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This problem comes from Bosch’s Commutative algebra and Algebraic geometry, Exercise 1.5.10.

Let R be the ring of all polynomials $f \in \mathbb{Q}[X]$ such that $f(\mathbb{Z}) \subset \mathbb{Z}$. Consider the polynomials $\; f_n (X)= \frac{1}{n!} X(X-1) \cdots (X-n+1) \in \mathbb{Q}[X], n \in \mathbb{N}$, and show :

(A) The system $(f_n)_{n \in \mathbb{N}}$ is a $\mathbb{Q}$-vector space basis of $\mathbb{Q}[X]$.

(B) All $f_n$ belong to R and $(f_n)_{n \in \mathbb{N}}$ is a free generating system of R as a $\mathbb{Z}$-module.

(C) If $n$ is a prime, $f_n$ does not belong to $(f_1, \cdots , f_{n-1})$, the ideal generated by $f_1, \cdots, f_{n-1}$ in R. In particular, R cannot be Noetherian.

Hint: The $\mathbb{Q}$-linear map $\Delta : \mathbb{Q}[X] \rightarrow \mathbb{Q}[X], f(X) \mapsto f(X+1)-f(X)$, satisfies $\Delta(f_n) = f_{n-1}$ for $n \in \mathbb{N}$ where $f_{-1} = 0$.

I have no problem with part (a) and (b), but I’m confused with part (c). (Probably, I could have missed something important regarding (a) and (b), which makes me confused with (c).)

Here is what I thought about part (c) :

Assume that R is Noetherian, then, since all $f_n$ belong to R, the ascending chain of ideals in R

$I_1 = (f_1) \subset I_2 = (f_1, f_2) \subset \cdots I_{n-1} = (f_1, f_2, \cdots, f_{n-1}) \subset I_n = (f_1, f_2, \cdots f_n) \subset \cdots$

should stabilize at some point, say $n$, which is the smallest positive integer such that $I_{n-1} \ne I_n$ and $I_n = I_{n+1} = I_{n+2} = \cdots$.

And we know that $\Delta$ sends the above chain of ideals to a chain of ideals in R, and $\Delta(f_n)=f_{n-1}$,

This implies that $\Delta(I_n) = I_{n-1}$, then $I_{n-1} = \Delta(I_n) = \Delta(I_{n+1}) = I_n$,

Which contradicts to the minimality of $n$.

So R is not Noetherian.

And the following is my question about this problem:

(1) I don’t know why the author assumed that $n$ is a prime. Is there a reason which I’m missing?

(2) I’m not sure whether what I’ve done is right or not. Could you give me a feedback?

Thank you in advance.

Community
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glimpser
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1 Answers1

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HINT:

Say we have $$\sum_{k=1}^{p-1} p_k f_k = f_p$$ Evaluating at $x=p$ produces $1$ on RHS. and an integer divisible by $p$ on LHS , contradiction.

orangeskid
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  • I see. with your argument, we can produce infinitely ascending chain of ideals in $R$. So $R$ is not Noetherian. Right? – glimpser Apr 01 '20 at 07:07
  • @glimpser: Yes, indeed. About this ring, it would be interesting to see what are the structure constants, that is how $f_m \cdot f_n$ is expressed as an integer combination of $f_k$'s . – orangeskid Apr 01 '20 at 07:16
  • Ok, then is the assumption of the minimality of $n$ wrong? I’d like to know whether my argument is right or not. – glimpser Apr 01 '20 at 07:19
  • @glimpser Your attempt does not even prove that $f_n\notin(f_1,\ldots,f_{n-1})$ for $n$ prime. – Angina Seng Apr 01 '20 at 08:44
  • @AnginaSeng Ah, No, it doesn't. I should have specified what I mean. Does my argument prove that R is not Noetherian, apart from part c? – glimpser Apr 01 '20 at 08:51