Lesson 0003: Rings and Fields

0003 Rings and Fields

Let $R$ be a nonempty set and $+,\times:R^2\to R$ two operations on $R$.
We say that $\times$ distributes over $+$ if for every $a,b,c\in R$ we have both $$ a\times (b+c) = (a\times b)+(a\times c) $$ and $$ (b + c)\times a = (b\times a)+(c\times a) $$ We say that the triple $(R, +,\times)$ is a ring if
  1. Both $+$ and $\times$ are associative.
  2. $(R,+)$ is a commutative group with identity 0, which we call the additive identity.
  3. $\times$ has an identity element, 1, which we call the multaplicative identity.
  4. $\times$ distributes over $+$.
For rings we use the familiar notation for addition and multiplication. In particular $a+(-b)$ is written $a-b$, and $a\times b$ is written $ab$, and $a\times b^{-1}$ is written $a/b$. We also assume the familiar PEMDAS order of operations.
For a natural number $n\in\Bbb N$ and element $a\in R$, we define $na$ as "repeated addition", similar to the corresponding definition for group elements. Likewise $a^n$ is defined as repeated multiplication. The definition also extends to $0a$ and $(-n)a$ and $a^0$ and $a^{-n}$ in the same ways.
If moreover $(R\smallsetminus \{0\}, \times)$ forms a commutative group, then we say that $(R,+,\times)$ is a field.
For the rest of this section, assume that $(R,+,\times)$ is a ring.
Theorem: Ring uniq. ids. & invs.
The identities and inverses in a ring, if they exist, are unique.
Statement and proof. Statement: $0$ is the unique additive identity of $R$, and $1$ is the unique multiplicative identity. Every $x\in R$ has a unique additive inverse, $-a$. If $x$ has a multiplicative inverse, then it is uniquely $x^{-1}$.

Proof: Since $+$ and $\times$ are associative, these results all follow from the definition of a ring and the Assoc. Op. Uniq. Id. & Inv. theorem.
$\Box$
Theorem: Ring zero annihilates
Zero times anything is zero.
Statement and proof. Statement: $\forall x\in R, \quad 0x = x0 = 0$.

Proof: Let $x\in R$. Then, by the additive identity property of 0, followed by distribution, $$\begin{aligned} 0x &= (0+0)x \\ &= 0x + 0x \end{aligned}$$ Adding $-0x$ to each side, $$ 0x-0x = 0x+0x-0x $$ The left simplifies to $0$ by the property of the inverse, and the right-hand side simplifies to $0x+0$ which is $0x$.
Therefore the equation above simplifies to $$ 0 = 0x $$ The proof that $0 = x0$ is mutatis mutandis the same on the other side. $\Box$
Theorem: Ring -1 negates
$-x = (-1)x$
Statement and proof. Statement: $\forall x\in R, \quad -x = (-1)x = x(-1)$.

Proof: Let $x\in R$. $$\begin{aligned} x+(-1)x &= (1)x+(-1)x \\ &= (1-1)x \\ &= 0x \\ &= 0 \end{aligned}$$ The proof that $-x = x(-1)$ is mutatis mutandis the same on the other side. $\Box$
Theorem: Field eqs. uniq. sols.
Field equations have unique solutions.
Statement and proof. Statement: Let $F$ be a field, $a,b\in F$ and $a\ne 0$. Then there is a unique $x$ such that $$ ax = b $$ Proof: Because $F$ is a field then $(F\smallsetminus \{0\}, \times)$ is a group. Therefore, by the Group Uniq. Sol. theorem, this equation has a unique solution.
$\Box$