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Let $R$ be a Noetherian ring. How can one prove that the ring of the formal power series $R[[x]]$ is again a Noetherian ring?

It is well-known that the ring of polynomials $R[x]$ is Noetherian. I try imitating the standard proof of the fact by replacing "leading coefficients" by "lowest coefficients", but it does not work.

Eric Wofsey
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Pooya Ronagh
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    Also, see Matsumura, Commutative Ring Theory, p 16. You can see his argument at the preview: http://books.google.com/books/about/Commutative_Ring_Theory.html?id=yJwNrABugDEC – John Myers Jan 26 '13 at 04:48

3 Answers3

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Step 1: Let $f\in R[[x]]$ such that the constant term of $f$ is a unit. Then $f$ is a unit. It's an easy exercise to construct $f^{-1}$ term-by-term.

Step 2: Show that if $I$ is an ideal, the set of coefficients of lowest nonzero terms of elements of $I$ is an ideal of $R$.

Step 3: Consider an ascending chain of ideal. $I_1\subset I_2\subset\ldots$, and let $J_1\subset J_2\subset\ldots$ be the corresponding chain of ideals in $R$, where $J_i$ is the ideal of coefficients of lowest nonzero terms of elements of $I_i$.

Step 4: $J_1\subset J_2\subset\ldots$ can only be strict inclusion at finitely many places. The only other way to have strict inclusion in $I_1\subset I_2\subset\ldots$ is via the lowest degree in $x$ of nonzero terms, which can only decrease finitely many times.

Eric Wofsey
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Brett Frankel
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  • @BrettFrankel How does it follow that $I_i \subset I_{i+1}?$ – The Substitute May 21 '14 at 02:12
  • @TheSubstitute The ascending chain $I_i$ is given. – Brett Frankel May 21 '14 at 03:02
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    Why can't we have a strict inclusion $I_1 \subset I_2$ with both having the same lowest degree of nonzero elements and the same leading coefficients?(Say $3x^4+x^5 \in I_2 \setminus I_1$ but $3x^4+x^6 \in I_1$) – Emolga Feb 02 '16 at 19:19
  • How can you define "the set of the lowest nonzero terms of elements of $I$"? A typical element in $I$ is of the form $\sum_{i=-\infty}^\infty a_ix^i$. Indeed, there're elements that haven't "lowest nonzero term" (such as $a=\sum_{-\infty}^\infty x^i$). – J. Doe Sep 02 '19 at 15:21
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    @J.Doe $R[[x]]$ is the set of formal power series with non-negative powers by definition. – SVG Sep 17 '19 at 18:02
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    @Emolga At least if $R = K$ is a field with $\operatorname{char}(K) \neq 3$, the reason is that $3x^4 + x^5$ and $3x^4 + x^6$ are equal up to a unit - indeed, we have $3x^4 + x^5 = x^4 (3 + x)$ and $3x^4 + x^6 = x^4 (3 + x^2)$, and both $3 + x$ and $3 + x^2$ are units in $K[[x]]$, as $3$ is a unit in $K$. For a general $R$, I also don't understand this, since not every nonzero element is invertible. – Smiley1000 Jan 16 '25 at 09:50
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Hilbert's basis theorem can be adapted for formal power series. I found a .pdf on the internet that describes the process well, if you look at section 8.2.3: here.

Basically, you simply replace the degree of the polynomial with the lowest degree in the power series.

feynhat
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guest196883
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If $f\in R[[x]]$, we write $o(f)$ for the degree of the first nonzero term of $f$ and $c(f)$ for the first nonzero coefficient of $f$. Let $I\subseteq R[[x]]$ be an ideal and let $$I_n=\{c(f):f\in I, o(f)=n\}\cup\{0\}.$$ Then each $I_n$ is an ideal in $R$ and $$I_0\subseteq I_1\subseteq I_2\subseteq\dots.$$ Since $R$ is Noetherian, there is some $N$ such that $I_n=I_N$ for all $n\geq N$. Also, each $I_n$ is finitely generated. So we can pick finitely many elements $a_1,\dots,a_m\in I$ which witness finite generating sets of $I_n$ for each $n\leq N$. Let $J$ be the ideal generated by $a_1,\dots,a_m$. Then for any $f\in I$, there exists $g\in J$ such that $o(f)=o(g)$ and $c(f)=c(g)$ (if $o(f)\leq N$ this is by definition of $J$, and if $o(f)>N$ this follows from the case $o(f)=N$ by multiplying by $x^{o(f)-N}$ since $I_{o(f)}=I_N$). In other words, we can match the first nonzero term of any element of $I$ with an element of $J$.

We now iterate this to realize an entire element of $I$ as an element of $J$. We start with $f=f_0\in I$ and pick $g_0\in J$ such that $o(g_0)=o(f_0)$ and $c(g_0)=c(f_0)$. Now let $f_1=f_0-g_0$; by our choice of $g_0$ we have $o(f_1)>o(f_0)$. Choose $g_1$ such that $o(g_1)=o(f_1)$ and $c(g_1)=c(f_1)$, and let $f_2=f_1-g_1$. Repeating this process, we get a sequence of elements $f_n\in I$ and $g_n\in J$ with $f_{n+1}=f_n-g_n$ and $o(f_n)=o(g_n)\geq n$ for each $n$. In particular, this implies $f_n\to 0$ in the $x$-adic topology so $\sum_n g_n$ converges to $f$. Moreover, if we write $g_n=\sum b_{in} a_i$, each coefficient $b_{in}$ is divisible by $x^{n-N}$ since $o(a_i)\leq N$ for each $i$ and $o(g_n)\geq n$. It follows that $\sum_n b_{in}$ converges for each $i$ and we can write $f=\sum g_n=\sum_i(\sum_n b_{in})a_i$. Thus $f\in J$. Since $f\in I$ was arbitrary, this means $I=J$ so $I$ is finitely generated.

Eric Wofsey
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  • I am not sure about the definition of $J$ as the ideal of $R[[x]]$ generated by $a_1,...,a_m$ and trying to show that $I=J$, since the $a_i$ themselves are in general not in $I$. For example, in $R[[x]]=\Bbb Z[[x]]$, take the principal ideal $I$ generated by $f_0=2+3x$. Then every $f\in I$ has $c(f)$ is divisible by 2 and $I_0=I_1=\cdots=(2)$, but $f_0$ is not in the ideal of $\Bbb Z[[x]]$ generated by $2$, as the $x$ coefficient of $f_0$ is not even. Maybe one needs to instead take specific elements of $I$ that have $c(f)=a_i$ and $o(f)$ minimal, or something like that. – PatrickR Dec 09 '20 at 03:44
  • Oops, the $a_i$ were supposed to be in $I$, not in $R$ (i.e., what you suggested in your last sentence). – Eric Wofsey Dec 09 '20 at 04:34
  • I feel confused on the notation of indices like the $n$ used in $I_n$ and the $n$ used in $n\geq N$ everywhere. I was assuming the latter one can be replaced by new notation $i\geq N$ and $$I_0\leq I_1 \leq\cdots\leq I_{i}=I_{i+1}=\cdots$$@EricWofsey Is my understanding correct here? – N00BMaster Dec 21 '23 at 19:12
  • @N00BMaster: Sure, you can use a different letter instead of $n$ if you prefer. – Eric Wofsey Dec 21 '23 at 20:18