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*}

Lemma 10.6.3. Let $R \to S$ be a ring map of finite presentation. For any surjection $\alpha : R[x_1, \ldots , x_ n] \to S$ the kernel of $\alpha $ is a finitely generated ideal in $R[x_1, \ldots , x_ n]$.

Proof. Write $S = R[y_1, \ldots , y_ m]/(f_1, \ldots , f_ k)$. Choose $g_ i \in R[y_1, \ldots , y_ m]$ which are lifts of $\alpha (x_ i)$. Then we see that $S = R[x_ i, y_ j]/(f_ l, x_ i - g_ i)$. Choose $h_ j \in R[x_1, \ldots , x_ n]$ such that $\alpha (h_ j)$ corresponds to $y_ j \bmod (f_1, \ldots , f_ k)$. Consider the map $\psi : R[x_ i, y_ j] \to R[x_ i]$, $x_ i \mapsto x_ i$, $y_ j \mapsto h_ j$. Then the kernel of $\alpha $ is the image of $(f_ l, x_ i - g_ i)$ under $\psi $ and we win. $\square$


Comments (3)

Comment #1099 by Filip Chindea on

It is obvious that , but I don't know how to prove the reverse inclusion and I would be grateful for a hint. The proof in EGA is quite involved, and I would have liked to see an elementary alternative. Thank you.

Comment #1132 by on

For any ring and element we have . Does this help? Or am I misunderstanding the question?

Comment #1158 by Filip Chindea on

Sorry for that, it was obvious: just consider the projection to and see what happens after composition. Still the proof is amazingly short.


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