Re: R[X,Y] polynomial ring

In article <d72cdn$mg$1@xxxxxxxxxxxxxxxxxx>, Mike <Mike_jones@xxxxxxxxx> wrote:
>Sorry about the inconsistant use of x and X
>
>have another little query
>Am trying to show that Z[sqrt(-5)] is not a PID
>I'm having trouble actually constructing a non principle ideal
>The ideal i get is (1 + sqrt(-5), 2) but having trouble showing it
>a) is an ideal

Surely you are ->DEFINING<- it to be an ideal! That is:

(1+sqrt(-5), 2) = { a*(1+sqrt(-5)) + 2*b | a,b in Z[sqrt(-5)] }

which is an ideal.

>b) is principle

You want to show it is NOT principal, not that it is principal.

Suppose there is a number of the form a + b*sqrt(-5) which is a
generator for the ideal (call it I for ease), with a and b
integers. Then, since the ideal contains both 2 and 1+sqrt(-5), this
putative generator a+b*sqrt(-5) must divide both 2 and 1+sqrt(-5).

LEMMA: If a+b*sqrt(-5) divides r+s*sqrt(-5) (where a,b,r,s are
integers) in Z[sqrt(-5)], then a-b*sqrt(-5) divides r-s*sqrt(-5).

Proof: Suppose (a+b*sqrt(-5))*(x+y*sqrt(-5)) = r+s*sqrt(-5).
Then we must have ax-5by = r, and ay+bx = s.

Then (a-b*sqrt(-5))*(x-y*sqrt(-5)) = (ax - 5by) - (ay+bx)*sqrt(-5)
= r - s*sqrt(-5).

QED

So, suppose a+b*sqrt(-5) divides 2 and divides 1+sqrt(-5). Then

a-b*sqrt(-5) divides 2, and divides 1-sqrt(-5).

Since a+b*sqrt(-5) and a-b*sqrt(-5) divide 2, then their product must
divide 2*2 =4. That is,

(a+b*sqrt(-5))*(a-b*sqrt(-5)) = a^2 + 5b^2 divides 4.

And we also have that (a+b*sqrt(-5))*(a-b*sqrt(-5)) divides
(1+sqrt(-5))*(1-sqrt(-5)); that is,

a^2 + 5b^2 divides 1 + 5 = 6.

So the integer a^2 + 5b^2 must divide both 4 and 6, and therefore must
divide 6-4 = 2.

But then the integer a^2+5b^2 must be no larger than 2. The only
possibility then is for b=0, a=1 or b=0, a=-1. But that would mean
that a+b*sqrt(-5) = 1 or a+b*sqrt(-5) = -1. In either case, we see
that the only possible way in which the ideal I is principal is if I
is generated by 1, and therefore if I is equal to all of Z[sqrt(-5)].

But that would mean that we can write 1 as

1 = 2*(x+y*sqrt(-5)) + (1+sqrt(-5))*(r+s*sqrt(-5))
= (2x) + (2y*sqrt(-5)) + (r-5s) + (s+r)sqrt(-5)
= (2x+r-5s) + (2y + s + r)sqrt(-5),

where x, y, r, and s are integers. So we must have

2x + r - 5s = 1
2y + s + r = 0.

>From the second equation we have that r+s is even, and therefore that
r and s have the same parity (both odd or both even).

If they are both even, then r-5s is also even, so 2x+r-5s is even,
contradicting that it is equal to 1. If they are both odd, then r-5s
is also even, so again we have that 2x+r-5s must be even, but it must
also be equal to 1.

So we conclude that there is in fact no such x, y, r and s, so 1 is
not an element of I. Therefore, I cannot be principal (the only way it
could be principal is if I=(1), but we just saw that 1 is not in I).

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

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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