Re: Prime ideals in Z[x]

From: Trav (lzwnews_at_yahoo.com)
Date: 11/29/04 

Date: 28 Nov 2004 18:13:26 -0800

Arturo Magidin wrote:
>
> No, the Krull dimension is no guarantee; you have there an example of
> a UFD which has Krull dimension greater than 1; and Z[sqrt(-5)] is an
> example where the Krull dimension is 1 but the ring is not a UFD.
>
> I'm not entirely sure what he meant; you can get arbitrarily high
> Krull dimension and still have a UFD, simply by taking things like
> Z[x1,....,xn]. Then you have the chain
> 0< (x1) < (x1,x2) < (x1,x2,x3) < ... < (x1,...,xn) < (2,x1,...,xn)
>
> so the dimension is at least n+1. On the other hand, "all but one" of
> those prime ideals come from the "transcendence degree".
>

More generally, a given integral domain R is a UFD if and only if all
prime ideals of height 1 are principal. It is easy to show that any
prime ideal of height r>0 must be generated by at least r elements. The
above tells us is that the converse holds for r=1 <=> R is a UFD.