# Subring in $\mathbb{Z}$ that is not an ideal.

[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?

• What is an "ideal ring"? Sep 27, 2016 at 10:18
• I think he means subrings that are not ideals. Sep 27, 2016 at 10:49
• 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. Sep 27, 2016 at 13:39
• 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. Sep 29, 2016 at 6:06
• 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. Sep 29, 2016 at 6:14

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.

• That's what I thought but isn't there any counterexample?
– MUH
Sep 27, 2016 at 11:28
• No, at least not in $\mathbb{Z}$. Sep 27, 2016 at 11:35
• @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 $\mathbb Q$. Sep 27, 2016 at 13:41