# Congruence Classes

by Brimley
Tags: classes, congruence
 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 $\lambda$ and I know that $\lambda = (3+\sqrt{-3})/2$ $\in$ $Q[\sqrt{-3}]$. I also know that $\lambda$ is a prime in $Q[\sqrt{-3}]$. From here, I would like to prove that iff $\lambda$ divides $a$ for some rational integer $a$ in $Z$, it can be proven that 3 divides $a$. Can this is done? If so, could someone show me? Lastly (or as a second part to this), what are the congruence classes $(mod (3+\sqrt{3})/2)$ in $Q[\sqrt{-3}]$ ? I really appreciate the help on this everyone! *Note: I intentionally put $(mod (3+\sqrt{3})/2)$ with the $\sqrt{3}$, so it should not be negative for this part.
 PF Gold P: 1,059 If $$\lambda \mid a, then N(\lambda) = 3 \mid a^2,$$ or that 3 divides a. Conversely, of course $$\lambda \mid 3$$ It would seem there is no way you can arrive at $$\sqrt3$$ in this field since obviously it would not be $$R\sqrt-3$$, or the Eisenstein integers. The positive and negatives of the quadratic field are not interchangeable.
P: 77
 Quote by robert Ihnot If $$\lambda \mid a, then N(\lambda) = 3 \mid a^2,$$ or that 3 divides a. Conversely, of course $$\lambda \mid 3$$
Thank you robert!

Do you have an idea on the second part? (quoted below)
 What are the congruence classes $(mod (3+\sqrt{3})/2)$ in $Q[\sqrt{-3}]$ ?

PF Gold
P: 1,059
Congruence Classes

 Quote by Brimley Thank you robert! Do you have an idea on the second part? (quoted below)
It would seem there is no way you can arrive at $$\sqrt3$$ in this field since obviously it would not be with the $$\sqrt{ -3}$$ 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 $$\sqrt X$$ to generate the field. The next step is to define and look for the quadratic integers in this set up.
P: 77
 Quote by robert Ihnot It would seem there is no way you can arrive at $$\sqrt3$$ in this field since obviously it would not be with the $$\sqrt{ -3}$$ 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 $$\sqrt X$$ 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 $(mod (3+\sqrt{3})/2)$ in $Q[\sqrt{-3}]$ ?"
 PF Gold P: 1,059 A quadratic integer, Eisenstein, is of the form $$a+b\omega$$ where $$\omega = \frac{-1+\sqrt-3}{2}$$ Here a and b are integers and $$\omega^3=1$$. The form will satisfy an integral equation with the squared term unity. Here we have for the cube root of 1, $$1+\omega+\omega^2 = 0$$. The roots of our quadratic are $$a+b\omega$$ $$a+b\omega^2$$ 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.
P: 77
 Quote by robert Ihnot A quadratic integer of the Eisenstein form is of the form $$a+b\omega$$ where $$\omega = \frac{-1+\sqrt-3}{2}$$
I understood this, however I don't understand where you're going with this...
 PF Gold P: 1,059 I tried to make this clear that $$\sqrt3$$ is not an algebratic integer in this set, so that it is useless to consider residue classes. If you want to ajoin $$\sqrt3$$ to this set then you would no longer be talking about a quadratic integer.
P: 77
 Quote by robert Ihnot A quadratic integer, Eisenstein, is of the form $$a+b\omega$$ where $$\omega = \frac{-1+\sqrt-3}{2}$$ Here a and b are integers and $$\omega^3=1$$. The form will satisfy an integral equation with the squared term unity. Here we have for the cube root of 1, $$1+\omega+\omega^2 = 0$$. The roots of our quadratic are $$a+b\omega$$ $$a+b\omega^2$$ 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.
Okay, I just want to try and format your answer again to make sure I'm getting it right:
 A quadratic integer, Eisenstein, is of the form $a+b\omega$ where $\omega = \frac{-1+\sqrt-3}{2}$ Here a and b are integers and $$\omega^3=1$$. The form will satisfy an integral equation with the squared term unity. Here we have for the cube root of $1$, $$1+\omega+\omega^2 = 0$$. The roots of our quadratic are: Root1: $$a+b\omega$$ Root1: $$a+b\omega^2$$ 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.
So what you're saying is we cannot find that form because we don't have $\sqrt{-3}$ in our mod statement, rather we have $\sqrt{3}$ which will prevent us from getting the statement of: $X^2-3 = 0$ ?
 PF Gold P: 1,059 The question is how is the form arrived at. First we start with the rationals, then we adjoin $$\sqrt-3$$ to this form and generate an expanded set of numbers. But that does not give us the form of $$\sqrt3$$ After all, what is the point of trying to form "reside classes" of $$\pi$$ relative to the integers?

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