Let \(G\) be a group that acts on \(X\). Define

Definition
A Ring: \(R, +, \cdot\) is a \(R\)-set with \(+, \cdot\) operations on \(R\) such that
  • \(R, +\) is an abelian group with \(0\) as the identity and \(-a\) as the inverse of \(a \in R\).
  • Multiplication is associative so \((ab)c = a(bc)\) for all \(a,b,c \in R\).
  • Distributive Law: \((a + b)c = (ac) + (bc)\) and \((a(b + c) = (ab) + (ac)\).


Warning: we don’t know if \(a + b \cdot c\) should be \((a + b) \cdot c\) or \(a + (b \cdot c)\). The convention is to use the second (operator precedence).


More Terminology

  • Ring with identity: This means a rin with a multiplicative identity since a ring always has an additive identity but not necessarily a multiplicative identity. This is a ring \(R\) such that \(\exists 1 \in R\) so that \(1a = a1 = a\) for all \(a \in R\).
  • Commutative Ring: A ring \(R\) such that \(ab = ba\) for all \(a, b \in R\).
  • If \(R\) has a multiplicative identity, then \(a \in R\) is a unit if there exists a \(b\) in \(R\) such that \(ab = 1 = ba\). We call \(b\) an inverse of \(a\) and write \(b = a^{-1}\). (So a unit is an element with a multiplicative inverse)



Basic Facts in a Ring

  1. \(a0 = 0 = 0a\) for all \(a \in R\). To show this, use the distributive law to see that
    $$ \begin{align*} 0a = (0 + 0)a = 0a + 0a. \end{align*} $$
    But \(R\) is abelian with respect to addition so we can cancel \(0a\) from both sides to see that \(0 = 0a\).
  2. If \(1 \in R\), then \(-a = (-1)a\). So this says the additive inverse is -1 multiplied by \(a\). To see why, observe that
    $$ \begin{align*} (1 + (-1))a &= 1a + (-1)a \\ 0a &= 0. \end{align*} $$
  3. The multiplicative identity is unique if it exists.
  4. If \(1 \in R\) and \(a\) is unit in \(R\), then it has a unique inverse
  5. If \(1 \in R\), then define \((R^{\times},\cdot)\) as the set of units in \(R\) so \((R^{\times},\cdot) = \{\text{units }a \in R\}\). So this is set of the elements that have a multiplicative inverse. This set with the multiplication operation is a group. As a consequence, if \(a\) and \(b\) are units, then their product is in \(R^{\times}\). In fact, \((ab)^{-1} = b^{-1}a^{-1}\).



The Zero Ring

This is a ring \(R\) with element \(R = \{0\}\). \(0 + 0 = 0\) and \(0 \cdot 0 = 0\). This is a commutative ring with identity. The multiplicative identity is \(1 = 0\) in this ring. In fact, If \(R\) is a ring with 1, then \(1 = 0\) if and only if \(R = \{0\}\). Sometimes this ring is excluded from the definition of rings …


Fields

Definition
A Field is a commutative ring with 1 such that every nonzero element is a unit and \(1 \neq 0\).


So we’re excluding the zero ring here.


Examples

  • \(\mathbf{Z}\) with addition and multiplication is a commutative ring with identity.
  • \(\mathbf{Z}2 = \{2n \ | \ n \in \mathbf{Z}\}\) is a commutative ring but it doesn't have the identity \(1 \not\in \mathbf{Z}2\).
  • \(\mathbf{Q}, \mathbf{R}, \mathbf{C}\) are all fields.
  • For \(n \geq 1\), \(\mathbf{Z}_n\) is a commutative ring with \(1 = [1]_n\). It is a field if and only if \(n\) is a prime.
  • Let \(R\) be any ring. Then \(S = \text{Mat}_{n \times n}(R)\) the set of matrices with entries in \(R\) is also a ring with matrix addition/multiplication.
    • If \(1 \in R\) so \(R\) includes the multiplicative identity. Then, \(I \in S\).
    • If \(R\) is commutative. Then, \(S\) might not be commutative.
  • Rings of functions: if \(X\) is an arbitrary set and \(R\) is an arbitrary function, then \(F(X,R) = \{ \text{ all functions } \ | \ f: X \rightarrow R \ \}\) is a ring via a pointwise operation. What's a pointwise operation? Given two different functions \(f, g: X \rightarrow R\), then define
    $$ \begin{align*} (f + g)(x) &:= f(x) + g(x) \\ (fg)(x) &:= f(x)g(x) \quad \text{ for } x \in X \end{align*} $$
    As long as the target is a ring this works. No restriction on \(X\).



Subrings

A subring \(S\) of ring \(R\) is a subset, which is a ring via operation inherited from \(R\). This implies that if \(a, b \in S\), then \(a + b, ab \in S\). To show \(S\) is a subring, we have the following proposition

Proposition
\(R\) ring, a subset \(S \subseteq R\) is a subring if and only if
  • \(0 \in S\) or \(S \neq \emptyset\).
  • if \(a,b \in S\), then \(a + b, ab \in S\).
  • If \(s \in S\) then \(-a \in S\).


Examples

  • \(\mathbf{Z}\) is a ring and \(\mathbf{Z}2\) is a subring.
  • \(\mathbf{Z} \subseteq \mathbf{Q} \subseteq \mathbf{R} \subseteq \mathbf{C}\) are subrings.
  • \(R = \text{Mat}_{n \times n}(\mathbf{R})\) be the ring of 2 by 2 matrices with entries in \(\mathbf{R}\). Then, \(S = \big\{ \begin{pmatrix}a & -b \\ b & a \end{pmatrix} \big\}, a, b \in R\) is a subring. In fact, \(S \cong \mathbf{C}\) as rings

Warning: we can have a subring \(S \subseteq R\) such that \(1_S \in S\) and \(1_R \in R\) but \(1_S \neq 1_S\).


Complex Numbers

One way to describe the complex numbers is to say that the elements of \(\mathbf{C}\) are “formal expressions” \(a + bi\) where \(a, b \in \mathbf{R}\) and \(i\) is a new symbol. We add in the usual way and when we multiply, we use the identity \(i^2 = -1\).

Another way to describe the complex numbers is to say that the elements of \(C\) are vectors \((a,b)\) in \(\mathbf{R}^2\) where we have a special notation for \(1 = (1, 0)\) and \(i = (0, 1)\). Now, define \(+\) by vector addition and define \(\cdot\) by

$$ \begin{align*} (a, b) \cdot (a', b') = (aa' - bb', ab' + ba') \end{align*} $$

In terms of the older notation where \((a,b) = a+bi\), it is

$$ \begin{align*} (a + bi)(a' + b'i) = (aa' - bb') + (ab' + ba')i \end{align*} $$

\(\mathbf{C}\) is a commutative ring with 1. In fact, \(\mathbf{C}\) is a field. Why? because we have straight forward formula to finding the inverse. First observe what happens when we multiply by the complex conjugate

$$ \begin{align*} (a + bi)(a' - b'i) = (a^2 + b^2) + 0i. \end{align*} $$

Fact: if \(a, b \in \mathbf{R}\) and \((a,b) \neq (0,0)\), then \(a^2 + b^2 > 0\). From this we get the inverse formula

$$ \begin{align*} \frac{a}{a^2+b^2} + \frac{-b}{a^2+b^2}i = (a + bi)^{-1} \end{align*} $$

So every non-zero element has a multiplicative inverse and so it’s a field because we have this formula.


Example

Let \(R = \mathbf{Z}_3[i]\) is ring. This is the set of formal expressions \(a + bi\) where \(a, b \in \mathbf{Z}_3\). \(i\) is a symbol such that \(i^2 = [-1]_3 = -1\).

We can check very easily that \(R\) is a commutative ring with identity. Is \(R\) a field? It’s obvious but the answer is yes. So for any \(a, b \in \mathbf{Z}_3\) where \((a,b) \neq (0,0)\), then \(a^2 + b^2 \neq 0\). Why? observe that, \(0^2 = 0\), \(1^2 = 1\) but \(2^2 = 1\) in \(\mathbf{Z}_3\). So the only way to get \(a^2 + b^2 = 0\) is to have \(a = b = 0\). The formula for a multiplicative inverse is

$$ \begin{align*} (a + bi)^{-1} = \frac{a}{a^2+b^2} + \frac{-b}{a^2+b^2}i \end{align*} $$

Therefore, \(\mathbf{Z}_3[i]\) is a field with 9 elements. Call this \(\mathbf{F}_9\).

What about \(\mathbf{Z}_5[i]\)? is it a field? No. Because \((2 + i)\) doesn’t have a multiplicative in \(\mathbf{Z}_5[i]\). To see this, observe that

$$ \begin{align*} (2 + i) \cdot (2 - i) = 2^2 + i^2 + 5 = 0 \end{align*} $$




References