Algebraic Structures - Fields, Rings, and Groups

We will now look at some algebraic structures, specifically fields, rings, and groups:

# Fields

 Definition: A field is a set with the two binary operations of addition and multiplication, both of which operations are commutative, associative, contain identity elements, and contain inverse elements. The identity element for addition is 0, and the identity element for multiplication is 1. Given x, the inverse element for addition is -x, and the multiplicative inverse element for multiplication is 1/x (x ≠ 0). Furthermore, multiplication distributes over addition.

One example is the field of rational numbers $\mathbb{Q}$, that is all numbers q such that for integers a and b, $q = \frac{a}{b}$ where b ≠ 0. The definition of a field applies to this number set. We also note that the set of real numbers $\mathbb{R}$ is also a field (see Example 1). Since $\mathbb{Q} \subset \mathbb{R}$ (the rational numbers are a subset of the real numbers), we can say that $\mathbb{Q}$ is a subfield of $\mathbb{R}$. Alternatively we can say that $\mathbb{R}$ is an extension of $\mathbb{Q}$.

## Example 1

Explain why $\mathbb{R}$ is a field.

Suppose that $a, b, c, d \in \mathbb{R}$. We know that $\mathbb{R}$ has addition and multiplication as binary operations since $(a + b) = c$ for some c, and $ab = d$ for some d. Furthermore, we know that addition and multiplication defined on real numbers is both commutative and associative.

Additionally, the identity element for addition is 0, since $\forall \: x \in \mathbb{R}$, $x + 0 = x$, and the identity element for multiplication is 1, since $1x = x$.

Lastly, the inverse element for addition is -x, since $x + (-x) = 0$ (0 being the identity for addition), and the inverse element for multiplication 1/x since $x \cdot \frac{1}{x} = 1$ when x ≠ 0.

## Example 2

Explain why $\mathbb{Z}$ is NOT a field.

We note that for $x \in \mathbb{Z}$, our multiplicative inverse $x^{-1} = \frac{1}{x}$. Note that if $x ≠ \pm 1$, then our inverse $x^{-1} \not \in \mathbb{Z}$. For example, if x = 3, then our multiplicative inverse would be $\frac{1}{3}$, but 1/3 is not an integer. Therefore, $\mathbb{Z}$ is not a field.

## Example 3

Given that S is a set such that $S = \{ x \in \mathbb{R} : x > 0 \}$, explain why S is NOT a field.

Suppose that $x \in S$. The identity element for addition would be 0, and the inverse for addition $x^{-1} = -x$. We note that all elements in S are positive, hence all inverses for addition must be negative. However once again, $x^{-1} \not \in S$, therefore S is not a field.

# Rings

 Definition: A ring is a set with two binary operations of addition and multiplication. Both of these operations are associative and contain identity elements. The identity element for addition is 0, and the identity element for multiplication is 1. Addition is commutative in rings (if multiplication is also commutative, then the ring can be called a commutative ring). Given x, addition also contains the inverse element -x. Multiplication also distributes over addition.

We note that there are two major differences between fields and rings, that is:

• Rings do not have to be commutative. If a ring is commutative, then we say the ring is a commutative ring.
• Rings do not need to have a multiplicative inverse.

From this definition we can say that all fields are rings since every component of the definition of a ring is also in the definition of a field.

## Example 4

Explain why $\mathbb{Z}$ is a ring.

Suppose that $a, b, c \in \mathbb{Z}$. Both addition and multiplication are associative since $a + (b + c) = (a + b) + c$, and $a(bc) = (ab)c$. Furthermore, it follows that the identity element for addition is 0 since, $a + 0 = a$. The identity element for multiplication is 1 since $1x = x$. Addition is commutative too since $a + b = b + a$ (We note that multiplication is also commutative since $ab = ba$, so $\mathbb{Z}$ can be called a commutative ring). Furthermore, addition has the inverse of -a since $a + (-a) = 0$ (Note that multiplication does not need to have a multiplicative inverse and in fact doesn't, since the multiplicative inverse to return the multiplicative identity would be $x^{-1} = \frac{1}{x}$, but $x^{-1} \not \in \mathbb{Z}$). Lastly, multiplication also distributes over addition, that is $a(b + c) = ab + ac$.

## Example 5

What restrictions on n for $n x n$ square matrices form a commutative ring?

Given the $n x n$ matrices A and B, we note that in general, $AB ≠ BA$. If n = 1, then A and B are scalars and are commutative. If n > 1, then the ring of matrices A and B are not commutative.

## Example 6

Is $\mathbb{Z}_n$ a ring for modular congruence?

Suppose that $a, b, c, m \in \mathbb{Z}$. We say that $a \equiv b \pmod m$ if when a and b are both divided by m, their remainders are the same (alternatively we say that m | (a - b)). If $\mathbb{Z}_n$ is the list of elements 0, 1, 2, …, n-1 (all possible remainders from division by n). Addition is commutative since $a + b \equiv b + a \pmod m$. Furthmore, addition and multiplication are associative since $a + (b + c) \equiv (a + b) + c \pmod m$, and $a(bc) \equiv (ab)c \pmod m$. Multiplication is also distributive over addition, that is $a(b + c) \equiv ab + ac \pmod m$. The identity element for addition is 0 as $a \equiv a + 0 \pmod m$, and the identity element for multiplication is 1 as $a \equiv 1a \pmod m$. Lastly, addition has the inverse element -x since $a + (-a) \equiv 0 \pmod m$.

# Groups

 Definition: A group is a set with a binary operation that is associative, contains an identity element and inverse elements for that operation. If multiplication is commutative, then we say the group is an Abelian Group.

We note that groups only have one binary operation while fields and rings have two binary operations.

## Example 7

Explain why the invertible elements of multiplication are a group.

We will denote the set of invertible elements $R^* = \{ x \in \mathbb{R} : \exists y \in \mathbb{R}, xy = 1 \}$. We note that this operation is associative since $xy = yx = 1$. The identity element in this group is 1, and for each x, there exists an inverse element y.

# Comparison of Fields, Rings, and Groups

Commutativity Associativity Distributivity Identity Elements Inverse Elements
Groups (If commutative over multiplication, this group is Abelian) Associative over some Binary Operation - - - Some identity element $e$ Some inverse elements $x^{-1}$