Why $K = (X_1, X_2, ...)$, the ideal generated by $X_1, X_2, ...$ not finitely generated as a R-module? 
Let $R = \mathbb{Z}[X_1, X_2, \dots]$ be the ring of polynomials in countably many variables over $\mathbb Z$. Why $K = (X_1, X_2, ...)$, the ideal generated by $X_1, X_2, ...$ is not finitely generated as an $R$-module? 

The proof given is that since every polynomial contains only finitely many variables, K is not finitely generated. However, From what I understand, if K is finitely generated, say by $K_1, K_2, ... K_n$, then K can be written as $a_1 K_1 + a_2 K_2 +... a_n K_n$ where $a_i \in R$ and $K_i \in K$. If that is the case, since $a_i$ can contain any number of variables, why I can't generate K with a finite number of variables? I don't quite understand the proof.
An ideal which is not finitely generated

$\renewcommand{\phi}[0]{\varphi}$First note that it follows from the universal property of polynomial rings that for each $t$, there is a (unique) homomorphism of rings
$$
\phi_{t} : \mathbb{Z}[X_1, X_2, \dots] \to \mathbb{Z}[X_t, X_{t+1}, \dots]
$$
which maps an integer to itself, $X_{i}$ to zero, for $i < t$, and $X_{i}$ to itself, for $i \ge t$. 
Suppose $g_{1}, g_{2}, \dots , g_{m}$ are generators for $K$ as an $R$-modulo. Choose $t$ so that for $i \ge t$, no $X_{i}$ appears in the $g_{j}$.
Suppose there are $a_{i} \in R$ such that
$$
X_{t} = a_{1} g_{1} + \dots + a_{m} g_{m}.
$$
Now apply $\phi_{t}$. You obtain $$X_{t} = \phi_{t}(X_{t}) =0,$$ a contradiction.
