Proving Congruence Classes in Q[√-3] with Prime Divisor λ

[Brimley]

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.

 

[robert Ihnot]

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.

 

[Brimley]

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 $Q[\sqrt{-3}]$ ?

 

[robert Ihnot]

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.

 

[Brimley]

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 $Q[\sqrt{-3}]$ ?"

 

[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.

 

[Brimley]

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...

 

[robert Ihnot]

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.

 

[Brimley]

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: $X^2-3 = 0$ ?

 

[robert Ihnot]

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?