Show that the ring of polynomials with coefficients in a field, and in infinitely many variables, is not Noetherian
Show that the ring of polynomials with coefficients in a field, and in infinitely many variables, is not Noetherian
Show that the ring of polynomials with coefficients in a
field, and in infinitely many variables, is not Noetherian, that is, $R = k [x_i: igeq1]$ is not Noetherian.
I know that I need to exhibit an ideal of the ring that is not finitely generated, what could this ideal be? Could it be $(x_1,x_2,...,)$?
Or could I give the following chain of ideals that do not have a maximal element $(x_1)subset(x_1,x_2)subset(x_1,x_2,x_3)subset...$?How can all ideals that are not finitely generated be classified? What to do in the case where the number of variables is non-countable?
2 Answers
2
A ring is Noetherian if and only if it satisfies the ascending chain condition, i.e. every increasing chain of ideals terminates. Now you have a chain
$$(x_1)subsetneq(x_1,x_2)subsetneq(x_1,x_2,x_3)subsetneqcdots$$
that never terminates, so $k[x_i: ige 1]$ is not Noetherian.
You've got all the right ideas. You don't need to fully classify the non-finitely-generated ideals to prove that $(x_1,x_2,dots)$ is not finitely generated. One way to do it is to reason:
Suppose $I = (x_1,x_2,dots)$ were finitely generated. Then there would be a finite list of generators $f_1,dots,f_n$. Each $f_i$ would be a polynomial in finitely many variables (because that's what elements of this ring are). Thus, there would be some $N$ that is the maximum $N$ for which $x_N$ appears in any of the $f_i$'s. Then, supposedly, every element of $I$ would be expressible as a linear combination of the $f_i$'s with coefficients in the ring of polynomials. But how would you ever express $x_k$ this way when $k>N$?
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