1

Let $R$ be a reduced noetherian rings of dimension $d$, $S$ be a multiplicative set of all regular elements of $R$, and $K=S^{-1}R$ be localisation of $R$. Show that $K[T]$ (polynomial in 1 variable over $K$) is a product of principal ideal domain and therefore show that $K[T]$ is a principal ideal ring (not necessarily a domain).

(Note that some of the conditions may be superfluous because I am taking this question off from a book which has many parts, and I am only taking some of it)

I tried searching online for answers, but the closest I can get is the Samuel-Zariski theorem which shows the converse. That is a principal ideal ring can be decomposed into finite product of principal ideal domain or special principal ring.

So if I try to reduce the problem, it would be $K[T]$ is a noetherian ring, of dimension $d+1$ I think, but not sure if it is reduced but $K$ is integrally closed in $K[T]$.

So the problem/ conjecture probably would be: let $A\hookrightarrow B$ be an inclusion of rings, $B$ be a noetherian ring of dimension $d$ and $A$ is integrally closed in $B$, then there is a decomposition of $B$ into product of PIDs.

I was thinking along the chinese remainder theorem because if I can find a suitable $\frak{p}_{i}$ for $i=1,...,n$ that are relatively prime (i.e. ${\frak{p}_{i}+\frak{p}_{j}}=B$ if $i\neq j$), then $B/\prod{\frak{p}_{i}\cong\prod }B/\frak{p}_{i}$. I have got a domain, but I can't proceed further.

Possibly my approach is all wrong, and the fact that $B$ is a polynomial ring shouldn't be left out. So I am pratically stuck.

How should I do this problem?

Thanks!

EDIT: I have changed the statement so that $S$ becomes a multiplicative set of all regular elements.

enoughsaid05
  • 2,371
  • 18
  • 28
  • You should put in $S$ all regular elements of $R$. Otherwise, as noticed by Georges, there is no hope. – Cantlog Nov 10 '13 at 09:37
  • Thanks! I have edited the question accordingly. – enoughsaid05 Nov 10 '13 at 09:49
  • Now you can show that $K$ is a finite product of fields. – Cantlog Nov 10 '13 at 09:52
  • I am quite lost here. Can you make the hint a ~little~ bit more explicit? (Like which of the conditions I should use to decompose, or any theorems in comm alg that proves the decomposition. Thanks!) – enoughsaid05 Nov 10 '13 at 09:57
  • 2
    Let $p_1, \dots, p_n$ be the minimal prime ideals of $R$. Show that (1) $R_{p_i}$ is a field for any $i$; (2) the canonical map $R\to \prod_i R_{p_i}$ induces an isomorphism from $K$ to the RHS. Note that $S$ is equal to the complement of the union of the $p_i$. – Cantlog Nov 10 '13 at 10:01

2 Answers2

1

Now (after the edit) the answer to your question is positive. From this topic one can learn that the total ring of fractions of a reduced noetherian ring is a finite direct product of fields, let's say $K_1,\dots,K_n$. Then $K[x]\cong K_1[x]\times\cdots\times K_n[x]$, that is, $K[x]$ is a direct product of PIDs, and this is obviously a PIR.

Community
  • 1
0

The statement in your grey-displayed initial section is completely false:

Take for $R$ a noetherian domain of dimension $d\gt 0$ and $S=\{1\}$.
Then $K=R$ and $K[T]$ has dimension $d+1\gt 1 $ and so cannot be a principal ideal domain, which has dimension $1$.

Georges Elencwajg
  • 150,790
  • Hi @Georges, while $K[T]$ certainly is not a principal ideal domain, but it can be a product of principal ideal domain, so there shouldn't be a contradiction here, I guess? – enoughsaid05 Nov 10 '13 at 09:15
  • No, $K[T]=R[T]$ is not a product of any rings whatsoever because it is a domain and a product of rings is never a domain. – Georges Elencwajg Nov 10 '13 at 09:29
  • Hi @Georges, I have changed the question so that $S$ becomes a multiplicative set of all regular elements. Would this change anything anything as well? Thanks. – enoughsaid05 Nov 10 '13 at 09:50