0
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Doesn't your second sentence give an example of an infinite chain of prime ideals? $\endgroup$ – anon Dec 2 '17 at 17:37
  • $\begingroup$ @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. $\endgroup$ – Crystal Dec 2 '17 at 17:48
2
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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). $\endgroup$ – Jef L Dec 2 '17 at 17:57
  • $\begingroup$ 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 $\endgroup$ – neptun Dec 2 '17 at 18:39
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.