Re: Prime ideals in Z[x]

From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 11/28/04


Date: Sun, 28 Nov 2004 23:51:47 +0000 (UTC)

In article <codo2u$qof$1@news.onet.pl>, sirix <sirix@poczta.onet.pl> wrote:
>Arturo Magidin wrote:
>> In article <codmt2$col$1@news.onet.pl>, sirix <sirix@poczta.onet.pl> wrote:
>>
>>>Greetings!
>>>How do prime ideals in Z[x] look like? I know Z[x] is noetherian, but I
>>>couldn't find any prime ideal that would have more than two generators,
>>>so... maybe there isn't any? :-)
>>
>>
>> Here is a post with an answer to that question, from Bill Dubuque:
>>
>> http://groups.google.com/groups?selm=y8zr83oaofh.fsf%40nestle.ai.mit.edu
>>
>> You can safely ignore my quoted response, which is harder than it
>> should be.
>
>Thanks. Just because of curiosity: some time ago my teacher told me (as
>a curiosity, just to show what will we be working on after geting
>through Atiyah-McDonald) something like that: "If the ring has a big
>Krull dimension then it is not a UFD. As a matter of fact, bigger the
>Krull dim is, worse things may happen with any kind of uniqueness." What
>precisely did he mean? I thought that there is something like "Ring has
>Krull dim=1 iff Ring is UFD" but it's not the case, as in Z[x] (x)
>\subset (x, 2) and both (x) and (x,2) are prime...

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".

-- 
======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu

Quantcast