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non-euclidean subring of Q[Sqrt(-19)]

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Tobias Fritz

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May 2, 2004, 11:48:58 AM5/2/04
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I've heard that there is a subring of the field Q[Sqrt(-19)] that is a
principal ideal domain (why?) but cannot be made into an euclidean ring.
Can anyone elaborate on this?

TIA,
Tobias
--
there's a tiny little monkey
he lives inside my head

reverse my forename for mail! - saibot

Timothy Murphy

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May 2, 2004, 12:32:28 PM5/2/04
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Tobias Fritz wrote:

> I've heard that there is a subring of the field Q[Sqrt(-19)] that is a
> principal ideal domain (why?) but cannot be made into an euclidean ring.
> Can anyone elaborate on this?

(1) A sub-ring of a field is always an integral domain.

(2) I would guess that the question refers to the ring A of integers
in this field.
Since -19 = 1 mod 4, A = Z[w] where w = (1 + sqrt{-19})/2.

If you consider the sub-ring B = Z[sqrt{-19}]
you will probably find that factorisation into irreducibles
in not unique, so B cannot be euclidean.
Maybe consider 1 - x^2, where x = sqrt{-19}.

--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland

Tobias Fritz

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May 2, 2004, 1:18:08 PM5/2/04
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>
>> I've heard that there is a subring of the field Q[Sqrt(-19)] that is a
>> principal ideal domain (why?) but cannot be made into an euclidean ring.
>> Can anyone elaborate on this?
>
> (1) A sub-ring of a field is always an integral domain.
>
trivial. But my question was for a *principal* integral domain.

> (2) I would guess that the question refers to the ring A of integers
> in this field.
> Since -19 = 1 mod 4, A = Z[w] where w = (1 + sqrt{-19})/2.

OK. Why is this called the ring of integers?
Tonight I'll have more time, so I'll check if A is a PID and
non-euclidean...

>
> If you consider the sub-ring B = Z[sqrt{-19}]
> you will probably find that factorisation into irreducibles
> in not unique, so B cannot be euclidean.
> Maybe consider 1 - x^2, where x = sqrt{-19}.
>

But this also means that B is no PID.

Arturo Magidin

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May 2, 2004, 4:41:58 PM5/2/04
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In article <c7357p$i2fvv$1...@ID-222434.news.uni-berlin.de>,

Tobias Fritz <tob...@mad.scientist.com> wrote:
>I've heard that there is a subring of the field Q[Sqrt(-19)] that is a
>principal ideal domain (why?) but cannot be made into an euclidean ring.
>Can anyone elaborate on this?


Let a = (1+\sqrt{-19})/2. The ring in question is R=Z[a], the ring of
integers of Q[sqrt(-19)]. Let b= 1-a = (1-sqrt(-19))/w, the Galois
conjugate of a.

Define a norm function N:R -> Z, by

N(x+ya) = x^2 + xy + 5y^2; this is just the result of multiplying
together x+ya and x+yb.

N is in fact a Dedekind-Hasse norm: it is positive for all x and y
except for x=y=0; and for every nonzero a,b in R, either a is a
mulitple of b, or else there is a nonzero element in the ideal (a,b)
of norm strictly smaller than b; that is, either b divides a, or there
exist s and t in R such that 0 < N(sa-tb) < N(b).

An integral domain is a PID if and only if R has a Dedekind-Hasse
norm, so R is a PID.

However, R does not have universal side divisors (an element u in R
which is not a unit and not zero is a universal side divisor if and
only if for every x in R there is some element z, either a unit or
z=0, such that u divides x-z). In a Euclidean domain, there must be
universal side divisors: for if R is Euclidean with respect to some
norm M, then pick an element u in R which has minimal M-norm among all
non-zero nonunits; for any x in D, we may write x = qu+r for some r
with r=0 or N(r)<N(u); by minimality, r is either 0 or a unit, so u is
a universal side divisor.

To see that R does not have universal side divisors, note that the
only units in R are 1 and -1; if u is a universal side divisor, then
it must either divide 2, 2-1, or 2+1 in R; since u is not a unit, it
must divide iether 2 or 3. But the only divisors of 2 are 1, -1, 2,
and -2; and the only divisors of 3 are 1, -1, 3, and -3. So u is
either 2, -2, 3, or -3. But now taking x = a, it is easy to verify
that none of 2, -2, 3, or -3 divide any of a, a+1, or a-1, so no
element can be a universal side divisor in R; thus, R cannot be
Euclidean.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
mag...@math.berkeley.edu

Timothy Murphy

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May 2, 2004, 6:24:17 PM5/2/04
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Tobias Fritz wrote:

>>> I've heard that there is a subring of the field Q[Sqrt(-19)] that is a
>>> principal ideal domain (why?) but cannot be made into an euclidean ring.
>>> Can anyone elaborate on this?
>>
>> (1) A sub-ring of a field is always an integral domain.
>>
> trivial. But my question was for a *principal* integral domain.

Sorry, I misunderstood or rather mis-read the question,
which is much more difficult than I thought.

>> (2) I would guess that the question refers to the ring A of integers
>> in this field.
>> Since -19 = 1 mod 4, A = Z[w] where w = (1 + sqrt{-19})/2.
>
> OK. Why is this called the ring of integers?

These are the algebraic integers in the field.

(As you probably know, a complex number c is an algebraic integer
if it satisfies a polynomial equation of the form
x^n + a_1 x^{n-1} + ... + a_n = 0
with integral a_i.
The term "integer" is often used for "algebraic integer"
in algebraic number theory -
ordinary integers then being called "rational integers".)

I think the question in this case is to show firstly that A is
a principal ideal domain (and so a unique factorisation domain).

The simplest way to show an algebraic number ring A
(the ring of integers in an algebraic number field)
is a PID is to show that there exists a norm-like function f:A -> N
with f(ab) = f(a)f(b) and f(a) = 0 iff a = 0
such that one can always divide a by b (where b <> 0), say
a = qb + r
with f(r) < f(b).
In this case one can carry out the euclidean algorithm
and so show A is a PID.
Such a number ring is said to be euclidean.

Your problem, I think, is to show that A is a PID but not euclidean.
This is difficult.
You would find a proof, IIRC, in Hardy & Wright,
Introduction to Number Theory,
a very old book but still the best in its class, IMHO.

Tobias Fritz

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May 3, 2004, 2:54:21 AM5/3/04
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> (As you probably know, a complex number c is an algebraic integer
> if it satisfies a polynomial equation of the form
> x^n + a_1 x^{n-1} + ... + a_n = 0
> with integral a_i.
> The term "integer" is often used for "algebraic integer"
> in algebraic number theory -
> ordinary integers then being called "rational integers".)
>
Ah, this clears it up.

>
> Your problem, I think, is to show that A is a PID but not euclidean.

Exactly.

> This is difficult.
> You would find a proof, IIRC, in Hardy & Wright,
> Introduction to Number Theory,
> a very old book but still the best in its class, IMHO.
>

I'll have a close look at Arturo's answer now, thanks.

Arturo Magidin

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May 3, 2004, 9:43:36 AM5/3/04
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In article <c74q99$ibi5a$1...@ID-222434.news.uni-berlin.de>,
Tobias Fritz <tob...@mad.scientist.com> wrote:

>I'll have a close look at Arturo's answer now, thanks.

Feel free to ask me to expand or explain.

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