Suppose that $R$ is not a noetherian ring. Then can we have an ideal in $R[x_1,...,x_n]$ which is not finitely generated?

Question

I know that if $R$ is noetherian then the statement holds true by Hilbert basis theorem. However I am looking for a example where it doesn't hold true if $R$ is not noetherian.

I was specifically wondering about the ring of the algebraic integers defined in the first example of this answer. If my ring $R$ is the ring of algebraic integers as defined in the link what would be an infinitely generated ideal in that case?

Answer 1

In the ring of algebraic integers the ideal $I=\langle\sqrt{2},\sqrt[4]{2}, \sqrt[8]{2},\dots, \sqrt[2^n]{2},\dots \rangle$ is not finitely generated since it is the union of the ideals $I_n=\langle\sqrt[2^n]{2}\rangle$ which form a strictly ascending chain.
The ideal $I[x_1,\dots,x_n]$ (of polynomials in $x_1,\dots,x_n$ with coefficients in $I$) is also not finitely generated.

Answer 2

Consider $R=k[T_i \ : \ i \in \mathbb{N}]$ (i.e. the polynomial ring over a field $k$ in infinitely many variables $T_i$, this is one of the standard examples of rings that are not noetherian, see also What is an easy example of non-Noetherian domain?), then $R[x_1, \dots, x_n] $ is isomorphic to $R$ (the isomorphism is given by mapping $x_1 \mapsto T_1, \dots, x_n \mapsto T_n, T_j \mapsto T_{n+j}$).

For a concrete ideal, consider the one which is generated by all variables except $T_1$.

Answer 3

$R$ is noetherian iff $R[X]$ is noetherian, because $R$ is a quotient of $R[X]$. Explicitly, if $I$ is a not finitely generated ideal of $R$ then the preimage under $X \mapsto 0$, that is, $I + X R[X]$ is a not finitely generated ideal of $R[X]$.