The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

9.5 The characteristic of a field

In the category of rings, there is an initial object $\mathbf{Z}$: any ring $R$ has a map from $\mathbf{Z}$ into it in precisely one way. For fields, there is no such initial object. Nonetheless, there is a family of objects such that every field can be mapped into in exactly one way by exactly one of them, and in no way by the others.

Let $F$ be a field. Think of $F$ as a ring to get a ring map $f : \mathbf{Z} \to F$. The image of this ring map is a domain (as a subring of a field) hence the kernel of $f$ is a prime ideal in $\mathbf{Z}$. Hence the kernel of $f$ is either $(0)$ or $(p)$ for some prime number $p$.

In the first case we see that $f$ is injective, and in this case we think of $\mathbf{Z}$ as a subring of $F$. Moreover, since every nonzero element of $F$ is invertible we see that it makes sense to talk about $p/q \in F$ for $p, q \in \mathbf{Z}$ with $q \not= 0$. Hence in this case we may and we do think of $\mathbf{Q}$ as a subring of $F$. One can easily see that this is the smallest subfield of $F$ in this case.

In the second case, i.e., when $\mathop{\mathrm{Ker}}(f) = (p)$ we see that $\mathbf{Z}/(p) = \mathbf{F}_ p$ is a subring of $F$. Clearly it is the smallest subfield of $F$.

Arguing in this way we see that every field contains a smallest subfield which is either $\mathbf{Q}$ or finite equal to $\mathbf{F}_ p$ for some prime number $p$.

Definition 9.5.1. The characteristic of a field $F$ is $0$ if $\mathbf{Z} \subset F$, or is a prime $p$ if $p = 0$ in $F$. The prime subfield of $F$ is the smallest subfield of $F$ which is either $\mathbf{Q} \subset F$ if the characteristic is zero, or $\mathbf{F}_ p \subset F$ if the characteristic is $p > 0$.

It is easy to see that if $E \subset F$ is a subfield, then the characteristic of $E$ is the same as the characteristic of $F$.

Example 9.5.2. The characteristic of $\mathbf{F}_ p$ is $p$, and that of $\mathbf{Q}$ is $0$.


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