# Show that the ring of formal power series in a commutative ring $R$, $R[[x]]$ is noetherian.

Yes, I am aware that this has been answered (If $R$ is a Noetherian ring then $R[[x]]$ is also Noetherian), but the answers given did not answer my specific question regarding this:

In my notes from a class I am taken, there is a proof of the hillberts basis theorem that is done by showing the contrapositive, that is, that $$R[x]$$ is not noetherian implies that $$R$$ is not noetherian. One does this by inductively constructing a sequence of polynomials $$f_i$$ with leading coefficients $$a_i$$ and then shows that the chain

$$(a_1) \subset (a_1,a_2) \subset \ldots$$ does not stabilize, by showing that we get a contradiction if it does stabilize.

That is, assuming that the chain above does stabilize, so that

$$(a_1,\ldots,a_n) = (a_1,\ldots,a_{n+1})$$ we get $$\exists r_1,\ldots,r_n \in R$$ so that

$$a_{n+1} = r_1a_1 + \ldots + r_na_n$$

and then we define $$g(x) = r_1f_1x^{d_{n+1}-d_1} + \ldots + r_nf_nx^{d_{n+1}-d_{n}}$$

where we find that the coefficient for $$x^{d_{n+1}}$$ is the same as in $$f_{n+1}$$

and we then find that

$$f_{n+1}-g \in I \setminus (f_1,\ldots,f_n)$$

and that $$\operatorname{deg}(f_{n+1}-g) < \operatorname{deg}(f_{n+1})$$

hence we get a contradiction to how we inductively created $$f_{n+1}$$ (it was choosen to be of minimal degree in $$I \setminus (f_1,\ldots,f_n)$$).

Now, to the question at hand:

If I instead of constructing polynomials of minimal degree, construct series $$f_i$$ with minimal non-zero coefficient, we can proceed similarly in the proof and we get that

$$f_{n+1}-g \in I \setminus (f_1,\ldots,f_n)$$ is a series with higher-degree term than $$f_{n+1}$$. But I don´t see what the contradiction here is?

Any help on how to prove that $$R[[x]]$$ is noetherian by this route would be appreciated.

The other way to solve this would be to construct

$$I_0 = \text{coefficients of the first term in a formal power series}$$ $$\vdots$$ $$I_n = \text{coefficients of the n:th term in a formal power series}$$

then I think we get that $$I_i$$ is an ideal in $$R$$ so finitely generated with generators $$(a_{1,i},\ldots,a_{i,n_{i}}).$$

We can then construct

$$S_0 = \text{finite set of formal power series with coefficients of the first term which generates} \ I_0$$ $$\vdots$$ $$S_n = \text{finite set of formal power series with coefficients of the n:th term which generate} \ I_n$$ and then take $$S = \bigcup_{i = 1}^{n} S_i.$$

Now, we have that $$S \subset I$$, we want to show that $$I \subset S$$.

So let $$f = \sum_{k = 0}^{\infty} a_kx^k \in I.$$

Now, maybe I can proceed as follows:

We know that the coefficients stabilize, that is $$I_n = I_{n+k}$$ for $$k \geq 0$$ where our $$n$$ is choosen so that this holds.

Now, by the well-ordering of $$\mathbb{N}$$ there is a smallest non-zero coefficient $$a_j$$ of $$f$$,say $$x^{j}$$. Now if $$j < n$$ we have that there is a series $$g_1 \in S$$ such that $$g_1$$ has $$a_j$$ as a coefficient in front of $$x^j$$.

We then get that $$f-g_jx^{j} = f_{j+1}$$ has a smallest non-zero coefficient attached to $$x^{j+1}$$.

We can continue onward up to $$j+k = n$$. Again, we find $$g_{j+k}x^{j+k}$$ such that $$f-g_{j+k}x^{j+k} = f_{j+k+1}$$ has smallest non-zero coefficient attached to $$x^{j+k+1}$$. Now we note that $$I_n = I_{j+k+1}$$ so that there is a series $$g_{j+k+1} \in S$$ with coefficient $$a_{j+k+1}$$ in front of $$x^n$$ (if I am not mistaken) so that $$f_{j+k+1} - g_{j+k+1}x^{(j+k+1)-n} = f_{j+k+2}$$ is such that it´s a series with smallest non-zero coeffecient being the coefficient of $$x^{j+k+2}$$. Inductively, I believe we can thus write every element in $$f$$ as a series where each term is a linear combination of elements in $$S$$.

If $$j \geq n$$ then we can proceed similarly as in the last step above, since $$I_n = I_{j}$$.

Hence we find that $$I \subset J$$ so that $$I = S$$. Since $$S$$ is a finite union of power series, it is finitely generated, hence $$I$$ is finitely generated.

Is this the correct idea?

• Does this answer your question? Commented Nov 27, 2023 at 1:02
• No, this was the answer I was referring to in the pre-amble. Commented Nov 27, 2023 at 1:39
• How is anyone supposed to know that? Answer: proactively link it Commented Nov 27, 2023 at 3:28
• The preamble (and the end-note) specifically ask how to do it via this proof-route. If one reads my post, it is clear that the link you gave does not answer this, unless you did not read my post. But you make a good point. I can add it as a link. Commented Nov 27, 2023 at 3:48