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Tag 0306

Chapter 10: Commutative Algebra > Section 10.30: Noetherian rings

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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 5693–5697 (see updates for more information).

    \begin{lemma}
    \label{lemma-Noetherian-power-series}
    If $R$ is a Noetherian ring, then so is the formal power
    series ring $R[[x_1, \ldots, x_n]]$.
    \end{lemma}
    
    \begin{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.
    \end{proof}

    Comments (2)

    Comment #343 by Fred on November 22, 2013 a 10:54 pm UTC

    I guess in the fourth line, ''... then $I_d=(a_{d,j})$'' means ''... then $I_d=(a_{d,1},a_{d,2},\dots,a_{d,n_d})$''? That notation is confusing imo.

    Comment #352 by Johan (site) on November 25, 2013 a 6:40 pm UTC

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

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