# 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

• A element in R = Z[x1,...,] is a finite sum of monomial multiplied with elements of a ring (here Z). But by definition a monomial in R is just a finite product of x_i !
– user171326
Apr 14, 2015 at 13:56

Let $X_N$ be the 'least' variable that does not occur in a given finite set of polynomials, $K_1,K_2,\ldots,K_n$. Can you find an expression for $X_N$ as $a_1K_1+\cdots +a_nK_n$, with $a_i$'s polynomials that can involve any variable?

• i can't do so, but I can't prove that it is impossible as well. In fact, that is my question. I don't understand why the proof given proves so. I can't see why I can just set all $x_i = 0$ and conclude that I can't. Why can I be sure that $a_i k_i$ won't give me an expression containing $x_N$ Apr 14, 2015 at 14:01
• @User136266 It might contain $X_N$ in there somewhere, but since $K_i$ doesn't have a constant term, the product $a_iK_i$ cannot contain any terms that are just $X_N$, possibly with a coefficient $r_j\in R$. Apr 14, 2015 at 14:15

$\renewcommand{\phi}{\varphi}$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.

• hi can i know why $X_t = 0$ gives a contradiction? Apr 15, 2015 at 5:02
• I can see a contradiction if i map$X_i$ to $1$ when$i \geq t$ which will give me 1 = 0. However I can't see why $X_t = 0$ also gives a contradiction Apr 15, 2015 at 5:32
• @user136266,$X_t$ is an indeterminate. Apr 15, 2015 at 6:34