Complex Plans and Quadratic Integers

**Samson** · July 27th 2010, 03:28 AM

Hello All,

Here is a problem that I'm having some trouble solving.

Part 1:
"In the complex plane, there exists three cube roots of one. Let X be the cube root of one which has positive imaginary part. Knowing this, it can be shown that X is a quadratic integer in Q[Sqrt(-3)] written in the form of (a+b*Sqrt(-3))/2 where a and be are either both even or both odd and are rational. Once written in this form, it can be written in the form: m+[(n/2)(1+Sqrt(-3))].

Part 2:

On a complex plane, how can one represent the quadratic integers in Q[Sqrt(-1)], Q[Sqrt(-3)], and Q[Sqrt(-5)]? I specifically need to note the points 0, 1, and sqrt(d) and which points are "units".

Any and all help is greatly appreciated

**Media_Man** · July 30th 2010, 09:49 AM

This seems like an interesting question, if I could understand it. Is this written exactly the way you received it?

So far here is what I gather: Define set $\mathbb{Q}[\sqrt{-d}]=\{m+\frac n2(1+\sqrt{-d})|m,n\in\mathbb{Q}\}$ . The real and complex parts of these elements are easily gotten: $a=m+\frac n2, b=\frac n2\sqrt{d}$

If we denote $(m,n)_Q$ as points in this set, then $0=(0,0)_Q, 1=(1,0)_Q, \sqrt{-d}=(-1,2)_Q$ . The number $\sqrt{d}$ is NOT in the set, because it corresponds to the complex number $\sqrt{d}+0i$ , and $a=m+\frac n2$ cannot be an irrational number if m,n are both rational.

**Samson** · July 30th 2010, 09:52 AM

Originally Posted by Media_Man

This seems like an interesting question, if I could understand it. Is this written exactly the way you received it?

So far here is what I gather: Define set $\mathbb{Q}[\sqrt{-d}]=\{m+\frac n2(1+\sqrt{-d})|m,n\in\mathbb{Q}\}$ . The real and complex parts of these elements are easily gotten: $a=m+\frac n2, b=\frac n2\sqrt{d}$

If we denote $(m,n)_Q$ as points in this set, then $0=(0,0)_Q, 1=(1,0)_Q, \sqrt{-d}=(-1,2)_Q$ . The number $\sqrt{d}$ is NOT in the set, because it corresponds to the complex number $\sqrt{d}+0i$ , and $a=m+\frac n2$ cannot be an irrational number if m,n are both rational.

Are you doing this just as a response to Part 1 of my question then?

Also yes, this is the way it was worded in my text. Lastly, what does that subscript Q mean again? I have seen the notation before on here I just forgot which thread it was in that someone explained it.

**undefined** · August 4th 2010, 10:01 AM

Originally Posted by Samson

Hello All,

Here is a problem that I'm having some trouble solving.

Part 1:
"In the complex plane, there exists three cube roots of one. Let X be the cube root of one which has positive imaginary part. Knowing this, it can be shown that X is a quadratic integer in Q[Sqrt(-3)] written in the form of (a+b*Sqrt(-3))/2 where a and be are either both even or both odd and are rational. Once written in this form, it can be written in the form: m+[(n/2)(1+Sqrt(-3))].

Part 2:

On a complex plane, how can one represent the quadratic integers in Q[Sqrt(-1)], Q[Sqrt(-3)], and Q[Sqrt(-5)]? I specifically need to note the points 0, 1, and sqrt(d) and which points are "units".

Any and all help is greatly appreciated

Note: In the course of writing this post, I've come across what seems to be an inconsistency in Wikipedia's notation, and without readily accessible number theory books to compare with, I'm not sure what the resolution is. Here $\omega$ for Eisenstein integers is defined as $\omega=\dfrac{1+\sqrt{-3}}{2}$ whereas here it is $\omega=\dfrac{-1+\sqrt{-3}}{2}$ , which seems to be more standard. (See also here.) As a result, I use both versions of $\omega$ in my post, and hopefully it won't be confusing.

Part 1

The part marked in red above seems silly to me. If we're talking about a and b being odd or even, then clearly they are integers, not just rationals.

The three cube roots of 1 are given here with explanation:

Math Forum - Ask Dr. Math

Also here.

The only cube root of 1 that has a positive imaginary part is $\displaystyle -\frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)i$ . Knowing this, it should be obvious that we can put a=-1 and b=1 to match with your definition. So this complex number is in the ring of Eisenstein integers $\displaystyle \mathcal{O}_{\mathbb{Q}(\sqrt{-3})} = \mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ and is also of course in $\mathbb{Q}(\sqrt{-3})$ .

To convert to the other form mentioned in part 1, we can do $m + \frac{1+\sqrt{D}}{2}\cdot n = \cdots = \frac{(2m+n)+n\sqrt{D}}{2}$ , giving

$n = b$

$a = 2m+n \implies m = \frac{a-b}{2}$

This highlights why a and b must have the same parity (both even or both odd) in order for the conversion to be possible in integers.

In this case, we find (m,n) = (-1,1).

Part 2

The quadratic integers in $\mathbb{Q}(\sqrt{-1})$ are the Gaussian integers and correspond with lattice points such that the real and imaginary parts are both integers. This is a square lattice.

The quadratic integers in $\mathbb{Q}(\sqrt{-3})$ are the Eisenstein integers and form a hexagonal lattice in the complex plane, see here.

The quadratic integers in $\mathbb{Q}(\sqrt{-5})$ form a rectangular lattice in the complex plane. This should be easy to see if you start plotting the points.

Originally Posted by Media_Man

The number $\sqrt{d}$ is NOT in the set, because it corresponds to the complex number $\sqrt{d}+0i$ , and $a=m+\frac n2$ cannot be an irrational number if m,n are both rational.

I will use uppercase D; what you wrote applies to $\sqrt{|D|}$ since it is common to let D be the value under the radical sign, i.e., in the examples provided here D is always negative.

The complex numbers 0, 1, and $\sqrt{D}$ are not hard to plot on the complex plane.

Originally Posted by Samson

...and which points are "units".

The fact that you put the word "units" in quotes indicates that you may not know what a unit is. See here.

It is not generally true that $\sqrt{D}$ is a unit. (For example, it is not true in Eisenstein integers.)

It is well documented that the units in Gaussian integers are $\{\pm1,\pm i\}$ and in Eisenstein integers are $\{\pm1,\pm\omega,\pm\omega^2 \}$ where $\omega=\dfrac{-1+\sqrt{-3}}{2}$ . These are the quadratic integers with norm 1 in their respective rings. Along these lines, I believe the only units in $\mathbb{Z}[5i]$ are 1 and -1, but someone correct me if I'm wrong.

**Samson** · August 5th 2010, 08:30 AM

Originally Posted by undefined

Part 2

The quadratic integers in $\mathbb{Q}(\sqrt{-1})$ are the Gaussian integers and correspond with lattice points such that the real and imaginary parts are both integers. This is a square lattice.

The quadratic integers in $\mathbb{Q}(\sqrt{-3})$ are the Eisenstein integers and form a hexagonal lattice in the complex plane, see here.

The quadratic integers in $\mathbb{Q}(\sqrt{-5})$ form a rectangular lattice in the complex plane. This should be easy to see if you start plotting the points.

The complex numbers 0, 1, and $\sqrt{D}$ are not hard to plot on the complex plane.

I followed your post well right up until I got to this point. Is there any place where I could find these Quadratic integers plotted on their respective plots? I followed your link (for Eisenstein Integers) but it was for a distinct example and the description was poor.

**undefined** · August 5th 2010, 08:54 AM

Originally Posted by Samson

I followed your post well right up until I got to this point. Is there any place where I could find these Quadratic integers plotted on their respective plots? I followed your link (for Eisenstein Integers) but it was for a distinct example and the description was poor.

Hmm, plotting points on the complex plane is very basic compared with algebraic number theory... but I suppose we all have gaps in knowledge.

The link for Eisenstein integers shows the lattice points in the upper right corner of the page, should grab your attention; full size image here.

For any complex number a+bi, plot it on the complex plane as (a,b) the same way you would plot (x,y) in R^2 normally. Of course $\sqrt{-5} = 5i$ . Where is the difficulty?

Edit: It's occurred to me that "and are rational" could be referring to the "rational integers". I assume based on previous threads that this is a translation from German, and as such, is possibly clearer in the original German, whereas here it seems closer to a typo.

**Samson** · August 9th 2010, 04:08 AM

Originally Posted by undefined

Hmm, plotting points on the complex plane is very basic compared with algebraic number theory... but I suppose we all have gaps in knowledge.

The link for Eisenstein integers shows the lattice points in the upper right corner of the page, should grab your attention; full size image here.

For any complex number a+bi, plot it on the complex plane as (a,b) the same way you would plot (x,y) in R^2 normally. Of course $\sqrt{-5} = 5i$ . Where is the difficulty?

Edit: It's occurred to me that "and are rational" could be referring to the "rational integers". I assume based on previous threads that this is a translation from German, and as such, is possibly clearer in the original German, whereas here it seems closer to a typo.

Yes, when I had said "and are rational" I meant that they are rational integers. As far as that plot goes, it totally miffs me. It has 3+2w --> 3+2i. So I should go 3 units to the right, and 2 units up and put a point. Why does it look like it is 2 units over and 2 units up? What am I doing wrong?

**undefined** · September 4th 2010, 12:19 PM

Originally Posted by Samson

Yes, when I had said "and are rational" I meant that they are rational integers. As far as that plot goes, it totally miffs me. It has 3+2w --> 3+2i. So I should go 3 units to the right, and 2 units up and put a point. Why does it look like it is 2 units over and 2 units up? What am I doing wrong?

using w as omega, easier to type

3 + 2w does not equal 3 + 2i

2w = -1 + i*sqrt(3)

Thus 3 + 2w = 2 + i*sqrt(3) and it is plotted correctly in the image on Wikipedia at point (2,sqrt(3)) in the complex plane.

**Samson** · September 5th 2010, 10:58 AM

Originally Posted by undefined

using w as omega, easier to type

3 + 2w does not equal 3 + 2i

2w = -1 + i*sqrt(3)

Thus 3 + 2w = 2 + i*sqrt(3) and it is plotted correctly in the image on Wikipedia at point (2,sqrt(3)) in the complex plane.

So could you provide the coordinates to the points in Q[Sqrt(-1)], Q[Sqrt(-3)], and Q[Sqrt(-5)] for points 0, 1, Sqrt(d) ? Which ones are units again?

Feel free to give them to me in the form of a+bw so I can apply the whole plotting process, then I'll upload an image and see if I drew them correct.

**undefined** · September 5th 2010, 11:21 AM

Originally Posted by Samson

So could you provide the coordinates to the points in Q[Sqrt(-1)], Q[Sqrt(-3)], and Q[Sqrt(-5)] for points 0, 1, Sqrt(d) ? Which ones are units again?

Feel free to give them to me in the form of a+bw so I can apply the whole plotting process, then I'll upload an image and see if I drew them correct.

This might sound bad, but you do realize this is like asking how to add two fractions with unlike denominators while doing an advanced calculus problem, right?

Plus, you've already received answers. The number 0 is at (0,0) the origin. The number 1 is at (1,0). Since d is negative, the number sqrt(d) is at (0,sqrt(-d)).

Units are given at the bottom of post #4.

**Samson** · September 5th 2010, 03:20 PM

Originally Posted by undefined

This might sound bad, but you do realize this is like asking how to add two fractions with unlike denominators while doing an advanced calculus problem, right?

Plus, you've already received answers. The number 0 is at (0,0) the origin. The number 1 is at (1,0). Since d is negative, the number sqrt(d) is at (0,sqrt(-d)).

Units are given at the bottom of post #4.

I understand your frustration, but I'm very frustrated myself. Could you state which post had mentioned the actual coordinates for each fields respectively? I seem to have got myself lost within this thread.

**undefined** · September 5th 2010, 04:47 PM

Originally Posted by Samson

I understand your frustration, but I'm very frustrated myself. Could you state which post had mentioned the actual coordinates for each fields respectively? I seem to have got myself lost within this thread.

I don't mean to be too harsh and I appreciate your enthusiasm, it's just a bit odd to be asking this kind of question while doing algebraic number theory.

Post #2 (by Media_Man) has the coordinates, and I just re-iterated in post #10.

Hope you don't get too frustrated.

**Samson** · September 7th 2010, 09:30 AM

Originally Posted by undefined

I don't mean to be too harsh and I appreciate your enthusiasm, it's just a bit odd to be asking this kind of question while doing algebraic number theory.

Post #2 (by Media_Man) has the coordinates, and I just re-iterated in post #11.

Hope you don't get too frustrated.

The only thing is it seems that i'm going to be plotting (0,0), (1,0), and (0,-Sqrt[d]) for all of Q[Sqrt(-1)], Q[Sqrt(-3)], and Q[Sqrt(-5)], however in post two the third point was listed as (-1,2). How is that possible if the third point is supposed to be (0,-Sqrt[d]) ?

**undefined** · September 7th 2010, 10:28 AM

Originally Posted by Samson

The only thing is it seems that i'm going to be plotting (0,0), (1,0), and (0,-Sqrt[d]) for all of Q[Sqrt(-1)], Q[Sqrt(-3)], and Q[Sqrt(-5)], however in post two the third point was listed as (-1,2). How is that possible if the third point is supposed to be (0,-Sqrt[d]) ?

Hmm maybe I should not have referenced post #2 since it's not as simple as possble. The point (-1,2) given in post #2 does not represent a point in the complex plane, which you will see if you read it carefully. The first two points happen to coincide though. Just use what I wrote in post #10, it's simpler:

"The number 0 is at (0,0) the origin. The number 1 is at (1,0). Since d is negative, the number sqrt(d) is at (0,sqrt(-d))."

Thread: Complex Plans and Quadratic Integers

LinkBack

Thread Tools

Display

Complex Plans and Quadratic Integers

Clarity

Similar Math Help Forum Discussions

Quadratic Integers Proof

Quadratic Integers

Relatively prime quadratic integers

congruence and quadratic integers.

Quadratic integers