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 10.30.2. If $R$ is a Noetherian ring, then so is the formal power series ring $R[[x_1, \ldots , x_ n]]$.

Proof. Since $R[[x_1, \ldots , x_{n + 1}]] \cong R[[x_1, \ldots , x_ n]][[x_{n + 1}]]$ it suffices to prove the statement that $R[[x]]$ is Noetherian if $R$ is Noetherian. Let $I \subset R[[x]]$ be a ideal. We have to show that $I$ is a finitely generated ideal. For each integer $d$ denote $I_ d = \{ a \in R \mid ax^ d + \text{h.o.t.} \in I\} $. Then we see that $I_0 \subset I_1 \subset \ldots $ stabilizes as $R$ is Noetherian. Choose $d_0$ such that $I_{d_0} = I_{d_0 + 1} = \ldots $. For each $d \leq d_0$ choose elements $f_{d, j} \in I \cap (x^ d)$, $j = 1, \ldots , n_ d$ such that if we write $f_{d, j} = a_{d, j}x^ d + \text{h.o.t}$ then $I_ d = (a_{d, j})$. Denote $I' = (\{ f_{d, j}\} _{d = 0, \ldots , d_0, j = 1, \ldots , n_ d})$. Then it is clear that $I' \subset I$. Pick $f \in I$. First we may choose $c_{d, i} \in R$ such that

\[ f - \sum c_{d, i} f_{d, i} \in (x^{d_0 + 1}) \cap I. \]

Next, we can choose $c_{i, 1} \in R$, $i = 1, \ldots , n_{d_0}$ such that

\[ f - \sum c_{d, i} f_{d, i} - \sum c_{i, 1}xf_{d_0, i} \in (x^{d_0 + 2}) \cap I. \]

Next, we can choose $c_{i, 2} \in R$, $i = 1, \ldots , n_{d_0}$ such that

\[ f - \sum c_{d, i} f_{d, i} - \sum c_{i, 1}xf_{d_0, i} - \sum c_{i, 2}x^2f_{d_0, i} \in (x^{d_0 + 3}) \cap I. \]

And so on. In the end we see that

\[ f = \sum c_{d, i} f_{d, i} + \sum \nolimits _ i (\sum \nolimits _ e c_{i, e} x^ e)f_{d_0, i} \]

is contained in $I'$ as desired. $\square$


Comments (2)

Comment #343 by Fred on

I guess in the fourth line, ''... then '' means ''... then ''? That notation is confusing imo.

Comment #352 by on

Yes. I am not sure what would be a better notation. Any suggestions?


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0306. Beware of the difference between the letter 'O' and the digit '0'.