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How does one prove that the ring of integer-valued polynomials $\text{Int}(\mathbb{Z})$ is not Noetherian?

I let $(1, f_1, ..., f_n,...)$ be the $\mathbb{Z}$-basis of $\text{Int}(\mathbb{Z})$, the ring of rational polynomials sending $\mathbb{Z}$ to $\mathbb{Z},$ where $f_1$, $f_2$, etc are the polynomials $X$, $X(X-1)/2$, etc. Then an infinite ascending chain of ideals is $(f_1) \subset (f_1, f_2) \subset \cdots \subset (f_1, f_2, ..., f_n) \subset \cdots$. How would you prove that $f_{n+1} \not \in (f_1, ..., f_n)$?

Is finding an infinite chain of ascending ideals generally a good method of showing that a ring is not Noetherian?

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    $\begingroup$ What is the meaning of the notation Int$\mathbb{Z}$? OK, I guess the set of functions that send integers to integers ? $\endgroup$ – wxu Jun 1 '13 at 4:58
  • $\begingroup$ Use fundamental theorem of symmetric polynomials $\endgroup$ – user23086 Jun 1 '13 at 5:04
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    $\begingroup$ It is the polynomials that send integers to integers (a subring of ${\mathbf Q}[x]$). $\endgroup$ – KCd Jun 1 '13 at 5:31
  • $\begingroup$ Fix a prime $p$ and look at the ideal in ${\rm Int}({\mathbf Z})$ that is generated by all $\binom{x}{p^r}$ for $r \geq 1$. $\endgroup$ – KCd Jun 1 '13 at 5:33
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So you know $\operatorname{Int}(\mathbb{Z})$ is a subring of $\mathbb{Q}[T]$ and $\operatorname{Int}(\mathbb{Z})$ a free abelian group with a $\mathbb{Z}$-basis $\binom{T}{i},i=0,1,\ldots$.

We now show that $\operatorname{Int}(\mathbb{Z})$ is not Noetherian. Write $f_{k}=\binom{T}{k}$ for $k=0,1,\ldots$, then $\operatorname{Int}(\mathbb{Z})=\mathbb{Z}[f_1,f_2,\ldots]$.

I won't prove that $f_{n+1}\notin (f_1,\ldots,f_n)$. Instead, I am going to show $f_p\notin (f_1,\ldots,f_{p-1})$ for any prime number. If not, $f_{p}=f_1g_1+\cdots+f_{p-1}g_{p-1}$ for some polynomials $g_i\in \operatorname{Int}(\mathbb{Z})$. Let $a_i=g_i(0)\in \mathbb{Z}$ for $i=1,\ldots,p-1$. We have $f_p-a_1f_1-a_2f_2-\ldots-a_{p-1}f_{p-1}=T^2g$ for some $g\in\mathbb{Q}[T]$. But the polynomial on the left hand side of the equation has no multiple roots at zero. Here is why: it equals to $Th(T)$ for an $h(T)\in \mathbb{Q}[T]$. If $h(0)=0$, then $\frac{(p-1)!}{p}\in \mathbb{Z}$ which is a contradiction.

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As in wxu's answer I will also show that $f_p$ is not in $(f_1, ... f_{p-1})$ for $p$ prime, but the argument is shorter: the leading coefficient of a polynomial in $(f_1, ... f_{p-1})$ can't have denominator divisible by $p$.

For general $n$ you can similarly show that the leading coefficient of a polynomial in $(f_1, ... f_{n-1})$ can't have denominator $n!$ (but must have denominator strictly dividing $n!$).

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  • $\begingroup$ Thanks for the pointer. I've worked out how to show this, although I haven't written it out. $\endgroup$ – byesac Jun 2 '13 at 3:56
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Another approach to showing that $\operatorname{Int}(\mathbb{Z})$ is non-noetherian can be found as Proposition 3 in the article What You Should Know About Integer-Valued Polynomials by Paul-Jean Cahen and Jean-Luc Chabert. There is it shown that the ideal $$ \mathfrak{m} = \{ f \in \operatorname{Int}(\mathbb{Z}) \mid \text{$f(0)$ is even} \} $$ is not finitely generated.

Step 1: Assumptions

Assume otherwise, so that $$ \mathfrak{m} = (f_1, \dotsc, f_n) $$ for some polynomials $f_1, \dotsc, f_n \in \mathfrak{m}$. Every polynomial $f \in \mathbb{Q}[X]$ can be written as $f = g/h$ for some polynomial $g \in \mathbb{Z}[X]$ and some non-zero integer $h$. Hence every generator $f_i$ can be written in this form as $$ f_i = \frac{g_i}{h_i} \,. $$ We can find a common denominator $h$ for all $f_i$, so that $$ f_1 = \frac{g_i}{h} \,, \quad \dotsc \,, \quad f_n = \frac{g_n}{h} $$ for some polynomials $g_1, \dotsc, g_n \in \mathbb{Z}[X]$ and some nonzero integer $h$. We decompose this integer $h$ as $$ h = 2^k d $$ where the factor $d$ is coprime to $2$ (i.e. odd).

Step 2: ???

We will now see that the value $f(2^{k+1})$ is again even for every polynomial $f \in \mathfrak{m}$.

We show this first for the generators $f_i$: The integer $$ f_i(0) = \frac{g_i(0)}{2^k d} $$ is by assumption even. This means that the integer $g_i(0)$—the constant term of $g_i$—must be divisible by $2^{k+1}$. It follows that the value $g_i(2^{k+1})$ is again divisible by $2^{k+1}$, and hence that $f_i(2^{k+1})$ is again even.

The claim now follows for every polynomial $f \in \mathfrak{m}$: We can write the polynomial $f$ as a linear combination $$ f = a_1 f_1 + \dotsb + a_n f_n $$ for some coefficients $a_1, \dotsc, a_n \in \operatorname{Int}(\mathbb{Z})$. Then $$ f(2^{k+1}) = a_1(2^{k+1}) f_1(2^{k+1}) + \dotsb + a_n(2^{k+1}) f_n(2^{k+1}) $$ where the values $a_i(2^{k+1})$ are integers and the values $f_i(2^{k+1})$ are even. The expression on the right hand side is again even.

Step 3: Contradiction

The polynomial $f(X) = \binom{X}{2^{k+1}}$ is an element of $\operatorname{Int}(\mathbb{Z})$ with $f(0) = 0$, and it is thus contained in the ideal $\mathfrak{m}$. But $$ f(2^{k+1}) = \binom{2^{k+1}}{2^{k+1}} = 1 $$ is odd, contradicting the previous step.

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