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From: Leonid Lenov on 22 Jan 2010 10:48
Hello,
Let p>2 be a prime. In Z[sqrt(-p)] the following holds:
(1-sqrt(-p))(1+sqrt(-p))=1+p=2t
so there is no unique factorization in Z[sqrt(-p)]. Is that correct?
Thanks.
From: Arturo Magidin on 22 Jan 2010 13:23
On Jan 22, 9:48 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
> Hello,
> Let p>2 be a prime. In Z[sqrt(-p)] the following holds:
>    (1-sqrt(-p))(1+sqrt(-p))=1+p=2t
> so there is no unique factorization in Z[sqrt(-p)]. Is that correct?

Yes; the only units in Z[sqrt(-p)] are 1 and -1, so 2 cannot be an
associate of either (1-sqrt(-p)) or (1+sqrt(-p)). A norm argument
shows that 2 is irreducible, and clearly 2 does not divide either 1-
sqrt(-p) nor 1+sqrt(-p).

Keep in mind, though, that Z[sqrt(-p)] may not be a number field: if
p=3 (mod 4), then the number ring of Q(sqrt(-p)) is Z[(1+sqrt(-p))/2].
So, for example, the number ring of Q(sqrt(-3)) *is* a UFD; that ring
is Z[(1+sqrt(-3))/2]. That said, there are only finitely many number
rings of imaginary quadratic number fields that are UFDs.

--
Arturo Magidin
From: Bill Dubuque on 22 Jan 2010 15:28
Arturo Magidin <magidin(a)member.ams.org> wrote:
> On Jan 22, 9:48 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
>>
>> Let p>2 be a prime. In Z[sqrt(-p)] the following holds:
>> -sqrt(-p))(1 +sqrt(-p))=1= 1+p = 2t
>> so there is no unique factorization in Z[sqrt(-p)]. Is that correct?
>
> Yes; the only units in Z[sqrt(-p)] are 1 and -1, so 2 cannot be an
> associate of either (1-sqrt(-p)) or (1+sqrt(-p)).

That's not needed. It suffices to show 2 is non-prime irreducible.
Norms => irreducible, and the factorization shows its non-prime.

> A norm argument shows that 2 is irreducible, and clearly 2 does not
> divide either 1-sqrt(-p) nor 1+sqrt(-p).
>
> Keep in mind, though, that Z[sqrt(-p)] may not be a number field: if
> p=3 (mod 4), then the number ring of Q(sqrt(-p)) is Z[(1+sqrt(-p))/2].
> So, for example, the number ring of Q(sqrt(-3)) *is* a UFD; that ring
> is Z[(1+sqrt(-3))/2]. That said, there are only finitely many number
> rings of imaginary quadratic number fields that are UFDs.

From: Achava Nakhash, the Loving Snake on 22 Jan 2010 16:48
On Jan 22, 7:48 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
> Hello,
> Let p>2 be a prime. In Z[sqrt(-p)] the following holds:
>    (1-sqrt(-p))(1+sqrt(-p))=1+p=2t
> so there is no unique factorization in Z[sqrt(-p)]. Is that correct?
> Thanks.

In order to properly understand the other replies, here is some
information you should know. First of all, when you are in the ring Z
[sqrt(-p)], prime means prime in this ring, and being a prime among
the set of integers does not make you a prime in this ring. For
instance,
p = -(sqrt(-p)))^2, and so p is not a prime in this ring.

Second of all, unique factorization is always up to unit factors. A
unit is defined as an element u of the ring such that there exists an
element v of the ring with u*v = 1. In other words, the units are the
multiplicatively invertible elements of the ring. With the integers,
the only units are 1 and -1, so there is no real issue. In your ring Z
[sqrt(-p)], it looks like there are no units except 1 and -1.
However, if you had used Z{sqrt(p)] instead. there would have been
infinitely many units.

Third of all, number theorists don't generally consider the rings Z
[sqrt(-p)]. Instead they consider all the elements of Q[sqrt(-p)]
which satisfy monic polynomials over Z. In your cases, this is the
same as Z[sqrt(-p)] as long as p = 1 (mod 4). However, when p = 3
(mod 4), then instead of Z[sqrt(-p)] you get everything of the form (a
+ b*sqrt(-p))/2 where a and b are integers that are either both even
or both odd. This contains, but is larger than Z[sqrt(-p)]. When p =
3, you actually get units doing this when a = b = 1. For other p
there are no units.

You can never get unique factorization out of Z[sqrt(-p)] when p = 3
(mod 4). When you do it the standard way that I described above, you
get unique factorization for precisely 8 values of p. Unfortunately I
won't have access to a number theory book to look up which one's they
are until I get home from work which wll be maybe 4 hours from now.
However, I know that 3 is among them. In fact I think that they are
all = 3 (mod 4). You can adjoin sqrt(-1) also and that gives you a
unique factorization domain whose units are 1, -1, i, -i. You will
hear in some quarters about the 9 numbers of Gauss which include -1
and -p for all the different p that give you unique factorization
domains doing it the way I did. I am pretty sure that
-19 and -163 are among these numbers.

Regards,
Achava
From: Andrew Usher on 22 Jan 2010 18:53
On Jan 22, 9:48 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
> Hello,
> Let p>2 be a prime. In Z[sqrt(-p)] the following holds:
>    (1-sqrt(-p))(1+sqrt(-p))=1+p=2t
> so there is no unique factorization in Z[sqrt(-p)]. Is that correct?
> Thanks.

Yes, and p need not be prime. As others have pointed out, if you allow
the half-integer values when p=3 mod 4, you have additional UFDs: p=3,
7, 11, 19, 43, 67, 163.

Andrew Usher
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