Example of subring that is not an ideal If $I$ is ideal, then $I$ is a subring.
But, if $R$ is a subring, $R$ is not necessarily an ideal.
In other words, the converse of the first statement does not always hold.
Examples of ideal in $\mathbb{Z}_6 = \{0,1,2,3,4,5\}$, where $I$ denoted an ideal:


*

*$I = \{ \overline{0} \}$

*$I = \{ \overline{0} , \overline{2} , \overline{4} \}$

*$I = \{\overline{0} , \overline{3} \}$
I have two questions here:


*

*How to find an example of a subring that is not necessarily an ideal? Are there any examples in $\mathbb{Z}$?

*How do we prove that $I = \{ \overline{0}, \overline{3} \}$ is an ideal of $\mathbb{Z}_6$, above?

 A: You can’t find any examples in $\Bbb Z$ of subrings that aren’t ideals, ’cause every subgroup of $\Bbb Z$ is an ideal.
As for examples of subrings that aren’t ideals, take any nonzero subring of any field. Like $\Bbb Z\subset\Bbb R$.
A: To answer the first question, take the ring $R = \mathbb{Z} \times \mathbb{Z}$. Consider the subring $S = \{ (n,n) : n \in \mathbb{Z} \}$. This is not an ideal, because $(1,0) \cdot (1,1) = (1,0) \not\in S$ even though $(1,0) \in R$ and $(1,1) \in S$.
Although the concrete example above should help you in visualising when a subring fails to be an ideal, do note that this example can be generalised easily. If $R$ is any nontrivial ring with unity, then $R \times R$ contains the subring $T = \{ (r,r) : r \in R \}$, which is not an ideal because $(1,0) \cdot (1,1) = (1,0) \not\in T$.

To answer the second question, note that an ideal must first of all be a subgroup. So, to search for the ideals you can first write down all the subgroups of $\mathbb{Z}_6$ and then verify whether or not each of them are ideals. In this way, it is natural to consider $\{ 0, 3 \}$ as a potential candidate for being an ideal of $\mathbb{Z}_6$, because $\{ 0, 3 \}$ is a subgroup of $\mathbb{Z}_6$.
A: Take a proper subring $S$ of a ring with identity $R$ so that $S$ and $R$ share the identity. Then $S$ cannot be an ideal. Can you find examples? (Caveat: $\mathbb{Z}$ and its quotient rings don't work.)
As another example, consider $R$ as a subring of $R[x]$ (polynomial ring in one variable), where $R$ has at least two elements.
