Congruence Class Relationships in \mathbb{Q}[\sqrt{3}]

[Brimley]

Hello PhysicsForums!

I had been reading up on a congruence classes when I came across an example that miffed me. A lot of information was given but I can't seem to make sense of it: Here it is:

Assume \lambda = (3+\sqrt{-3})/2 \in \mathbb{Q}[\sqrt{3}]. We know that \lambda is prime and that there are three congruence classes (mod \lambda), one with 0, one with 1, and the last with -1.

The following three relationships exist:
A. If x \equiv 1 (mod \lambda), then x^3 \equiv 1 (mod \lambda)^3.
B. If x \equiv -1 (mod \lambda), then x^3 \equiv -1 (mod \lambda)^3.
C. If x \equiv 0 (mod \lambda), then x^3 \equiv 0 (mod \lambda)^3.

Can someone help explain to me why these three relationships exist? I don't know if factoring x^3 - 1 in \mathbb{Q}[\sqrt{d}] completely into linear factors can help, but its an idea.

 

[hochs]

A. follows from x^3 - 1 = (x-1)(x - w)(x - w^2) where w is a primitive 3rd root of unity, i.e. 1 - w = (3 + sqrt(-3))/2 = lambda.

B. similar.
C. trivial.

Yes, factoring x^3 - 1 completely helps.

 

[Brimley]

A. follows from x^3 - 1 = (x-1)(x - w)(x - w^2) where w is a primitive 3rd root of unity, i.e. 1 - w = (3 + sqrt(-3))/2 = lambda.

B. similar.
C. trivial.

Yes, factoring x^3 - 1 completely helps.

Okay, so let me make sure I'm getting this right:

A. x^3 - 1 = (x - 1)(x - w)(x - w^2) where w is a primitive third root of unit, or w = (-1-\sqrt{-3})/2. This can be written as also as 1 - w = (3+\sqrt{-3})/2 = 1 - \lambda, however I don't see the significance of writing it in this form.

Does the following above truly prove why that relationship exists? I want to be sure I understand A before I try to explain my interpretation of B and C.

 

[hochs]

You mean 1 - w = (3 + sqrt(-3))/2 = lambda.

If x = 1 (mod lambda), then x = 1 + a(1 - w) + b(1 - w)^2 for some a,b integers (the ring of integers in this cyclotomic field is Z[1 - w]). Note that all of x-1, x-w, and x-w^2 are now divisible by 1-w = lambda.

 

[Brimley]

You mean 1 - w = (3 + sqrt(-3))/2 = lambda.

If x = 1 (mod lambda), then x = 1 + a(1 - w) + b(1 - w)^2 for some a,b integers (the ring of integers in this cyclotomic field is Z[1 - w]). Note that all of x-1, x-w, and x-w^2 are now divisible by 1-w = lambda.

I didn't know why you intentionally wrote it in the form of (1 - w) as opposed to w, however I see why now. I understand why this makes the relation on the left side true, but I don't see how it explains the right side, x^3 \equiv 1 (mod \lambda)^3. Could you explain this?

 

[hochs]

I didn't know why you intentionally wrote it in the form of (1 - w) as opposed to w, however I see why now. I understand why this makes the relation on the left side true, but I don't see how it explains the right side, x^3 \equiv 1 (mod \lambda)^3. Could you explain this?

Same story, except with x^3 + 1 = (x+1)(x + w)(x + w^2) now.

 

[Brimley]

Same story, except with x^3 + 1 = (x+1)(x + w)(x + w^2) now.

I'm just trying to make the connection between the translation you made from the right side of the first equation and the second (shown below again):

RHS1: 1 + a(1 - w) + b(1 - w)^2.
RHS2: (x + 1)(x + w)(x + w^2).

The differences in form and the omission of a and b are confusing me.

 

[hochs]

I'm just trying to make the connection between the translation you made from the right side of the first equation and the second (shown below again):

RHS1: 1 + a(1 - w) + b(1 - w)^2.
RHS2: (x + 1)(x + w)(x + w^2).

The differences in form and the omission of a and b are confusing me.

... just do the exact same thing with x = -1 + a*(1 - w) + b*(1 - w)^2.

 

[Brimley]

... just do the exact same thing with x = -1 + a*(1 - w) + b*(1 - w)^2.

I'm totally confused as to what you mean by this (as I was trying to explain in my previous post) simply because I'm not making the relation to what you wrote afterwards for x^3 \equiv 1 (mod \lambda)^3.

 

[hochs]

I'm totally confused as to what you mean by this (as I was trying to explain in my previous post) simply because I'm not making the relation to what you wrote afterwards for x^3 \equiv 1 (mod \lambda)^3.

For the first one, if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2. Then x-1, x-w, x-w^2 are all divisible by lambda, so clearly x^3 - 1 = (x-1)(x-w)(x-w^2) is divisible by lambda^3.

Do the same for part B.

Part C is trivial.

 

[Brimley]

For the first one, if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2. Then x-1, x-w, x-w^2 are all divisible by lambda, so clearly x^3 - 1 = (x-1)(x-w)(x-w^2) is divisible by lambda^3.

Do the same for part B.

Part C is trivial.

How do you reach the conclusion I have bolded above? I'm trying to manipulate this algebraically but I don't see it.

 

[hochs]

How do you reach the conclusion I have bolded above? I'm trying to manipulate this algebraically but I don't see it.

if x = 1 + a*(1 - w) + b*(1 - w)^2 then x - 1 = a*(1 - w) + b*(1 - w)^2 = (1-w)*(a + b*(1 - w)) is clearly divisible by 1-w = lambda. It's exactly the same trivial algebraic verifications for the others.

 

[Brimley]

if x = 1 + a*(1 - w) + b*(1 - w)^2 then x - 1 = a*(1 - w) + b*(1 - w)^2 = (1-w)*(a + b*(1 - w)) is clearly divisible by 1-w = lambda. It's exactly the same trivial algebraic verifications for the others.

Okay, I see what you did there, but how did you get from:

(1 - w)*(a + b*(1-w))

to

(x-1)(x-w)(x-w^2)?

 

[hochs]

Okay, I see what you did there, but how did you get from:

(1 - w)*(a + b*(1-w))

to

(x-1)(x-w)(x-w^2)?

I didn't. Just show that x-1, x-w, and x-w^2 are each divisible by 1-w. I can't give more hints than this. I feel like I'm doing your homework 100%.

 

[Brimley]

I didn't. Just show that x-1, x-w, and x-w^2 are each divisible by 1-w. I can't give more hints than this. I feel like I'm doing your homework 100%.

This isn't homework at all. I'm just trying to understand that. I guess I'm just confused in your translation from using a,b,w to using solely x and w.

 

[hochs]

This isn't homework at all. I'm just trying to understand that. I guess I'm just confused in your translation from using a,b,w to using solely x and w.

You just need to show that each of x-1, x-w, and x-w^2 are divisible by lambda to show that x^3 - 1 = (x-1)(x-w)(x-w^2) is divisible by lambda^3.

The above sentence should've been super clear. If not, think more.

So it suffices to show that each of x-1, x-w, and x-w^2 are divisible by lambda. This is what you need to show.

 

[Brimley]

You just need to show that each of x-1, x-w, and x-w^2 are divisible by lambda to show that x^3 - 1 = (x-1)(x-w)(x-w^2) is divisible by lambda^3.

The above sentence should've been super clear. If not, think more.

So it suffices to show that each of x-1, x-w, and x-w^2 are divisible by lambda. This is what you need to show.

So I should have the following then (for a truly complete explanation):

A. x \equiv 1 (mod \lambda), then x^3 \equiv 1(mod \lambda)^3.

If x = 1 (mod \lambda), then x = 1 + a*(1 - w) + b*(1 - w)^2.
x - 1 = a*(1 - w) + b*(1 - w)^2
= (1 - w)*(a + b*(1 - w)). Clearly, this is divisible by 1 - w = \lambda. From here, we can see that:

x^3 - 1 = (x - 1)(x - w)(x - w^2) is also divisible by 1-w = \lambda, and hence also divisible by \lambda. Since this is divisible by \lambda, this means it is also divisble by \lambda^3 (This last line seems weak).

Afterwards I'll just explain what I think the mods for B and C should be.

 

[hochs]

So I should have the following then (for a truly complete explanation):

A. x \equiv 1 (mod \lambda), then x^3 \equiv 1(mod \lambda)^3.

If x = 1 (mod \lambda), then x = 1 + a*(1 - w) + b*(1 - w)^2.
x - 1 = a*(1 - w) + b*(1 - w)^2
= (1 - w)*(a + b*(1 - w)). Clearly, this is divisible by 1 - w = \lambda. From here, we can see that:

x^3 - 1 = (x - 1)(x - w)(x - w^2) is also divisible by 1-w = \lambda, and hence also divisible by \lambda. Since this is divisible by \lambda, this means it is also divisble by \lambda^3 (This last line seems weak).

Afterwards I'll just explain what I think the mods for B and C should be.

Just wow... are you a native English speaker?

I told you, you need to show separately that x-1, x-w, and x-w^2 are all divisible by lambda. That's 3 statements to show.

my god..

 

[hochs]

Alright I'm wasting way too much time on this triviality. Here's the complete ****ing proof, leave me alone now:

if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2 for some integers a,b.

Then x-1 = a*(1 - w) + b*(1 - w)^2 is divisible by 1-w clearly.

x-w = 1 - w + a*(1 - w) + b*(1 - w)^2 = (1-w)(1 + a + b*(1 - w)) is divisible by 1-w clearly.

Finally,
x-w^2 = 1 - w^2 + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w) + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w + a + b(1 - w)) is divisible by 1-w.

So all of x-1, x-w, and x-w^2 are divisible by lambda. Do you not see that (x - 1)(x - w)(x - w^2) is now divisible by lambda^3 ?

 

[Brimley]

Just wow... are you a native English speaker?

I told you, you need to show separately that x-1, x-w, and x-w^2 are all divisible by lambda. That's 3 statements to show.

my god..

I'm sorry about your frustration hochs! I'm really trying here! (Btw, it takes a ton of time to write it in LaTeX, but I think its easier on the eyes of those reading it herein). I'm putting in the time and effort to understand this, so I do appreciate the patience!

If you remember, I told you in several posts that I had trouble understanding your translation from using a,b, and w to just x and w. I'm strill trying to make that connection, then I could better attempt explaining why \lambda divides x-1,x-w, and x-w^2.

 

[Brimley]

if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2 for some integers a,b.

Then x-1 = a*(1 - w) + b*(1 - w)^2 is divisible by 1-w clearly.

x-w = 1 - w + a*(1 - w) + b*(1 - w)^2 = (1-w)(1 + a + b*(1 - w)) is divisible by 1-w clearly.

Finally,
x-w^2 = 1 - w^2 + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w) + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w + a + b(1 - w)) is divisible by 1-w.

So all of x-1, x-w, and x-w^2 are divisible by lambda. Do you not see that (x - 1)(x - w)(x - w^2) is now divisible by lambda^3 ?

I do see it now. I really appreciate it, but please don't get mad at me. I'm just trying to understand what you were doing, and I had most of it but was confused in the translation. Seeing that performed makes sense of it. I do appreciate it and I understand your frustration, but please don't get mad at someone who is just trying to learn mathematics.

I take it to alter this for part B, all I would need to do is flip the + and - signs for part A's?
Is there a quick explanation on why part C's is trivial?

 

[stoolie77]

I take it to alter this for part B, all I would need to do is flip the + and - signs for part A's?

interesting thread. I've tried my best to follow along and i believe that this is all you need to do. hochs should confirm it first though as he's beek working with you this whole time.

 

[Brimley]

interesting thread. I've tried my best to follow along and i believe that this is all you need to do. hochs should confirm it first though as he's beek working with you this whole time.

Can anyone help me finish this? Stoolie I appreciate your input, but yes, I'd really appreciate if someone could confirm this or give me a hand finishing this.