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.
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\sqrt-3 [/tex], or the Eisenstein integers. The positive and negatives of the quadratic field are not interchangeable.
It would seem there is no way you can arrive at [tex] \sqrt3 [/tex] in this field since obviously it would not be with the [tex] \sqrt{ -3} [/tex] or the Eisenstein integers. The positive and negatives of the quadratic field are not interchangeable. What 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.
Is that the same if you treat this as a separate problem entirely? 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] ?"
A quadratic integer, Eisenstein, is of the form [tex]a+b\omega[/tex] where [tex]\omega = \frac{-1+\sqrt-3}{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-(2a-b)X+a^2-ab+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^2-3 = 0. You can try to find that.
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.
Okay, I just want to try and format your answer again to make sure I'm getting it right: So what you're saying is we cannot find that form because we don't have [itex]\sqrt{-3}[/itex] in our mod statement, rather we have [itex]\sqrt{3}[/itex] which will prevent us from getting the statement of: [itex]X^2-3 = 0 [/itex] ?
The question is how is the form arrived at. First we start with the rationals, then we adjoin [tex]\sqrt-3[/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?