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I'm doing this exercise in the Dummit and Foote textbook and got no clue for it. Hope some one can help me solve this. Thanks

Prove that a polynomial ring in a infinitely many variables with coefficients in any commutative ring contains ideals that are not finitely generated.

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Hint: Consider the ideal $I=\langle x_1,x_2,x_3... \rangle$.

Following what anon said. You can apply the same trick one below to prove what annon is asking you.

Alternate: Let us prove that $R'=R[x_1,...]$ is not noetherian. Take the chain $(x_1)\subset (x_1,x_2)\subset\cdots$. If this stabilizes, then for some $n$ we have $(x_1,...,x_n)=(x_1,..,x_{n+1})$. Then $x_{n+1}=h_1x_1+\cdots+h_nx_n=f$. Evaluate $f$ at the point that is $0$ in every coordinate except $n+1$, here say it is $1$. Then you get $1=0$. A contradiction.

Remember that a different characterization of Noetherian is that every ideal is finitely generated. Then by above since it is not noetherian, we have that there must be an ideal that is not finitely generated.

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  • $\begingroup$ Oh, how can I prove that this ideal is not finitely generated. The ideal is all the polynomial without free coefficient, I know, but how to prove the problem??? $\endgroup$ – le duc quang Sep 22 '13 at 3:47
  • $\begingroup$ Sorry for saying this but I know nothing about Noetherian ring. Anyway, thanks for your answer. You're so kind. $\endgroup$ – le duc quang Sep 22 '13 at 4:26
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Finding an infinitely generated ideal in particular means picking an infinite number of elements to do the generating. What is the most obvious infinite set to pick? The fact that there are a specifically infinite number of variables suggests an immediate answer! This is your "clue."

Here's a hint for how to prove this ideal is not finitely generated. Suppose $(x_1,x_2,\cdots)$ were finitely generated by a finite number of multivariable polynomials $\{\pi_1,\cdots,\pi_m\}$. How many variables appear in this finite generating set of polynomials? A finite number, say $\{x_s:s\in S\}$ with $S$ a finite subset of $\Bbb N$. So is $x_r\in(\pi_1,\cdots,\pi_m)$ possible if $r\not\in S$?

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  • $\begingroup$ Yes, but the real difficulty that I meet is how to prove it is not finitely generated, like how to prove that this ideal (recommended by Daniel Montealegre in the previous answer) $I = <x_{1}, x_{2}, ...>$ is not finitely generated. $\endgroup$ – le duc quang Sep 22 '13 at 4:01
  • $\begingroup$ @leducquang When writing a question it is helpful to communicate your own thoughts and work on a problem - in particular outline what difficulties you have. In theory this should yield warmer and better tailored responses and upvotes on your question. In particular, here it would have signaled that identifying such an ideal was not your issue, but proving it was in fact not finitely generated. $\endgroup$ – anon Sep 22 '13 at 4:16
  • $\begingroup$ Oh really. You mean that we can't define a polynomial which has infinitely components, like $p = \sum_{i \in S}{x_{i}}$ is unacceptable, right? $\endgroup$ – le duc quang Sep 22 '13 at 4:18
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    $\begingroup$ Yes. A polynomial ring in infinitely many variables is defined as a direct limit $R[x_1,x_2,\cdots]$ $=\varinjlim R[x_1,\cdots,x_n]$, so each element has only finitely many variables present. In order to obtain elements with infinitely many variables (but bounded degree in each variable) we would need to take the inverse limit $\varprojlim R[x_1,\cdots,x_n]$ (where the projection maps truncate later variables by evaluating them to $0$). And for power series with no bound on number of variables or degree in any variable, one completes the inverse limit at the prime ideal $(x_1,x_2,\cdots)$. $\endgroup$ – anon Sep 22 '13 at 4:20
  • $\begingroup$ Well, I even don't know the definition of infinitely many variables polynomial ring like you present above. Seem I have to research more to get it. Thanks anyway Anon... $\endgroup$ – le duc quang Sep 22 '13 at 4:25

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