# Rings with Identity

Recall from that Rings page that a ring $(R, +, *)$ is essentially an abelian group $(R, +)$ with the extra binary operation $*$ that is required to be associative and distributive with $+$. We do not require the existence of an identity element for $*$ though. Of course, for many such rings an identity element exists for the second operation of the ring and we define such rings as rings with identity.

Definition: Let $(R, +, *)$ be a ring. Then $(R, +, *)$ is said to be a Ring with Identity if there exists an element $1 \in R$ such that for all $x \in R$ we have that $x * 1 = 1 * x = x$. |

Many of the rings that we have already examined are indeed rings with identity. For example, $(\mathbb{R}, +, *)$, $(\mathbb{Z}, +, *)$, and $(\mathbb{Q}, +, *)$ are all rings with identity (namely the number $1$).

Of course, not all rings have an identity though.

For example, consider the following set:

(1)Note that if $M_{22}$ denotes the set of all $2 \times 2$ matrices then $H \subset M_{22}$. We already know that $(M_{22}, +, *)$ is a ring and for $(H, +, *)$ to be a ring we only need to show that $H$ is closed under $+$, for every element $a \in H$ there exists an element $-a \in H$ such that $a + (-a) = 0$ ($where [[$ 0$ is the identity for $+$), and $H$ is closed under $*$.

Clearly $H$ is closed under $+$ since for all $\begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix}, \begin{bmatrix} c & d\\ 0 & 0 \end{bmatrix} \in H$ we have that:

(2)Also, for every element $\begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix} \in H$ we have that $\begin{bmatrix} -a & -b\\ 0 & 0 \end{bmatrix} \in H$ and:

(3)Lastly, $H$ is closed under $*$ since for all $\begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix}, \begin{bmatrix} c & d\\ 0 & 0 \end{bmatrix} \in H$ we have that;:

(4)So indeed $(H, +, *)$ is a subgroup of $(M_{22}, +, *)$. However there is no identity for $*$ (why?)