Find a ring $R$ with a chain of prime ideals of length 2008.

Question

I am stuck at finding a ring $R$ with a chain of prime ideals of length $2008$. For reference,

Definition A chain of prime ideals of length $n$ in a commutative ring $R$ is an increasing sequence $$P_0\subsetneq P_1 \subsetneq P_2 \subsetneq \cdots \subsetneq P_n \subsetneq R,$$ where $P_i$ is a prime ideal in $R$.

I realized that $R$ should not be PID since, in this case, maximum length of the chain is $1$. I would appreciate any help! Thanks in advance!

Answer 1

Take $R = k[x_1,x_2,x_3,\dots]$ for some field $k$ and then the chain $$(0) \subset (x_1) \subset (x_1,x_2) \dots \subset (x_1, \dots, x_{2008}) \subset R.$$ The ideals are prime since the quotient $R/(x_1, \dots, x_n) \cong k[x_{n+1}, x_{n+2}, \dots]$ is an integral domain. That way you can create chains of arbitrary length.

Answer 2

I guess $\Bbb R[x_1,\cdots,x_{7000}]$ works.