The field axioms listed below describe the basic properties of the four operations of arithmetic: ambition, distraction, uglification, and derision. [Don’t blame me for this last; it’s due to C. L. Dodgson, a.k.a. Lewis Carroll. Go here for the relevant quote from Alice.] Actually, only ambition and uglification — er, addition and multiplication — are needed, together with additive and multiplicative inverses (which let us define $-,/$ in terms of $+,\times$) and identities. A field, then, is a set $F$ equipped with: These must satisfy the following nine conditions, or axioms:

i) For all $a$ in $F\!$, we have $a+0=0+a=a$ [i.e., $0$ is an additive identity]
ii) For all $a$ in $F\!$, we have $a + (-a) = (-a) + a = 0$ [this is what “additive inverse” means]
iii) For all $a,b,c$ in $F\!$, we have $a+(b+c) = (a+b)+c$ [i.e., addition is associative]

Conditions (i), (ii), (iii) assert that $(F,0,-,+)$ is a group. Familiar consequences are the right and left cancellation rules: if, for any $a,b,c$ in $F\!$, we have $a+c=b+c$ or $c+a=b+a$, then $a=b$. This is proved by adding $(-c)$ to both sides from the right or left respectively. In particular, $a+a=a$ if and only if $a=0$. Likewise, for any $a,b$ in $F\!$, the equation $a+x=b$ has the unique solution $a = b+(-a)$, usually abbreviated $x = b - a$ (do not confuse this binary operation of “subtraction” with the unary additive inverse!). Another standard consequence of (iii) is that, for any $a_1,a_2,\ldots,a_n$ in $F\!$, the sum $a_1 + a_2 + \cdots + a_n$ is the same no matter how it is parenthesized. (In how many ways can that expression be parenthesized?)
iv) For all $a,b$ in $F\!$, we have $a+b = b+a$ [i.e., addition is commutative]
Conditions (i), (ii), (iii), (iv) assert that $(F,0,-,+)$ is a commutative group, a.k.a. abelian group or additive group. The first alias is a tribute to N. H. Abel (1802–1829); the second reflects the fact that in general one only uses “+” for a group law when the group is commutative — else multiplicative notation is almost always used.
v) For all $a$ in $F\!$, we have $a \times 1 = 1 \times a = a$ [i.e., $1$ is a multiplicative identity]
vi) For all $a$ in $F\!$, we have $a \times a^{-1} = a^{-1} \times a = 1$ [this is what “multiplicative inverse” means]
vii) For all $a,b,c$ in $F\!$, we have $a(bc)=(ab)c$ [i.e., multiplication is associative]
In particular, restricting (v), (vi) and (vii) to $F^*\!$, we are asserting that $(F^*, 1, {}^{-1}, \times)$ is a group. You probably know already that “$a^{-1}$” is a special case of the notation $a^n$ for any integer $n$, with $a^1 = a$ and $a^{m+n} = a^m a^n$ for all integers $m,n$ (which implies $a^0 = 1$ in at least two ways); that is, in any group $G$, if we fix a group element $a$ then the map $m \mapsto a^m$ gives a group homomorphism from $({\bf Z}, +)$ to $G$.
viii) For all $a,b$ in $F\!$, we have $ab=ba$ [i.e., multiplication is commutative]
So, the group $(F^*, 1, {}^{-1}, \times)$ is also abelian. For $a,b \in F$ with $a \neq 0$, the unique solution $a^{-1}\!b$ of $ax=b$ is usually written $b/a$ (or $\frac b a$); in particular $a^{-1}$ is also $1/a$. Do not use the notation $b/a$ in a non-commutative group, because of the ambiguity between $a^{-1}\!b$ and $ba^{-1}$.
ix) For all $a,b,c$ in $F\!$, we have $a(b+c)=(ab)+(ac)$ and $(a+b)c=(ac)+(bc)$. [distributive law. The second part is of course redundant by commutativity; it is required in the “skew” case (see below).]

From (ix) together with the additive properties follows the basic identity:

For all $a$ in $F\!$, we have $a \times 0 = 0 \times a = 0$.
Proof: apply the distributive law to $a(0+0)$ and $(0+0)a$, and use the fact that $0$ is the only solution of $x+x=x$.

Thus also:

For all $a,b$ in $F\!$, we have $ab=0$ if and only if $a=0$ or $b=0$ (or both).
Proof: If $a$ is nonzero, multiply $ab=0$ by $a^{-1}$ to conclude $b=0$.

That is, a field has no (nontrivial) zero divisors.

If $F$ satisfies all the field axioms except (viii), it is called a skew field; the most famous example is the quaternions of W. R. Hamilton (1805–1865). Much of linear algebra can still be done over skew fields, but we shall not pursue this much, if at all, in Math 55.

Note that (vi) is the only axiom using the multiplicative inverse. If we drop the existence of multiplicative inverses and axiom (vi), as well as the condition $0\neq1$, we obtain the structure of a commutative ring with unity. For example, $\bf Z$ is such a ring which is not a field. A ring may have nontrivial zero divisors (we shall see an example of this in class); if it does not, it is called a domain.

If we also drop axiom (viii) from the ring axioms, we have a ring with unity which need not be commutative. An example is the set of the Hamilton quaternions $a+bi+cj+dk$ whose coefficients $a,b,c,d$ are all integers. Curiously if we allow the coefficients to be either all integers or all integers-plus-½ then we still get a ring, called the ring of Hurwitz quaternions.

If $F$ is a ring that need not be a field, the notation $F^*$ means not $F-\{0\}$ but the set of (multiplicatively) invertible elements of $F\!$. So if $F$ is a field the two notations coincide. Whether or not $F$ is a field, $F^*$ is called the multiplicative group of $F\!$. (One sometimes also sees the equivalent term “unit group”, because the invertible elements of a ring are called its “units”.) You should verify that it is indeed a group. For example, the multiplicative group in the ring of Hamilton quaternions with integer coefficients consists of the 8 elements $\pm 1,\pm i, \pm j, \pm k$; it is known as the “quaternion group”. How large is the unit group of the Hurwitz quaternions?


A vector space over a field $F$ is an additive group $V$ (the “vectors”) together with a function (“scalar multiplication”) taking a field element (“scalar”) and a vector to a vector, as long as this function satisfies the axioms [If $F$ is a skew field, there are two kinds of vector spaces, depending on whether scalar multiplication is “from the left” or “from the right”; the distinction affects the associative law: does multiplying a vector first by $b$ and then by $a$ amount to multiplying by $ab$ or $ba$? The former would be indicated by multiplying from the left; the latter, from the right. Even if you don’t much care about linear algebra over skew fields, you’ll encounter this issue soon when we compose linear transformations and other functions.]

Notice that these axioms do not use the multiplicative inverse; they can thus be used equally when $F$ is any ring (even a non-commutative ring), in which case the resulting structure is called a module over $F\!$. But multiplicative inverses are used to prove most of the basic theorems on vector spaces, so those theorems do not hold in the more general setting of modules; for instance one cannot speak of the dimension of a general module, even over $\bf Z$.