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*}

Example 10.134.7. Let $n , m \geq 1$ be integers. Consider the ring map

\begin{eqnarray*} R = \mathbf{Z}[a_1, \ldots , a_{n + m}] & \longrightarrow & S = \mathbf{Z}[b_1, \ldots , b_ n, c_1, \ldots , c_ m] \\ a_1 & \longmapsto & b_1 + c_1 \\ a_2 & \longmapsto & b_2 + b_1 c_1 + c_2 \\ \ldots & \ldots & \ldots \\ a_{n + m} & \longmapsto & b_ n c_ m \end{eqnarray*}

In other words, this is the unique ring map of polynomial rings as indicated such that the polynomial factorization

\[ x^{n + m} + a_1 x^{n + m - 1} + \ldots + a_{n + m} = (x^ n + b_1 x^{n - 1} + \ldots + b_ n) (x^ m + c_1 x^{m - 1} + \ldots + c_ m) \]

holds. Note that $S$ is generated by $n + m$ elements over $R$ (namely, $b_ i, c_ j$) and that there are $n + m$ equations (namely $a_ k = a_ k(b_ i, c_ j)$). In order to show that $S$ is a relative global complete intersection over $R$ it suffices to prove that all fibres have dimension $0$.

To prove this, let $R \to k$ be a ring map into a field $k$. Say $a_ i$ maps to $\alpha _ i \in k$. Consider the fibre ring $S_ k = k \otimes _ R S$. Let $k \to K$ be a field extension. A $k$-algebra map of $S_ k \to K$ is the same thing as finding $\beta _1, \ldots , \beta _ n, \gamma _1, \ldots , \gamma _ m \in K$ such that

\[ x^{n + m} + \alpha _1 x^{n + m - 1} + \ldots + \alpha _{n + m} = (x^ n + \beta _1 x^{n - 1} + \ldots + \beta _ n) (x^ m + \gamma _1 x^{m - 1} + \ldots + \gamma _ m). \]

Hence we see there are at most finitely many choices of such $n + m$-tuples in $K$. This proves that all fibres have finitely many closed points (use Hilbert's Nullstellensatz to see they all correspond to solutions in $\overline{k}$ for example) and hence that $R \to S$ is a relative global complete intersection.

Another way to argue this is to show $\mathbf{Z}[a_1, \ldots , a_{n + m}] \to \mathbf{Z}[b_1, \ldots , b_ n, c_1, \ldots , c_ m]$ is actually also a finite ring map. Namely, by Lemma 10.37.5 each of $b_ i, c_ j$ is integral over $R$, and hence $R \to S$ is finite by Lemma 10.35.4.


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