• Jul 21st 2010, 04:14 AM
Samson
Hello All,

I forgot to throw this into the QFields and Norms thread, and I'm not sure if I should exhume it considering I already gave out thanks and all.

Anyways, my book says that an element exists in Q[Sqrt(-1)] that is not a quadratic integer but its norm in Q[Sqrt(-1)] is an integer.

I tried taking it back to fundamentals like had previously been suggested, but here is what I get.
x=Sqrt(-1)
(x^2)=-1
(x^2)-1=0

I also know that if x = Sqrt(-1), then X = i, so then I have have (i^2)-1=0 from my last statement. I don't see how that helps me prove what I need to, but its a start.

All help is appreciated!
• Jul 21st 2010, 04:19 AM
undefined
Quote:

Originally Posted by Samson
Hello All,

I forgot to throw this into the QFields and Norms thread, and I'm not sure if I should exhume it considering I already gave out thanks and all.

Anyways, my book says that an element exists in Q[Sqrt(-1)] that is not a quadratic integer but its norm in Q[Sqrt(-1)] is an integer.

I tried taking it back to fundamentals like had previously been suggested, but here is what I get.
x=Sqrt(-1)
(x^2)=-1
(x^2)-1=0

I also know that if x = Sqrt(-1), then X = i, so then I have have (i^2)-1=0 from my last statement. I don't see how that helps me prove what I need to, but its a start.

All help is appreciated!

(Transferred from PM conversation, hence the time stamps being so close together.)

Where did you use the definition of quadratic integer?

Quadratic integer - Wikipedia, the free encyclopedia

Note that D = -1 means we have $\alpha = a + bi$ and if a and b are integers then we have Gaussian integers for which all elements are quadratic integers

Gaussian integer - Wikipedia, the free encyclopedia

So in particular we know that a and b can't both be integer, at least one must be of the form p/q where q does not divide p. Furthermore if say b is non-integer but a is integer then a^2 + b^2 can't be integer, so a and b must both be non-integer.

So we're looking for a^2 + b^2 is an integer and a and b are both non-integers, subject also to the condition that a + bi must not be a quadratic integer. So while looking for a pair (a,b) to satisfy, it seems Pythagorean triples could help, for example a = 3/5, b = 4/5 where 3 and 4 are from a Pythagorean triple, with the result that a^2 + b^2 = 1. Now all we need to do is check that this is not a quadratic integer. So notice that for quadratic equation

$x^2 = Ax^2 + Bx + C$

both solutions are given by

$x = \dfrac{-B \pm \sqrt{A^2-4AC}}{2A}$

Setting A = 1 as per definition of quadratic integer, we get

$x = \dfrac{-B \pm \sqrt{1-4C}}{2}$

There is no way a number of this form can be

$\displaystyle \frac{3}{5}+ \frac{4}{5}i$

(notice the denominator 5 cannot be reduced to denominator 2), so we are done.
• Jul 21st 2010, 04:21 AM
Samson
Okay, I followed your example right down to the end, but I failed to see if you were or if it was a direct relation to my problem. I saw you said D=-1 at the top, but I missed anything else thereafter about Q[Sqrt(-1)] and norms in Q[Sqrt(-1)]. Could you provide additional clarity? Thank you!
• Jul 21st 2010, 04:23 AM
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Quote:

Originally Posted by Samson
Okay, I followed your example right down to the end, but I failed to see if you were or if it was a direct relation to my problem. I saw you said D=-1 at the top, but I missed anything else thereafter about Q[Sqrt(-1)] and norms in Q[Sqrt(-1)]. Could you provide additional clarity? Thank you!

The norm N(a+bi) in this quadratic field is a^2 - b^2D = a^2 - b^2(-1) = a^2 + b^2.

The question you posed is: an element exists in Q[Sqrt(-1)] that is not a quadratic integer but its norm in Q[Sqrt(-1)] is an integer; find such an element.

So, my whole post was working towards finding such an element, which, unless I'm mistaken, I accomplished.
• Jul 21st 2010, 04:30 AM
Samson
Quote:

Originally Posted by undefined
The norm N(a+bi) in this quadratic field is a^2 - b^2D = a^2 - b^2(-1) = a^2 + b^2.

The question you posed is: an element exists in Q[Sqrt(-1)] that is not a quadratic integer but its norm in Q[Sqrt(-1)] is an integer; find such an element.

So, my whole post was working towards finding such an element, which, unless I'm mistaken, I accomplished.

So is the element 3/5 ? or are the elements 3/5 and 4/5 or neither?
• Jul 21st 2010, 04:32 AM
undefined
Quote:

Originally Posted by Samson
So is the element 3/5 ? or are the elements 3/5 and 4/5 or neither?

$\alpha = a + bi$

a = 3/5

b = 4/5

We've been using the same notation over and over.. surely you can get used to it by now? The element is neither 3/5 nor 4/5 but is rather (3/5) + (4/5)i .

Note: 4/5 + (3/5)i also works, as do their conjugates
• Jul 21st 2010, 04:34 AM
Samson
Quote:

Originally Posted by undefined
$\alpha = a + bi$

a = 3/5

b = 4/5

We've been using the same notation over and over.. surely you can get used to it by now? The element is neither 3/5 nor 4/5 but is rather (3/5) + (4/5)i .

Note: 4/5 + (3/5)i also works, as do their conjugates

Okay, so we refer to alpha then as "the element" which is in the form of a+bi, correct?
• Jul 21st 2010, 04:36 AM
undefined
Quote:

Originally Posted by Samson
Okay, so we refer to alpha then as "the element" which is in the form of a+bi, correct?

Well the letters we choose are arbitrary, but in this case, yes, we want alpha.
• Jul 21st 2010, 04:40 AM
Samson
Thank you! I really appreciate all the help!