0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 ! $\endgroup$ – user171326 Apr 14 '15 at 13:56
1
$\begingroup$

$\renewcommand{\phi}[0]{\varphi}$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.

$\endgroup$
  • $\begingroup$ hi can i know why $X_t = 0$ gives a contradiction? $\endgroup$ – user10024395 Apr 15 '15 at 5:02
  • $\begingroup$ 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 $\endgroup$ – user10024395 Apr 15 '15 at 5:32
  • $\begingroup$ @user136266,$X_t$ is an indeterminate. $\endgroup$ – Andreas Caranti Apr 15 '15 at 6:34
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $x_N$ $\endgroup$ – user10024395 Apr 14 '15 at 14:01
  • 1
    $\begingroup$ @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 $r_j\in R$. $\endgroup$ – Arthur Apr 14 '15 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.