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?
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 $\mathbb{Z}_6$.