
#1
Oct1810, 07:04 PM

P: 77

Hello PhysicsForums!
I have been reading up on congruence classes and working out some examples. I came across one example that I seem to struggle understanding. I've solved for [itex]\lambda[/itex] and I know that [itex]\lambda = (3+\sqrt{3})/2[/itex] [itex]\in[/itex] [itex]Q[\sqrt{3}][/itex]. I also know that [itex]\lambda[/itex] is a prime in [itex]Q[\sqrt{3}][/itex]. From here, I would like to prove that iff [itex]\lambda[/itex] divides [itex]a[/itex] for some rational integer [itex]a[/itex] in [itex]Z[/itex], it can be proven that 3 divides [itex]a[/itex]. Can this is done? If so, could someone show me? Lastly (or as a second part to this), what are the congruence classes [itex] (mod (3+\sqrt{3})/2) [/itex] in [itex] Q[\sqrt{3}] [/itex] ? I really appreciate the help on this everyone! *Note: I intentionally put [itex] (mod (3+\sqrt{3})/2) [/itex] with the [itex] \sqrt{3} [/itex], so it should not be negative for this part. 



#2
Oct1910, 09:54 AM

PF Gold
P: 1,059

If [tex]\lambda \mid a, then N(\lambda) = 3 \mid a^2, [/tex] or that 3 divides a. Conversely, of course [tex]\lambda \mid 3[/tex]
It would seem there is no way you can arrive at [tex]\sqrt3[/tex] in this field since obviously it would not be [tex]R\sqrt3 [/tex], or the Eisenstein integers. The positive and negatives of the quadratic field are not interchangeable. 



#3
Oct1910, 12:03 PM

P: 77

Do you have an idea on the second part? (quoted below) 



#4
Oct1910, 11:19 PM

PF Gold
P: 1,059

Congruence ClassesWhat happens is that we begin with the rationals and add the [tex]\sqrt X [/tex] to generate the field. The next step is to define and look for the quadratic integers in this set up. 



#5
Oct1910, 11:49 PM

P: 77

Perhaps if I word it like this it will be different (if not just say no): "What are the congruence classes [itex] (mod (3+\sqrt{3})/2) [/itex] in [itex] Q[\sqrt{3}][/itex] ?" 



#6
Oct2010, 12:31 AM

PF Gold
P: 1,059

A quadratic integer, Eisenstein, is of the form [tex]a+b\omega[/tex] where [tex]\omega = \frac{1+\sqrt3}{2}[/tex] Here a and b are integers and [tex]\omega^3=1[/tex]. The form will satisfy an integral equation with the squared term unity. Here we have for the cube root of 1, [tex] 1+\omega+\omega^2 = 0 [/tex]. The roots of our quadratic are [tex]a+b\omega[/tex] [tex] a+b\omega^2 [/tex]
This gives then the form of X^2(2ab)X+a^2ab+b^2. If we let a=1,b=2, we arrive at X^2+3 = 0. The question is can we arrive at the form X^23 = 0. You can try to find that. 



#7
Oct2010, 12:35 AM

P: 77





#8
Oct2010, 12:59 AM

PF Gold
P: 1,059

I tried to make this clear that [tex]\sqrt3[/tex] is not an algebratic integer in this set, so that it is useless to consider residue classes.
If you want to ajoin [tex]\sqrt3[/tex] to this set then you would no longer be talking about a quadratic integer. 



#9
Oct2010, 01:28 AM

P: 77





#10
Oct2010, 07:24 PM

PF Gold
P: 1,059

The question is how is the form arrived at. First we start with the rationals, then we adjoin [tex]\sqrt3[/tex] to this form and generate an expanded set of numbers. But that does not give us the form of [tex]\sqrt3[/tex]
After all, what is the point of trying to form "reside classes" of [tex] \pi [/tex] relative to the integers? 


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