# Prove that Associates Have the Same Norm

1. Aug 9, 2007

### Frillth

1. The problem statement, all variables and given/known data

I need to prove that any two quadratic integers that are associates must also have the same norm.

2. Relevant equations

If α = a + b√d, the norm of α is N(α) = a^2 - b^2*d.

If two quadratic integers α and β are associates, α divides β, β divides α, and α/β and β/α both equal some unit, although each may be equal to a different unit.

N(unit) = ±1

3. The attempt at a solution

This is what I've done so far:

α/β = ε
N(α/β) = N(ε)
N(α/β) = ±1
N(α)/N(β) = ±1
N(α) = ±N(β)

From here, I guess I need to show that it is impossible to have N(α) = -N(β), but I'm not sure how to do that. Any ideas?

Last edited: Aug 9, 2007
2. Sep 16, 2007

### Frillth

Sorry to go gravedigging here, but I just had a new idea. Is it even necessary that associates have the same norm?

α and β are defined as associates if α = β*ε

Thus, we can take any quadratic integer α in any quadratic field that has units that have negative norms, and multiply α by some unit ε with a negative norm to get β. Then when we take the norm we have:

N(α) = N(β*ε)
N(α) = N(β)*N(ε)
N(α) = -N(β)

Is all of my logic here correct? Because if so, it seems I may have been trying to prove a statement that is actually false.

3. Sep 16, 2007

### HallsofIvy

Staff Emeritus
Can you actually find two such associates?

By the way, have you actually proved either $N(\alpha /\beta)= N(\alpha)/N(\beta)$ or $N(\alpha\beta)= N(\alpha)N(\beta)$?

4. Sep 19, 2007

### Frillth

In Q[sqrt(2)], 7 - 5*sqrt(2) has norm -1, which makes it a unit. I chose α = 2 + 2*sqrt(2) to test my initial assumption:

β = (2 + 2*sqrt(2))(7 - 5*sqrt(2))
β = -6 + 4*sqrt(2)

N(α) = -4
N(β) = 4

Did I do something wrong, or did I prove that my original assumption was incorrect?