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

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?

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.

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 $n!$).