# 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 ?

• Doesn't your second sentence give an example of an infinite chain of prime ideals?
– anon
Dec 2, 2017 at 17:37
• @anon : Yes, it does and I find it odd that my lecturer asks this trivial question. This is confusing. I expected something more tricky. So maybe I just conclude that it has infinite dimension and then it is fine. Dec 2, 2017 at 17:48

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
Dec 2, 2017 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 Dec 2, 2017 at 18:39

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

For anyone interested in such things:

The Krull dimension of $$R=k[x_0, x_1, x_2, \dots ]$$ is at least $$\mathfrak{c}=|\mathbb{R}|$$.

Proof: Let $$\mathcal{C}$$ be a linear chain of subsets of $$\mathbb{N}$$ of cardinality $$\mathfrak{c}$$. Then the ideals $$\mathfrak{P}_S:=(x_n\;|\; n \in S),\;\; S \in \mathcal{C}$$ are prime ideals since each of the quotients $$R/\mathfrak{P}_S \simeq k[x_n\;|\; n \notin S]$$ is a domain, clearly $$\mathfrak{P}_S$$ are distinct and form a linear chain of cardinality $$\mathfrak{c}$$. $$\square$$

(At least in the case when $$k$$ is at most countable, the dimension is exactly $$\mathfrak{c}$$ since the cardinality of all subsets of $$R$$ is $$\mathfrak{c}$$.)