• Oct 20th 2006, 12:24 AM
beta12
1) Which of the following are quadratic integers in Q[sqrt(-5)]?
a) 3/5
b) (3+8sqrt(-5))/5
c) i*sqrt(-5)

Prove your result in each case.

2) Find an element of Q[sqrt(-1)] tht is not a quadratic integer, and yet its norm in Q[sqrt(-1)] is an integer.
• Oct 20th 2006, 03:59 AM
ThePerfectHacker
[QUOTE=beta12;24319]1) Which of the following are quadratic integers in Q[sqrt(-5)]?
a) 3/5
b) (3+8sqrt(-5))/5
c) i*sqrt(-5)

Again, use what I said before, namely that,
$\{1,i\sqrt{5}\}$ forms a basis for the field $\mathbb{Q}(\sqrt{-5})$ viewed as a vector space over $\mathbb{Q}$.
We note that,
$\mathbb{Q}\leq \mathbb{Q}(\sqrt{-5})$
Meaning, that the rationals are contained in this simple extension.

a)Yes, as just explained all rationals are contained thin this field.

b)Yes, it is $\frac{3}{5}+\frac{8}{5}\sqrt{-5}$ so if you select $a=3/5,b=8/5 \in \mathbb{Q}$ as a linear combination for $a+b\sqrt{-5}$ then it works!

c)No, unless you are asking for $i\sqrt{5}$.
• Oct 20th 2006, 05:22 AM
beta12
Hi Perfecthacker,

What about (3+5sqrt(2))/2 ? It can be written as 3/2 + 5/2*sqrt(2) in the form of a +bsqrt(d). So is it a quadratic integer?
• Oct 20th 2006, 07:44 AM
ThePerfectHacker
Quote:

Originally Posted by beta12
Hi Perfecthacker,

What about (3+5sqrt(2))/2 ? It can be written as 3/2 + 5/2*sqrt(2) in the form of a +bsqrt(d). So is it a quadratic integer?

Yes.

But is it not a quadradic integer in $\mathbb{Q}(\sqrt{-5})$
• Oct 20th 2006, 08:36 AM
beta12
Quote:

Originally Posted by ThePerfectHacker
Yes.

But is it not a quadradic integer in $\mathbb{Q}(\sqrt{-5})$

Hi Perfecthacker,

I see.

Can you teach me how to solve the second question?
2) Find an element of Q[sqrt(-1)] tht is not a quadratic integer, and yet its norm in Q[sqrt(-1)] is an integer.

Thank you very much.
• Oct 20th 2006, 09:42 AM
ThePerfectHacker
Quote:

Originally Posted by beta12
Can you teach me how to solve the second question?
2) Find an element of Q[sqrt(-1)] tht is not a quadratic integer, and yet its norm in Q[sqrt(-1)] is an integer.

Not sure about this one. See, I never studied algebraic number theory (noob :( ). However, I was able to answer those questions you had because I studied field theory. The following is going to be a guess, though I am highly certain about it.

The field $\mathbb{Q}(i)$ are the Gaussian integers (rationals is more proper). Hence, any can be expressed as a linear combination,
$a+bi\, \exists a,b\in \mathbb{Q}$
The number,
$\frac{ \sqrt{2} }{2}+i\frac{ \sqrt{2} }{2}$ is not Gaussian. Yet its norm is 1 which is Gaussian.
• Oct 21st 2006, 12:25 AM
beta12
I see. But you are still great at Quadratic field. :)