Rings with Identity

# 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)
\begin{align} \quad H = \begin{Bmatrix} \begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix} : a, b \in \mathbb{R} \end{Bmatrix} \end{align}

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 Hwe have that: (2) \begin{align} \quad \begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix} + \begin{bmatrix} c & d\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} a + c & b + d\\ 0 & 0 \end{bmatrix} \in H \end{align} 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 Hand: (3) \begin{align} \quad \begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix} + \begin{bmatrix} -a & -b\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \end{align} 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 Hwe have that;: (4) \begin{align} \quad \quad \begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix} * \begin{bmatrix} c & d\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} ac & ad\\ 0 & 0 \end{bmatrix} \in H \end{align} So indeed(H, +, *)$is a subgroup of$(M_{22}, +, *)$. However there is no identity for$*\$ (why?)