Show that the ring of polynomials with coefficients in a field, and in infinitely many variables, is not Noetherian

Question

Show that the ring of polynomials with coefficients in a field, and in infinitely many variables, is not Noetherian, that is, $R = k [x_i: i\geq1]$ 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?

Answer 1

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 $$(x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots$$ that never terminates, so $k[x_i: i\ge 1]$ is not Noetherian.

Answer 2

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 $k>N$?