# If $R$ is a Noetherian ring then $R[[x]]$ is also Noetherian

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.

• Also, see Matsumura, Commutative Ring Theory, p 16. You can see his argument at the preview: books.google.com/books/about/… Commented Jan 26, 2013 at 4:48

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.

• @BrettFrankel How does it follow that $I_i \subset I_{i+1}?$ Commented May 21, 2014 at 2:12
• @TheSubstitute The ascending chain $I_i$ is given. Commented May 21, 2014 at 3:02
• 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 $3x^4+x^6 \in I_1$) Commented Feb 2, 2016 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 $a=\sum_{-\infty}^\infty x^i$). Commented Sep 2, 2019 at 15:21
• @J.Doe $R[[x]]$ is the set of formal power series with non-negative powers by definition. Commented Sep 17, 2019 at 18:02

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.

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.

• 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. Commented Dec 9, 2020 at 3:44
• Oops, the $a_i$ were supposed to be in $I$, not in $R$ (i.e., what you suggested in your last sentence). Commented Dec 9, 2020 at 4:34
• 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? Commented Dec 21, 2023 at 19:12
• @N00BMaster: Sure, you can use a different letter instead of $n$ if you prefer. Commented Dec 21, 2023 at 20:18