The first place where ordinary arithmetic begins to fracture is not with a complicated polynomial, but with the innocent equation $2x=1$. In $\mathbb{Q}$ it has the solution

\begin{align*} x=\frac{1}{2}. \end{align*}

In $\mathbb{Z}$ it has no solution. The difference is not that $\mathbb{Z}$ lacks addition or multiplication; it has both. The missing operation is division by a nonzero element.

Fields are the algebraic worlds in which that obstruction has been removed. They are rings where every nonzero scalar can be inverted, and that single condition makes linear equations, polynomial division, vector spaces, determinants, and much of algebraic number theory behave with their familiar strength. A field is not merely a set with arithmetic. It is a place where multiplication by a nonzero element is reversible.

[example: The Equation $2x=1$] In the ring $\mathbb{Z}$, the equation $2x=1$ asks for an integer $x$ whose double is $1$. If $x\in\mathbb{Z}$, then $2x$ is even, while $1$ is odd, so no integer $x$ satisfies $2x=1$. In $\mathbb{Q}$, the element $2$ has inverse $\frac{1}{2}$ because \begin{align*} 2\cdot \frac{1}{2}=1. \end{align*} Multiplying both sides of $2x=1$ by $\frac{1}{2}$ gives \begin{align*} \frac{1}{2}\cdot 2x=\frac{1}{2}\cdot 1. \end{align*} Using associativity and the identity property, \begin{align*} \left(\frac{1}{2}\cdot 2\right)x=\frac{1}{2}. \end{align*} Since $\frac{1}{2}\cdot 2=1$, this becomes \begin{align*} x=\frac{1}{2}. \end{align*} Substitution checks the solution: \begin{align*} 2\cdot \frac{1}{2}=1. \end{align*} This is the structural difference: multiplication by $2$ on $\mathbb{Q}$ has inverse multiplication by $\frac{1}{2}$, so every rational target $y$ is hit by $x=\frac{y}{2}$, while multiplication by $2$ on $\mathbb{Z}$ hits only the even integers and therefore misses $1$. [/example]

The example raises the organizing question for the chapter: what algebraic hypotheses guarantee that every nonzero scalar can be divided by, and what consequences follow from that guarantee?

## Definition

To make division into a permanent algebraic operation, we start with a ring and impose a condition on its nonzero elements. The question is not whether some elements have inverses, but whether every element except zero does.

[definition: Field] A field is a nonzero commutative ring $k$ with identity $1_k$ such that every nonzero element has a multiplicative inverse in $k$. Thus for every $a \in k \setminus \{0\}$, there exists $b \in k$ such that \begin{align*} ab = 1_k. \end{align*} Such an invertible element is called a unit; the next section isolates that word in the broader setting of rings. [/definition]

The letter $k$ is often used for a base field, while $K/k$ denotes a [field extension](/page/Field%20Extension). The slash in $K/k$ does not denote a quotient; it records that $K$ contains a copy of $k$ as a subfield. The definition also assumes the standard commutative-algebra convention that fields are commutative; division rings, where multiplication need not commute, are a separate noncommutative variant.

## Units and First Examples

### Units in Rings

The field axiom depends on the word unit, and the next question is what it means for multiplication by an element of a ring to be reversible inside that same ring. Naming this concept lets us compare fields with rings such as $\mathbb{Z}$, where only a few elements can be undone multiplicatively.

[definition: Unit] Let $R$ be a ring with identity $1_R$. An element $u \in R$ is a unit if there exists $v \in R$ such that \begin{align*} uv = vu = 1_R. \end{align*} The set of all units of $R$ is denoted $R^\times$. [/definition]

Thus the defining condition for a field can be written compactly as $k^\times = k \setminus \{0\}$. This notation is useful because $k^\times$ carries a group structure under multiplication, while $k$ itself carries both addition and multiplication.

### First Examples

A definition should immediately separate the familiar systems that support division from the familiar systems that do not. The next example asks where the field axiom is already present in standard arithmetic.

[example: Standard Fields] The usual arithmetic laws make $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ commutative rings with identity, so the field axiom is the point to check. If $a \in \mathbb{Q}\setminus\{0\}$, write $a=\frac{m}{n}$ with $m,n\in\mathbb{Z}$ and $m\ne 0$, $n\ne 0$. Then $\frac{n}{m}\in\mathbb{Q}$, and \begin{align*} \frac{m}{n}\cdot\frac{n}{m}=\frac{mn}{nm}=1. \end{align*} Thus every nonzero rational number has a rational inverse. If $x\in\mathbb{R}\setminus\{0\}$, its real reciprocal $\frac{1}{x}$ satisfies \begin{align*} x\cdot\frac{1}{x}=1. \end{align*} If $z=a+bi\in\mathbb{C}\setminus\{0\}$, then $a,b\in\mathbb{R}$ are not both $0$, so $a^2+b^2\ne 0$. The inverse candidate \begin{align*} \frac{a-bi}{a^2+b^2} \end{align*} satisfies \begin{align*} (a+bi)\frac{a-bi}{a^2+b^2}=\frac{a^2-abi+abi-b^2i^2}{a^2+b^2}. \end{align*} Since $i^2=-1$, this becomes \begin{align*} \frac{a^2+b^2}{a^2+b^2}=1. \end{align*} Hence $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ are fields. For a prime number $p$, the residue class ring $\mathbb{Z}/p\mathbb{Z}$ is also a field. Let $\bar{a}\ne\bar{0}$. Then $p\nmid a$, and since $p$ is prime this gives $\gcd(a,p)=1$. By *Bezout's identity*, there exist integers $r,s\in\mathbb{Z}$ such that \begin{align*} ra+sp=1. \end{align*} Reducing this equality modulo $p$ gives \begin{align*} \overline{ra+sp}=\bar{1}. \end{align*} Using compatibility of reduction with addition and multiplication, \begin{align*} \bar{r}\bar{a}+\bar{s}\bar{p}=\bar{1}. \end{align*} Since $\bar{p}=\bar{0}$ in $\mathbb{Z}/p\mathbb{Z}$, we get \begin{align*} \bar{r}\bar{a}+\bar{s}\bar{0}=\bar{1}. \end{align*} Because $\bar{s}\bar{0}=\bar{0}$, this is \begin{align*} \bar{r}\bar{a}=\bar{1}. \end{align*} Thus every nonzero residue class modulo $p$ has a multiplicative inverse, so $\mathbb{Z}/p\mathbb{Z}$ is a field. [/example]

The previous example suggests that fields are common, but the definition is restrictive. The next question is what fails in the most important ring that is not a field.

[example: The Ring $\mathbb{Z}$ Is Not a Field] In $\mathbb{Z}$, a unit is an integer $n$ for which there is an integer $m$ satisfying $mn=1$. Since $mn=1\ne 0$, neither $m$ nor $n$ is $0$. If $|n|\ge 2$, then $|m|\ge 1$, so \begin{align*} |mn|=|m||n|\ge 2. \end{align*} But $mn=1$ gives $|mn|=1$, a contradiction. Therefore $|n|=1$, so $n=1$ or $n=-1$. Conversely, \begin{align*} 1\cdot 1=1. \end{align*} \begin{align*} (-1)(-1)=1. \end{align*} Thus the only units of $\mathbb{Z}$ are $1$ and $-1$. In particular, $2\in\mathbb{Z}\setminus\{0\}$ is not a unit, because it is neither $1$ nor $-1$. Hence $\mathbb{Z}$ is not a field: the nonzero element $2$ has no multiplicative inverse in $\mathbb{Z}$. The same obstruction appears in equations such as $2x=3$. If $x\in\mathbb{Z}$, then $2x$ is even, while $3$ is odd, so no integer $x$ satisfies $2x=3$. Polynomial division over $\mathbb{Z}[x]$ has the same defect: dividing by a leading coefficient such as $2$ would require the missing integer inverse of $2$. [/example]