## 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 5723–5727 (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}
```

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