Basic Theorems Regarding Rings

# Basic Theorems Regarding Rings

Recall from the Rings page that a ring is a (nonempty) set $R$ with two (closed) binary operations $+$ and $\cdot$ such that:

- $+$ is associative, $R$ contains an additive identity element $0$ under $+$, $R$ contains an additive inverse element for each element in $R$, and $+$ is commutative.

- $\cdot$ is associative, and $R$ contains a multiplicative identity element $1$ under $\cdot$.

- $\cdot$ distributes over $+$.

When first looking at groups we looked at some basic results on the Basic Theorems Regarding Groups page. Since every ring is a group with respect to the operation $+$ (more precisely - an abelian group), we note that the results on that page also hold for rings. We will now look at some more basic results.

Proposition 1: Let $(R, +, \cdot)$ be a ring and let $1$ be the identity of $\cdot$. The the identity $1$ is unique. |

**Proof:**Suppose that $1$ and $i$ are both identities on elements in $R$ of $\cdot$. Then:

\begin{align} \quad 1 = 1 \cdot i = i \end{align}

- The first equality comes from the fact that $e$ is an identity of $\cdot$, while the second equality comes from the fact that $1$ is an identity of $\cdot$. Therefore $1 = i$ and so the identity $1$ of $\cdot$ is unique. $\blacksquare$

Proposition 2: Let $(R, +, \cdot)$ be a ring with $0$ be the identity of $+$, and $1$ be the identity of $\cdot$. Then for all $a \in R$ we have that $a \cdot 0 = 0$ and $0 \cdot a = 0$. |

**Proof:**Let $a \in R$. Then:

\begin{align} \quad a \cdot 0 = a \cdot (1 + (-1)) = (a \cdot 1) + a \cdot (-1) = a + (-a) = 0 \end{align}

- Similarly:

\begin{align} \quad 0 \cdot a = (1 + (-1)) \cdot a = (1 \cdot a) + ((-1) \cdot a) = a + (-a) = 0 \quad \blacksquare \end{align}

Proposition 3: Let $(R, +, \cdot)$ be a ring with $1$ be the identity of $\cdot$. Then $a \cdot (-b) = (-a) \cdot b = -(a \cdot b)$. |

**Proof:**Let $a, b \in R$. Then:

\begin{align} \quad a \cdot (-b) = a \cdot ((-1) \cdot b) = (a \cdot (-1)) \cdot b = (-a) \cdot b = (-1) \cdot (a \cdot b) = -(a \cdot b) \quad \blacksquare \end{align}