Sets and Fields

Four units of this course have used the phrase without defining it. Dimension is relative to the field. A matrix is diagonalizable over a field. One-sided invertibility is two-sided for square matrices over a field. Each states a genuine dependence, and none says what the dependence is on.

The answer is short. A field is a set where addition, subtraction, multiplication and division by anything nonzero all work and obey the usual rules. That is the entire content, and everything linear algebra does with scalars, dividing by a pivot, normalising a vector, cancelling a common factor, is one of those four operations.

What makes the definition worth stating precisely is that plausible systems fail it. The integers are closed under addition and multiplication and still are not a field. Arithmetic modulo 6 has all four operations available and still is not a field, for a reason visible in a single equation. Locating exactly which axiom breaks, and exhibiting the element that breaks it, is the skill this unit builds.

The set notation comes first, since the axioms are written in it.

Sets and their notation. A set is determined by its members alone, so $\{1,2,3\}=\{3,1,2,2\}$: order and repetition record nothing. The notation in use throughout:

Membership and inclusion are different relations and the distinction matters: $1\in \{1,2\}$ is true while $1\subseteq \{1,2\}$ is not even well formed, since $1$ is not a set. Two sets are equal exactly when each is a subset of the other, and that double inclusion is the standard way to prove a set identity.

The Cartesian product is what builds the objects of linear algebra: ${\mathbb{R}}^2=\mathbb{R}\times \mathbb{R}$ is the set of ordered pairs, and ${\mathbb{R}}^n$ the set of $n$-tuples.

The field axioms. A field is a set $F$ with operations $+$ and $\cdot$ such that, for all $a,b,c\in F$:

1. Closure. $a+b\in F$ and $ab\in F$.
2. Associativity. $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$.
3. Commutativity. $a+b=b+a$ and $ab=ba$.
4. Identities. There are $0,1\in F$ with $a+0=a$ and $a\cdot 1=a$, and $1\ne 0$.
5. Additive inverses. For each $a$ there is $-a$ with $a+(-a)=0$.
6. Multiplicative inverses. For each $a\ne 0$ there is $a^{-1}$ with $a{a}^{-1}=1$.
7. Distributivity. $a(b+c)=ab+ac$.

Axiom 6 is the one that carries the weight, and its restriction to $a\ne 0$ is not an oversight. If $0$ had an inverse then $0\cdot 0^{-1}=1$, while $0\cdot b=0$ for every $b$, forcing $1=0$ and contradicting axiom 4. So the exclusion is required for consistency, not convenience.

Axiom 7 is the only one that mentions both operations. Everything connecting addition to multiplication, including $0\cdot a=0$, which is a theorem rather than an axiom, is derived from it.

Each of these satisfies most of the axioms. Naming the one that fails, and the element witnessing it, is the whole exercise.

$\mathbb{N}=\{0,1,2,\dots \}$ — fails axiom 5. Closed under both operations, associative, commutative, distributive, with both identities present. But $2$ has no additive inverse: no natural number $b$ satisfies $2+b=0$. Subtraction is unavailable, and axiom 6 fails too.

$\mathbb{Z}$ — fails axiom 6 only. Every other axiom holds, which is what makes it the instructive case. The witness is $2$: there is no integer $b$ with $2b=1$, since $b$ would have to lie strictly between $0$ and $1$. This single missing axiom is why elimination over the integers cannot divide by an arbitrary pivot. Adjoining the missing inverses produces $\mathbb{Q}$.

${\mathbb{Z}}_6$ — fails axiom 6, and visibly. Modular arithmetic supplies closure, associativity, commutativity, distributivity and both identities. But

and two nonzero elements have multiplied to zero. Neither can have an inverse: if $2^{-1}$ existed, multiplying $2\cdot 3=0$ by it would give $3=0$, which is false. Computing the products confirms that $2$, $3$ and $4$ all lack inverses, while only $1$ and $5$ have them ( ).

The set of $2\times 2$ real matrices — fails axioms 3 and 6. Multiplication is not commutative, and a nonzero matrix such as is singular, so has no inverse. It is also a second source of zero divisors, since that matrix times is the zero matrix.

$\{0\}$ — fails axiom 4. Every other axiom holds vacuously, but $1=0$ here, which axiom 4 forbids. The exclusion exists precisely to rule this out; without it the trivial set would be a field and every vector space over it would collapse.

Closure is present in all five. It is the cheapest axiom and it decides nothing on its own. The failures live in the inverses, in commutativity, or in the requirement that the two identities be distinct.

A field has no zero divisors. If $ab=0$ and $a\ne 0$, then $a^{-1}$ exists by axiom 6, and

So a product vanishes only when one of its factors does. Every step used an axiom: the identity, the inverse, associativity, and the theorem $0\cdot a=0$ that distributivity supplies.

Cancellation follows. If $ac=bc$ with $c\ne 0$, then $(a-b)c=0$, so $a-b=0$ and $a=b$. Cancellation is not a separate assumption. It follows from the absence of zero divisors, and it is used silently whenever a common factor is removed from an equation.

The converse fails. $\mathbb{Z}$ has no zero divisors and is still not a field. A commutative ring with identity and no zero divisors is an integral domain; every field is one, and $\mathbb{Z}$ is the standard example of one that is not a field.

${\mathbb{Z}}_n$ is a field exactly when $n$ is prime. Both directions are short.

If $n$ is composite, write $n=ab$ with . Then $a$ and $b$ are nonzero in ${\mathbb{Z}}_n$ but $ab=n=0$, producing a zero divisor, so by the first principle ${\mathbb{Z}}_n$ is not a field.

If $n=p$ is prime and $a\not\equiv 0$, then $gcd(a,p)=1$, so integers $x,y$ exist with $ax+py=1$. Reducing modulo $p$ gives $ax\equiv 1$, so $x$ is the inverse of $a$. Every nonzero element is invertible and ${\mathbb{Z}}_p$ is a field.

The criterion, in short form: composite means zero divisors, prime means inverses. Checking $n$ for primality settles the question without computing a single product table.

$\mathbb{Q}$, the rationals. The smallest field containing $\mathbb{Z}$. Every nonzero $p/q$ has inverse $q/p$, which is where $\mathbb{Z}$ fell short. Closure under division is exactly what was added.

$\mathbb{R}$, the reals. The field linear algebra defaults to. It carries an order compatible with its arithmetic, which is why real eigenvalues can be sorted and real quadratic forms can be called positive, but the order is extra structure, not a field axiom.

$\mathbb{C}$, the complex numbers. A field with no compatible order. Inverses come from the conjugate: for $z\ne 0$,

which is defined precisely because $|z{|}^2$ is a nonzero real. It is also algebraically closed, which $\mathbb{R}$ is not, and that is why characteristic polynomials always split over it.

${\mathbb{Z}}_5$, a finite field. Five elements, and every nonzero one invertible:

Checking: $2\cdot 3=6=1$, $4\cdot 4=16=1$, and $1\cdot 1=1$, all modulo 5. No two nonzero elements multiply to zero, as the prime criterion guarantees. Linear algebra runs over this field unchanged, matrices, elimination, determinants and eigenvalues all work, with arithmetic done modulo 5.

$\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b\in \mathbb{Q}\}$. Closed under multiplication, since

for instance $(1+2\sqrt{2})(3-\sqrt{2})=-1+5\sqrt{2}$. Inverses come from the same conjugate trick as $\mathbb{C}$: multiplying by $a-b\sqrt{2}$ gives the rational norm $a^2-2b^2$, so

and multiplying back returns exactly $1$. The norm is never zero for a nonzero element, because $\sqrt{2}$ is irrational, so no rational $a/b$ can satisfy $a^2=2b^2$.

These five differ in size, in whether they are ordered, and in whether polynomials split, and all satisfy the same seven axioms. That is what makes results stated "over a field" useful: they hold for every one of them at once.

The method is the same each time: assume nothing from resemblance to $\mathbb{R}$, take the axioms in order, and stop at the first failure with a witness in hand.

1. Is ${\mathbb{Z}}_7$ a field?

Seven is prime, so the criterion settles it immediately. The inverses: For each nonzero $a$, find $b$ with $ab\equiv 1(\mathrm{mod}7)$:

Checking three of them: $2\cdot 4=8=1$, $3\cdot 5=15=14+1=1$, $6\cdot 6=36=35+1=1$, all modulo 7. Every nonzero element is invertible.

Answer: yes, a field with seven elements.

2. Is ${\mathbb{Z}}_8$ a field?

Eight is composite, $8=2\cdot 4$, so the criterion predicts a zero divisor. Confirming: $2\cdot 4=8=0(\mathrm{mod}8)$, with both factors nonzero.

That single equation kills axiom 6 for both. If $2^{-1}$ existed, multiplying $2\cdot 4=0$ through by it would give $4=0$, false in ${\mathbb{Z}}_8$. The same argument rules out an inverse for 4.

Answer: no. Failing axiom: multiplicative inverses. Witness: $2\cdot 4=0$ with $2,4\ne 0$.

Note what was not claimed, that no inverse was found by searching. The argument shows none can exist.

3. Is $E=\{x\in \mathbb{Z}:x\text{ is even}\}$ a field?

Closure holds: even plus even is even, even times even is even. Associativity, commutativity and distributivity are inherited from $\mathbb{Z}$. Additive inverses are present, since $-x$ is even whenever $x$ is.

The failure is earlier than expected. Axiom 4 requires a multiplicative identity $1\in E$, and $1$ is odd, so $E$ does not contain it. No element of $E$ acts as an identity either: if $e\cdot x=x$ for all even $x$, taking $x=2$ forces $e=1\notin E$.

Answer: no. Failing axiom: multiplicative identity. This is the case where checking axioms in order matters, jumping to inverses would miss that the identity they are defined against is absent.

4. Is $\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b\in \mathbb{Q}\}$ a field?

Closure under multiplication needs the expansion:

which has the required form since $\mathbb{Q}$ is closed. Concretely $(1+2\sqrt{2})(3-\sqrt{2})=(3-4)+(-1+6)\sqrt{2}=-1+5\sqrt{2}$.

For inverses, multiply by the conjugate $a-b\sqrt{2}$:

a rational number. It is nonzero whenever $a+b\sqrt{2}\ne 0$: if $a^2=2b^2$ with $b\ne 0$ then $(a/b{)}^2=2$, making $\sqrt{2}$ rational. So

For $1+2\sqrt{2}$: the norm is $1-8=-7$, giving $(1-2\sqrt{2})/(-7)=-\frac{1}{7}+\frac{2}{7}\sqrt{2}$. Verifying, $(1+2\sqrt{2})(-\frac{1}{7}+\frac{2}{7}\sqrt{2})=(-\frac{1}{7}+\frac{8}{7})+(\frac{2}{7}-\frac{2}{7})\sqrt{2}=1$.

Answer: yes. The structure of the argument is identical to the one used for $\mathbb{C}$, with the norm $a^2-2b^2$ playing the role of $|z{|}^2$.