Field Axioms

From the Real Number Line to Algebraic Rules

The previous tutorial introduced \(\mathbb{R}\) as the set of real numbers represented by points on the number line. We now focus on the arithmetic structure carried by that set. Rather than treating familiar rules such as associativity or distributivity as automatic, we list the basic rules that make the real numbers a field. Later results about addition and multiplication can then be traced to these axioms.

A field is a set equipped with two operations, called addition and multiplication, that satisfy specific rules. The field axioms apply to familiar number systems such as \(\mathbb{Q}\) and \(\mathbb{R}\), but not to every set of numbers. In particular, the integers have addition and multiplication but do not have multiplicative inverses for all their nonzero elements.

The Field Axioms

Here are the rules in full. For any \(x,y,z\in F\), addition and multiplication satisfy:

1. Closure: \(x+y\in F\) and \(xy\in F\).
2. Commutativity: \(x+y=y+x\) and \(xy=yx\).
3. Associativity: \((x+y)+z=x+(y+z)\) and \((xy)z=x(yz)\).
4. Additive identity: There is an element \(0\in F\) such that \(x+0=x\).
5. Additive inverse: For each \(x\in F\), there is an element \(-x\in F\) such that \(x+(-x)=0\).
6. Multiplicative identity: There is an element \(1\in F\) such that \(x1=x\).
7. Multiplicative inverse: For each \(x\in F\) with \(x\ne0\), there is an element \(x^{-1}\in F\) such that \(xx^{-1}=1\).
8. Distributivity: \(x(y+z)=xy+xz\).

The displayed identities and inverses work on either side of an element because the operations are commutative. Thus, for example, \(0+x=x\), \((-x)+x=0\), and \(x^{-1}x=1\) when \(x\ne0\). In the field of real numbers, the operations are the usual addition and multiplication, and the elements \(0\) and \(1\) are the familiar real numbers. In this tutorial we take the fact that these operations on \(\mathbb{R}\) satisfy the field axioms as part of the algebraic description of the real numbers.

Subtraction and division need not be listed as separate operations. They are defined using the field operations and inverses: \(x-y\) means \(x+(-y)\), and \(x/y\) means \(xy^{-1}\) when \(y\ne0\). Division by zero is not defined, since the multiplicative inverse axiom applies only to nonzero elements.

This example points to a useful distinction: having addition and multiplication is not enough to be a field. The requirement that every nonzero element have a multiplicative inverse is essential. Likewise, \(\mathbb{N}_0\) fails to be a field, since, among other failures, a positive integer does not have an additive inverse in \(\mathbb{N}_0\).

Uniqueness of Identities and Inverses

The axioms require identities and inverses to exist. They do not need to include uniqueness as a separate rule: uniqueness follows from the rules already given. The next theorem proves this directly.

Proof. Suppose \(0\) and \(0'\) are both additive identities. Since \(0'\) is an additive identity, \(0+0'=0\). Since \(0\) is an additive identity, \(0+0'=0'\). Hence \(0=0'\).

Suppose \(1\) and \(1'\) are both multiplicative identities. Using first that \(1'\) is an identity and then that \(1\) is an identity gives \(1\cdot1'=1\) and \(1\cdot1'=1'\), so \(1=1'\).

Now let \(x\in F\), and suppose \(u\) and \(v\) are both additive inverses of \(x\). Then \(x+u=0\) and \(x+v=0\). By associativity, commutativity, and the identity and inverse rules, \[ u=u+0=u+(x+v)=(u+x)+v=(x+u)+v=0+v=v. \] Thus the additive inverse is unique.

Finally, let \(x\ne0\), and suppose \(u\) and \(v\) are both multiplicative inverses of \(x\). Then \(xu=1\) and \(xv=1\). Using associativity, commutativity, and the identity and inverse rules, \[ u=u\cdot1=u\cdot(xv)=(ux)v=(xu)v=1\cdot v=v. \] Thus the multiplicative inverse is unique. This proves all four claims. \(\square\)

Because the inverses are unique, the notation \(-x\) and \(x^{-1}\) refers to definite elements, rather than to arbitrary choices. The condition \(x\ne0\) remains important for \(x^{-1}\): the theorem does not provide, and the axioms do not define, a multiplicative inverse for zero.

Cancellation and the Zero-Product Property

Several familiar algebraic rules are consequences of the field axioms, not additional assumptions. Two important examples are cancellation and the zero-product property. Cancellation permits removal of a common additive term; for multiplication, the factor being cancelled must be nonzero.

Proof. First suppose \(x+z=y+z\). Add the additive inverse \(-z\) to both sides. By associativity and the inverse and identity rules, \[ (x+z)+(-z)=(y+z)+(-z) \quad\Longrightarrow\quad x+(z+(-z))=y+(z+(-z)) \quad\Longrightarrow\quad x+0=y+0. \] Therefore \(x=y\).

Next suppose \(z\ne0\) and \(xz=yz\). The multiplicative inverse \(z^{-1}\) exists. Multiply both sides by it and use associativity: \[ (xz)z^{-1}=(yz)z^{-1} \quad\Longrightarrow\quad x(zz^{-1})=y(zz^{-1}) \quad\Longrightarrow\quad x\cdot1=y\cdot1. \] Therefore \(x=y\).

For any \(a\in F\), distributivity gives \(a0=a(0+0)=a0+a0\). Since \(0+a0=a0\), additive cancellation applied to \(0+a0=a0+a0\) yields \(0=a0\). For the final claim, suppose \(xy=0\). If \(x=0\), the conclusion already holds. If \(x\ne0\), multiply both sides by \(x^{-1}\). Then \[ x^{-1}(xy)=x^{-1}\cdot0 \quad\Longrightarrow\quad (x^{-1}x)y=0 \quad\Longrightarrow\quad 1\cdot y=0, \] so \(y=0\). In either case, \(x=0\) or \(y=0\), as claimed. \(\square\)

What the Field Axioms Do—and Do Not—Say

The field axioms provide a precise foundation for algebraic manipulation. They explain why parentheses can be regrouped in sums and products, why terms can be cancelled under the stated conditions, and why a product equal to zero must have a zero factor. When using these rules, the hypotheses matter: cancelling a multiplicative factor is justified only when that factor is nonzero.

A field structure alone does not specify which numbers are positive or how numbers are ordered. The real numbers have an order as well as field operations, but order properties are separate from the field axioms. Similarly, the field axioms do not by themselves characterize the real numbers among all fields: the rational numbers form a field too. Further properties of \(\mathbb{R}\) will be needed to describe what distinguishes the real number system.