Notion of Congruence

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Hello PhysicsForums!

I had been reading some examples of notions of congruence and I came across one that miffed me. I was hoping that someone could help me define this notion of congruence as described below.

If [itex]\alpha[/itex] is a quadratic integer in [itex]Q[\sqrt{-d}][/itex], a notion of congruence [itex](mod \alpha)[/itex] can be defined. Additionally, the [itex]+,-,[/itex] and [itex]x[/itex] can be defined for for congruence classes.

Can someone help me define this notion of congruence as well as the [itex]+,-,[/itex] and [itex]x[/itex] for congruence classes?

Much Thanks, --Brim
 
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Answers and Replies

  • #2
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You just mod the ring of integers by the ideal generated by element \alpha.
 
  • #3
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You just mod the ring of integers by the ideal generated by element \alpha.
Could someone maybe expand/elaborate on hochs post as well as address the cases I had mentioned above? Thanks --Brim
 
  • #4
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Could someone maybe expand/elaborate on hochs post as well as address the cases I had mentioned above? Thanks --Brim
I am not sure but I think the ideal of [tex]\alpha = A + B\sqrt{-D}[/tex] is [tex]A^2 +B^2*D[/tex] which is found by multiplying [tex]\alpha[/tex] by its complement.
Therefore, the ring of integers would be {0,1,2,3 ... A^2 + B^2*D - 1}.
 
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  • #5
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I am not sure but I think the ideal of [tex]\alpha = A + B\sqrt{-D}[/tex] is [tex]A^2 +B^2*D[/tex] which is found by multiplying [tex]\alpha[/tex] by its complement.
Can anyone confirm this? And what is meant by modding the ring of integers by this [itex]\alpha[/itex] ?
 
  • #6
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Therefore, the ring of integers would be {0,1,2,3 ... A^2 + B^2*D - 1}.
If this is true, how to we modify this to define the [itex] + , -, x [/itex] notions of congruence for congruence classes? --Brim
 
  • #7
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Simply define +,-,x on the representatives (as usual), then check that it's well-defined. The equivalence relation is two algebraic integers x,y in the field are equivalent iff x - y = \alpha * k for some algebraic integer k in the field.

And no, I didn't mean take the norm of alpha and look at Z/(norm of alpha) (which is what the above posts are doing).

In particular, the number of elements in the congruence class is |Norm of alpha|.
 
  • #8
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Simply define +,-,x on the representatives (as usual), then check that it's well-defined. The equivalence relation is two algebraic integers x,y in the field are equivalent iff x - y = \alpha * k for some algebraic integer k in the field.

And no, I didn't mean take the norm of alpha and look at Z/(norm of alpha) (which is what the above posts are doing).

In particular, the number of elements in the congruence class is |Norm of alpha|.
Okay, I'm confused between what hochs and ramsey2879 are saying here. Hochs said above he didn't mean to take the norm of alpha and look at Z/(norm of alpha).

I essentially don't know what he's saying, nor what he means by "simply define +,-,x on the representatives (as usual), then check that it's well-defined " or how to use this equivalence relation.

Can someone help explain this and put this together? (if possible take it from the top???)
 
  • #9
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Okay, I'm confused between what hochs and ramsey2879 are saying here. Hochs said above he didn't mean to take the norm of alpha and look at Z/(norm of alpha).

I essentially don't know what he's saying, nor what he means by "simply define +,-,x on the representatives (as usual), then check that it's well-defined " or how to use this equivalence relation.

Can someone help explain this and put this together? (if possible take it from the top???)
Tell me, do you know some abstract algebra? Do you know what integral closure is? How much algebra and number theory do you know? I'm asking this so I can explain in accordance to your level.
 
  • #10
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Tell me, do you know some abstract algebra? Do you know what integral closure is? How much algebra and number theory do you know? I'm asking this so I can explain in accordance to your level.
I had taken abstract algebra and number theory years ago, however this is brushing up. I totally understand what you and ramsey are saying individually, but I'm not seeing how they coexist and mesh with one another. If you or someone could make one accurate explanation that flows from one piece to another, I'd really appreciate it. At the moment, I'm lost trying to combine yours and ramsey's posts together into something conclusive.
 
  • #11
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I had taken abstract algebra and number theory years ago, however this is brushing up. I totally understand what you and ramsey are saying individually, but I'm not seeing how they coexist and mesh with one another. If you or someone could make one accurate explanation that flows from one piece to another, I'd really appreciate it. At the moment, I'm lost trying to combine yours and ramsey's posts together into something conclusive.
I personally feel that what Ramsey is saying is irrelevant to your question, since you're asking how to define (mod alpha) for the ring of integers, O_K with K = your quadratic field.

For analogy How would you define (mod n) for the integers? You mod the ring Z by the ideal generated by n, nZ, to get Z/nZ the congruence group mod n.

By analogy if you have \alpha in the ring of integers O_K, you can form O_K / (\alpha), which is the ring of integers in K mod the ideal generated by \alpha. This is a perfectly good ring and is what people usually mean by (mod \alpha) in ring of integers.

In some literature you may find people saying x mod p, where p is a rational prime. What they mean is that K is galois over Q, so that every residue field of any prime lying above p in K is the same, and you look at the ring of integers in K modulo any prime ideal lying above p.
 
  • #12
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I personally feel that what Ramsey is saying is irrelevant to your question, since you're asking how to define (mod alpha) for the ring of integers, O_K with K = your quadratic field.
That is what I was saying, I couldn't piece the two of yours together.
You mod the ring Z by the ideal generated by n, nZ, to get Z/nZ the congruence group mod n.
Could you explain this in a little more detail?

By analogy if you have [itex]\alpha[/itex] in the ring of integers [itex]O_{K}[/itex], you can form [itex]O_{K} / (\alpha)[/itex], which is the ring of integers in K mod the ideal generated by [itex]\alpha[/itex]. This is a perfectly good ring and is what people usually mean by [itex] (mod \alpha) [/itex] in ring of integers.
So summing it up, what will the notion of congruence [itex] mod \alpha [/itex] be defined as?
 
  • #13
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Could you explain this in a little more detail?
Um, Z is a ring, n is an element of Z, nZ = {nx : x is integer} is an ideal of Z, so you can form a factor ring Z/nZ, which has n elements. This is just the ring of integers modulo n. You should already be very familiar with this from your response.

So summing it up, what will the notion of congruence [itex] mod \alpha [/itex] be defined as?
I already explained! If O_K is the ring of integers in K, then mod \alpha is just O_K / (alpha * O_K)!
 
  • #14
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I already explained! If O_K is the ring of integers in K, then mod \alpha is just O_K / (alpha * O_K)!
So how does one define the [itex] +,-,[/tex] and [itex]x[/itex] congruence classes? Are the [itex]+,-[/itex] just modifying your statement above by [itex]+\alpha[/itex] and [itex]-\alpha[/itex]?
 
  • #15
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So how does one define the [itex] +,-,[/tex] and [itex]x[/itex] congruence classes? Are the [itex]+,-[/itex] just modifying your statement above by [itex]+\alpha[/itex] and [itex]-\alpha[/itex]?
The same way you define +,-, x in R/I for any ring R and ideal I. You said you already know abstract algebra, and this is a very standard construction.
 
  • #16
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The same way you define +,-, x in R/I for any ring R and ideal I. You said you already know abstract algebra, and this is a very standard construction.
I said I took a course in it years ago and I could use some refreshment. Could you please expand on this? (In terms of time utility, it would be more efficient for this example to be finished than for me to dig up my textbooks, read 4 chapters, and flood this forum with questions as opposed to finishing this example here and now).
 
  • #17
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I said I took a course in it years ago and I could use some refreshment. Could you please expand on this? (In terms of time utility, it would be more efficient for this example to be finished than for me to dig up my textbooks, read 4 chapters, and flood this forum with questions as opposed to finishing this example here and now).
I had already given you all the data you need to define this on your own. I don't know why you expect people to spoon-feed trivial verifications, but I'll play along:

define an equivalence class on the integers O_K, by x ~ y iff x - y = \alpha * k for some k in O_K. Check that this is an equivalence relation.

Denote [x] by the class of elements z in O_K that is equivalent to x (so it's a set).

define + and [x] + [y] = [x+y]. You need to check that this is well-defined, i.e. if x ~ x' and y ~ y' then x+y ~ x' + y'.

define - by [x] - [y] = [x-y]. Same Story.

define * by [x]*[y] = [xy]. Same story.

And no, you're going to have to check these yourself.
 
  • #18
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The only thing that I don't understand is how you define y. Is y an additional set in O_K akin to x ?
 
  • #19
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The only thing that I don't understand is how you define y. Is y an additional set in O_K akin to x ?
?? y is just any other element of O_K, then [y] is another class. [x] is a class and [y] is a class. Now define [x] + [y] by [x+y]. You need to check that this is well-defined. I don't know what/why you're asking this. There's nothing to define for y. That's just an element of O_K.
 
  • #20
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Denote [x] by the class of elements z in O_K that is equivalent to x (so it's a set).
This is what confused me. I think you meant to say "that is equivalent to [x] (so it's a set)".

Is this correct?
 
  • #21
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This is what confused me. I think you meant to say "that is equivalent to [x] (so it's a set)".

Is this correct?
[x] = {z in O_K : z ~ x}
 
  • #22
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Part of the misunderstanding might be that hochs's original statement, while correct, recommended modding the "ring of integers" by [tex] \alpha [/tex], which Brimley seems to have interpreted to mean [tex] \mathbb{Z} / (\alpha) [/tex]--which is not well-defined, and which could, therefore, have understandably caused confusion. To clarify, [tex] O_k [/tex] denotes the ring of quadratic integers in the number field [tex] \mathbb{Q}(\sqrt{d}) [/tex], and the ideal [tex] (\alpha) [/tex] consists of all multiples of [tex] \alpha [/tex] in [tex] O_k [/tex]. The quotient [tex] O_k / (\alpha) [/tex] then consists, as hochs said, of all equivalence classes modulo the relation [tex] x \equiv y \iff x - y \in (\alpha) [/tex]. In other words, two quadratic integers [tex] x [/tex] and [tex] y [/tex] are equivalent "mod [tex] \alpha [/tex]" if [tex] x - y [/tex] is a multiple of [tex] \alpha [/tex].

An interesting follow-up question to think about is the question of which ideals in [tex] O_k [/tex] are maximal. If you don't remember what this means, just think about primes in [tex] \mathbb{Z} [/tex]: The integers mod [tex] p [/tex] form a field whenever [tex] p [/tex] is prime (i.e., every nonzero residue class mod [tex] p [/tex] is invertible). Is this always true for "primes" in [tex] O_k [/tex]? What does "prime" even mean there? In the integers, reducing "mod [tex] m [/tex] and mod [tex] n [/tex]" is always equivalent to reducing mod the gcd of [tex] m [/tex] and [tex] n [/tex]. Can a similar construction always work in quadratic integer rings? In particular, when does a meaningful division algorithm exist in [tex] O_k [/tex]?
 
  • #23
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Part of the misunderstanding might be that hochs's original statement, while correct, recommended modding the "ring of integers" by [tex] \alpha [/tex], which Brimley seems to have interpreted to mean [tex] \mathbb{Z} / (\alpha) [/tex]--which is not well-defined, and which could, therefore, have understandably caused confusion. To clarify, [tex] O_k [/tex] denotes the ring of quadratic integers in the number field [tex] \mathbb{Q}(\sqrt{d}) [/tex], and the ideal [tex] (\alpha) [/tex] consists of all multiples of [tex] \alpha [/tex] in [tex] O_k [/tex]. The quotient [tex] O_k / (\alpha) [/tex] then consists, as hochs said, of all equivalence classes modulo the relation [tex] x \equiv y \iff x - y \in (\alpha) [/tex]. In other words, two quadratic integers [tex] x [/tex] and [tex] y [/tex] are equivalent "mod [tex] \alpha [/tex]" if [tex] x - y [/tex] is a multiple of [tex] \alpha [/tex].
Yes, thank you.

An interesting follow-up question to think about is the question of which ideals in [tex] O_k [/tex] are maximal. If you don't remember what this means, just think about primes in [tex] \mathbb{Z} [/tex]: The integers mod [tex] p [/tex] form a field whenever [tex] p [/tex] is prime (i.e., every nonzero residue class mod [tex] p [/tex] is invertible). Is this always true for "primes" in [tex] O_k [/tex]? What does "prime" even mean there? In the integers, reducing "mod [tex] m [/tex] and mod [tex] n [/tex]" is always equivalent to reducing mod the gcd of [tex] m [/tex] and [tex] n [/tex]. Can a similar construction always work in quadratic integer rings? In particular, when does a meaningful division algorithm exist in [tex] O_k [/tex]?
I'm guessing this is intended for Brimley. The last question regarding division algorithm is, of course, a difficult one.
 
  • #24
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I had taken abstract algebra and number theory years ago, however this is brushing up. I totally understand what you and ramsey are saying individually, but I'm not seeing how they coexist and mesh with one another. If you or someone could make one accurate explanation that flows from one piece to another, I'd really appreciate it. At the moment, I'm lost trying to combine yours and ramsey's posts together into something conclusive.
Please ignore what I said. Back to the drawing board! Sorry I confused the norm with the Ideal, but I am 64 years old and never had math beyond differential and intergral calculas for engineers and I don't remember that much.
 
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