Given a ring $R$, a subset $Q \subset R$ is called a subring of $R$ if it inherits the ring structure from $R$. That is, $Q$ must contain both the $0$ and $1$ (additive and multiplicative identities) of $R$ and be closed under the ring operations of multiplication, addition and additive inverse-taking.


Consider the ring $R = \mathbb{Z} \times \mathbb{Z}$ of ordered pairs of integers with coordinatewise operations, i.e. $(a, b) + (c, d) = (a + c, b + d)$ and $(a, b) \cdot (c, d) = (ac, bd)$. Then the diagonal ring $D = \{(a, a) \mid a \in \mathbb{Z}\}$ is a subring of $R$: it contains the additive identity $(0, 0)$, the multiplicative identity $(1, 1)$ and is closed under multiplication and addition.


The notion of a subring is slightly more subtle than that of a subgroup. Suppose that $R$ is a commutative ring with an idempotent element $i$ other than $0$ and $1$, i.e. $i$ is a solution to the equation $i^2 = i$. Consider the principle ideal $I = Ri = \{a \in R \mid \exists b, a = bi\}$. As an ideal, this set is closed under addition and multiplication and contains the additive identity of $R$. Moreover, this ideal is a ring with multiplicative identity $i$: $i \cdot bi = bi^2 = bi$ for every $b \in R$, so $i\cdot a = a$ for every $a \in I$. However, it is not a subring of $R$ because it does not contain the multiplicative identity of $R$. (Otherwise $1 \in I$ and there is some $j \in R$ such that $ij = 1$, so $i^2j = i$ but also $i^2j = ij  = 1$, and we assumed $i \neq 1$, a contradiction.)

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