Krull dimension of $F[x_1, x_2, \ldots]$

Question

Let $F$ be a field and $F[x_1, x_2, \ldots]$ be the polynomial ring in countably many variables.

I found out that $F[x_1, x_2, \ldots]$ is a unique factorization domain, but not a Noetherian ring, since the chain of ideals $(x_1) \subsetneq (x_1, x_2) \subsetneq (x_1, x_2, x_3) \subsetneq \cdots$ cannot become stationary.

I know that the Krull dimension is the supremum of the lengths of all chains of prime ideals. For instance, $\dim(F[x_1, \ldots, x_n]) = n$. The polynomial ring in infinitely many variables $F[x_1, x_2, \ldots]$ has infinite dimension. I guess that it is the same for the polynomial ring in countably many variables, but I am not sure.

Does the polynomial ring in countably many variables have infinite dimension? If not, how can I determine it ?

Thank you for your help.

Answer 1

Yes, it is infinite. As you noted yourself, the ring $F[x_1,...,x_n]$ has dimension $n$ for every $n$. And all those are contained in your ring. You also gave an infinite chain of prime ideals explicitly!

Answer 2

For each $i $, $F [x_1,x_2,...]/\left<x_1,...,x_i\right>\cong F [x_{i+1},x_{i+2},...] $, which is a domain. Hence, for each $i $, $\left<x_1,...,x_i\right>$ is a prime ideal of $F [x_1,x_2,...]$. Thus, $\left<x_1\right>\subsetneq \left<x_1,x_2\right>\subsetneq \cdots\subsetneq \left<x_1,...,x_i\right>\subsetneq\cdots$ is an infinite chain of prime ideals. Hence, the Krull dimension of $F [x_1,x_2,...]$ is not finite.