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

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 ?

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!
• Although the answer is correct, it's maybe important to notice that if $S$ is a subring of $R$ then the dimension of $S$ is not necessarily smaller than that of $R$ (take eg Z in Q). – Jef L Dec 2 '17 at 17:57
• Yes, but in this case we are only adding transcendental elements with no further relations, unlike the case with $\mathbb Q$ and $\mathbb Z$, which makes it much more intuitive – neptun Dec 2 '17 at 18:39
For each $i$, $F [x_1,x_2,...]/<x_1,...,x_i>\cong F [x_{i+1},x_{i+2},...]$, which is a domain. Hence, for each $i$, $< x_1,...,x_i>$ is a prime ideal of $F [x_1,x_2,...]$. Thus, $< x_1>\subsetneq < x_1,x_2>\subsetneq ...\subsetneq < x_1,...,x_i>\subsetneq...,$ is an infinite chain of prime ideals. Hence, the krull dimension of $F [x_1,x_2,...]$ is not finitely.