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*}

Theorem 10.34.11. Let $k$ be a field. Let $S$ be a $k$-algebra generated over $k$ by the elements $\{ x_ i\} _{i \in I}$. Assume the cardinality of $I$ is smaller than the cardinality of $k$. Then

  1. for all maximal ideals $\mathfrak m \subset S$ the field extension $k \subset \kappa (\mathfrak m)$ is algebraic, and

  2. $S$ is a Jacobson ring.

Proof. If $I$ is finite then the result follows from the Hilbert Nullstellensatz, Theorem 10.33.1. In the rest of the proof we assume $I$ is infinite. It suffices to prove the result for $\mathfrak m \subset k[\{ x_ i\} _{i \in I}]$ maximal in the polynomial ring on variables $x_ i$, since $S$ is a quotient of this. As $I$ is infinite the set of monomials $x_{i_1}^{e_1} \ldots x_{i_ r}^{e_ r}$, $i_1, \ldots , i_ r \in I$ and $e_1, \ldots , e_ r \geq 0$ has cardinality at most equal to the cardinality of $I$. Because the cardinality of $I \times \ldots \times I$ is the cardinality of $I$, and also the cardinality of $\bigcup _{n \geq 0} I^ n$ has the same cardinality. (If $I$ is finite, then this is not true and in that case this proof only works if $k$ is uncountable.)

To arrive at a contradiction pick $T \in \kappa (\mathfrak m)$ transcendental over $k$. Note that the $k$-linear map $T : \kappa (\mathfrak m) \to \kappa (\mathfrak m)$ given by multiplication by $T$ has the property that $P(T)$ is invertible for all monic polynomials $P(t) \in k[t]$. Also, $\kappa (\mathfrak m)$ has dimension at most the cardinality of $I$ over $k$ since it is a quotient of the vector space $k[\{ x_ i\} _{i \in I}]$ over $k$ (whose dimension is $\# I$ as we saw above). This is impossible by Lemma 10.34.10.

To show that $S$ is Jacobson we argue as follows. If not then there exists a prime $\mathfrak q \subset S$ and an element $f \in S$, $f \not\in \mathfrak q$ such that $\mathfrak q$ is not maximal and $(S/\mathfrak q)_ f$ is a field, see Lemma 10.34.5. But note that $(S/\mathfrak q)_ f$ is generated by at most $\# I + 1$ elements. Hence the field extension $k \subset (S/\mathfrak q)_ f$ is algebraic (by the first part of the proof). This implies that $\kappa (\mathfrak q)$ is an algebraic extension of $k$ hence $\mathfrak q$ is maximal by Lemma 10.34.9. This contradiction finishes the proof. $\square$


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