The Stacks project

Example 9.3.1 (Rational numbers). The rational numbers form a field. It is called the field of rational numbers and denoted $\mathbf{Q}$.

Example 9.3.2 (Prime fields). If $p$ is a prime number, then $\mathbf{Z}/(p)$ is a field, denoted $\mathbf{F}_ p$. Indeed, $(p)$ is a maximal ideal in $\mathbf{Z}$. Thus, fields may be finite: $\mathbf{F}_ p$ contains $p$ elements.

Example 9.3.3. In a principal ideal domain, an ideal generated by an irreducible element is maximal. Now, if $k$ is a field, then the polynomial ring $k[x]$ is a principal ideal domain. It follows that if $P \in k[x]$ is an irreducible polynomial (that is, a nonconstant polynomial that does not admit a factorization into terms of smaller degrees), then $k[x]/(P)$ is a field. It contains a copy of $k$ in a natural way. This is a very general way of constructing fields. For instance, the complex numbers $\mathbf{C}$ can be constructed as $\mathbf{R}[x]/(x^2 + 1)$.

Example 9.3.4 (Quotient fields). Recall that, given a domain $A$, there is an imbedding $A \to F$ into a field $F$ constructed from $A$ in exactly the same manner that $\mathbf{Q}$ is constructed from $\mathbf{Z}$. Formally the elements of $F$ are (equivalence classes of) fractions $a/b$, $a, b \in A$, $b \not= 0$. As usual $a/b = a'/b'$ if and only if $ab' = ba'$. The field $F$ is called the quotient field, or field of fractions, or fraction field of $A$. The quotient field has the following universal property: given an injective ring map $\varphi : A \to K$ to a field $K$, there is a unique map $\psi : F \to K$ making

commute. Indeed, it is clear how to define such a map: we set $\psi (a/b) = \varphi (a)\varphi (b)^{-1}$ where injectivity of $\varphi$ assures that $\varphi (b) \not= 0$ if $b \not= 0$.

Example 9.3.5 (Field of rational functions). If $k$ is a field, then we can consider the field $k(x)$ of rational functions over $k$. This is the quotient field of the polynomial ring $k[x]$. In other words, it is the set of quotients $F/G$ for $F, G \in k[x]$, $G \not= 0$ with the obvious equivalence relation.

Example 9.3.6. Let $X$ be a Riemann surface. Let $\mathbf{C}(X)$ denote the set of meromorphic functions on $X$. Then $\mathbf{C}(X)$ is a ring under multiplication and addition of functions. It turns out that in fact $\mathbf{C}(X)$ is a field. Namely, if a nonzero function $f(z)$ is meromorphic, so is $1/f(z)$. For example, let $S^2$ be the Riemann sphere; then we know from complex analysis that the ring of meromorphic functions $\mathbf{C}(S^2)$ is the field of rational functions $\mathbf{C}(z)$.