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?
 A: 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.
A: 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 $n!$). 
A: 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.
