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?

  • $\begingroup$ Does this answer your question? $\endgroup$
    – rschwieb
    Commented Nov 27, 2023 at 1:02
  • $\begingroup$ No, this was the answer I was referring to in the pre-amble. $\endgroup$
    – Ben123
    Commented Nov 27, 2023 at 1:39
  • $\begingroup$ How is anyone supposed to know that? Answer: proactively link it $\endgroup$
    – rschwieb
    Commented Nov 27, 2023 at 3:28
  • $\begingroup$ 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. $\endgroup$
    – Ben123
    Commented Nov 27, 2023 at 3:48


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