\endgroup$
    – rschwieb
    Sep 27, 2016 at 13:39
  • $\begingroup$ Hint (for the original question): find a way to embed $\mathbb{Z}$ as a subring into $\mathbb{Z}\oplus\mathbb{Z}$ and then notice that this is not an ideal. $\endgroup$
    – user26857
    Sep 29, 2016 at 6:06
  • $\begingroup$ See this. Notice that many modern textbooks require in the definition of a ring that it has a multiplicative identity, i.e. all rings (and subrings) must be unitary. According to this definition, $\Bbb Z$ has no proper subring. $\endgroup$
    – Jean-Claude Arbaut
    Sep 29, 2016 at 6:14

1 Answer 1

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The additive subgroups of $\mathbb{Z}$ are precisely $n\mathbb{Z}$ for $n\in{\mathbb{N_0}}$. Every such subgroup is also an ideal, therefore the answer to your question is (if I understood it correctly):

No. Every subring of $\mathbb{Z}$ is also an ideal.

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  • $\begingroup$ That's what I thought but isn't there any counterexample? $\endgroup$
    – MUH
    Sep 27, 2016 at 11:28
  • $\begingroup$ No, at least not in $\mathbb{Z}$. $\endgroup$
    – Janik
    Sep 27, 2016 at 11:35
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    $\begingroup$ @MUH If you still need an example of a subring that isn't an ideal, $\mathbb Q$ has a subring $\mathbb Z$ that isn't an ideal of $\mathbb Q$. $\endgroup$
    – rschwieb
    Sep 27, 2016 at 13:41

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