Rings

A ring is a set R with two binary operations, $+, \cdot$ such that 1) $(R, +)$ is an abelian group 2) $\cdot$ is associative ($\forall x, y, z \in R, x \cdot (y \cdot x) = (x \cdot y) \cdot z$) 3) $\cdot$ distributes over $+$ ($\forall x, y, z \in R x\cdot (y+z) = x \cdot y + x \cdot z, (x+y)\cdot z = x \cdot z + y \cdot z$)

A ring is commutative if $\cdot$ is commutative ($\forall x, y \in R, x \cdot y = y \cdot x$). A ring is unitary if there is an identity for $\cdot$. (there exists $1 \in R$ such that $\forall x \in R, x \cdot 1 = 1 \cdot x = x$).

A note on notation: $ab = a \cdot b$ and parenthesis will be excluded if unambiguous.

Examples:

• $(\mathbb{Z}, +, \cdot)$ is a commutative ring
• $(\mathbb{Z}/n\mathbb{Z}, +, \cdot)$ is a commutative ring for all $n > 1$.
• $\{0\}$ is a commutative unitary ring
• ($M_n(\mathbb{R}), +, \cdot)$ which is the ring of $n \times n$ matrices is a unitary ring.
• Let R be a ring and X a set, then $R^x = \{f:X \rightarrow R\}$ forms a ring with pointwise addition and multiplication ($f+g = x \mapsto f(x)+g(x), f \cdot g = x \mapsto f(x)\cdot g(x)$).

A division ring (or skew field) is a unitary ring R such that for all $x \in R\backslash \{0\}$ there exists $y \in R, xy = yx = 1$.

A field is a commutative division ring. For example. $(\mathbb{Q}, +, \cdot), (\mathbb{R}, +, \cdot), (\mathbb{C}, +, \cdot)$ are fields.

Proposition: Let R be a ring 1) for all $x \in R, 0 \cdot 0 = 0$ 2) for all $x, y \in R, (-x)y = -xy = x(-y)$ 3) if R is unitary: multiplicative identity 1 is unique and $(-1)a = -a$.

Proof: 1) $) \cdot x = (0+0)\cdot x = 0\cdot x + 0 \cdot x$ so $0 = 0 \cdot x$ (because cancellation in (R,+)) $x \cdot 0 = x \cdot (0+0) = x \cdot 0+x \cdot 0$ so $0 = x \cdot 0$ 2) $0 = 0 \cdot y = (x+(-x))\cdot y = x \cdot y + (-x)y$ so $(-x)y = -xy$. $0 = x \cdot 0 = x \cdot (y+(-y)) = xy + x(-y)$ so $x(-y) = -xy$. 3) Let 1 and e be multiplicative identities. Then $1 = 1 \cdot e = e$ so the identity is unique. Also $(-1)\cdot a = -(1 \cdot a) = -a$.

Zero divisor: let R be a ring, an element $x \in R$ such that $y \in R \backslash \{0\} : xy = 0$ or $yx = 0$. For example, $\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}^2 = \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$ so $\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$ is a zero divisor.

Unit: let R be a unitary ring, $x \in R$ is a unit if there exists $y \in R : xy = yx = 1$. A division ring is unitary if every $x \in R \backslash \{0\}$ is a unit.

Proposition: if R is a unitary ring, then $R* = \{x \in R: \exists y \in R xy = yx = 1\}$ is a group.

Proof (non rigurous): 1 is the identity, R* is associative (due to ring axioms), and it is closed under inverses by definition, hence R* is a group.

Proposition: Let R be a unitary ring, $1 \neq 0$. Then no element is both a unit and a zero divisor.

Proof: assume $x, y, v \in R$ are nonzero such that xy = 0 and xv = 1 = vx. Then $y = 1 \cdot y = vxy = v \cdot 0 = 0$ which is a contradiction. If instead yx = 0 then $y = y \cdot 1 = yxv = 0 \cdot v = 0$ which is also a contradiction.

Proposition: let R be a ring. If a \in R is not a zero divisor then for all $b \in R$ if ab = ac then b = c, if ba = ca, then b = c (proof is left as an exercise to the reader).

A commutative unitary ring whose only zero divisor is 0 is called an integral domain. For example, every field is an integral domain, $\mathbb{Z}$ is an integral domain and $\mathbb{Z}/n\mathbb{Z}$ is both an integral domain and a field if n is prime.

Proposition: A finite integral domain is a field.

Proof: Let R be an integral domain. Pick $a \in R \backslash \{0\}$. Then $R \rightarrow R$ defined by $x \mapsto a\cdot x$ is injective. Since $\vert R \vert = \vert R \vert$ and both are finite, the map is surjective. Then there exists an $x \in R$ such that $a \cdot x = 1$.