Re: [math-fun] Simple finite group problem
Dan Asimov <dasimov@earthlink.net> wrote:
P.S. Now I'd like to know the same thing for the Gaussian integers Z[i] and the Eisenstein integers Z[w] where w = exp(2pi*i/3). These apparently have some complicated factor rings.
I'm not sure what you mean by modular arithmetic on Gaussian integers. * Modular in both real and imaginary, meaning that the complex plane is tessellated with rectangles of some integer side lengths, and the rectangles are all identified with each other? Equivalently, there's just one rectangle, and its opposite edges are connected, making a torus? * Modular just in the reals, with the imaginaries unbounded? * Modular just in the imaginaries, with the reals unbounded? I think modular in both real and imaginary sounds most interesting. Especially since if you think of it as a tessellation of rectangles, you can generalize to offset rectangles, like bricks in a wall. And if you think of it as a torus, you can generalize to a Klein bottle. (I wonder what the Mandelbrot set looks like in the modular complex plane.) Long ago I noticed that modulo a number that's one more than a square, negative one has a real square root. For instance mod 10, i = 3 and -i = 7. (Or vice versa if you prefer.) All Gaussian integers have consistent real integer values mod 10. For instance 2i is 6, 2i-5 is 1, etc. But this doesn't seem to get you anything new and interesting. I played around a lot with Gaussian integers. I calculated and plotted, by hand, the first hundred or so Gaussian primes. I also calculated and plotted Gaussian squares, and noticed that they formed interlocking parabolas. I never played with Eisenstein integers. Would you use a triangular tessellation for them when going modular? Are those the only two interesting ways of generalizing integers to the complex plane? What about the vertices of a regular hexagonal tessellation?
On 14/05/2016 04:55, Keith F. Lynch wrote:
I'm not sure what you mean by modular arithmetic on Gaussian integers.
a = b mod m iff (a-b) = mk where k is an integer. In this case you'd interpret "integer" as "Gaussian integer".
* Modular in both real and imaginary, meaning that the complex plane is tessellated with rectangles of some integer side lengths, and the rectangles are all identified with each other? Equivalently, there's just one rectangle, and its opposite edges are connected, making a torus?
That's what you get if you work mod m in the Gaussian integers and m is a real integer. The rectangles are always squares. If, e.g., m = 2+3i then the squares will no longer be axis-aligned; one of the squares will have its corners at 0, 2+3i, -3+2i, -1+5i.
I think modular in both real and imaginary sounds most interesting. Especially since if you think of it as a tessellation of rectangles, you can generalize to offset rectangles, like bricks in a wall. And if you think of it as a torus, you can generalize to a Klein bottle.
I think that if you do either of those things you're no longer doing modular arithmetic. (Though you might be doing some other interesting thing.)
Long ago I noticed that modulo a number that's one more than a square, negative one has a real square root.
The precise condition is that all the odd prime factors of the modulus should be 1 mod 4, and that there should be at most one factor of 2 in the modulus. (If p|m and m=n^2+1 then -1 is a square mod p and therefore p equals 2 or is 1 mod 4; and you never have 4 | square+1.)
I never played with Eisenstein integers. Would you use a triangular tessellation for them when going modular?
It might be better to think of the fundamental domain as a rhombus (two of your triangles glued together).
Are those the only two interesting ways of generalizing integers to the complex plane?
No. Let D be any square-free positive integer > 1 and consider numbers of the form p+q.sqrt(d) where p,q are rational; this forms a field; the algebraic integers in the field are either the numbers of the form p+q.sqrt(d) where p,q are integers, or the numbers of the form p+q(1+sqrt(d))/2 where p,q are integers, and these are by any reasonable criteria an "interesting way of generalizing integers to the complex plane". The Gaussian and Eisenstein integers are the nicest, though.
What about the vertices of a regular hexagonal tessellation?
I doubt there's any way to do arithmetic on those. Certainly not the "obvious" way -- they don't form a lattice. -- g
So another way to phrase what Gareth is saying is that modular arithmetic is the ring of Gaussian integers modulo an ideal. Right? If the (an) element of Z[i] generating the ideal is a + bi, a, b in Z, then the ideal will have a^2 + b^2 elements. Question: Is there a simple way to say which finite ring Z[i] / (a + bi) is isomorphic to? Like maybe it's always (isomorphic to) a product of rings of form Z/(n) ? I believe that if a^ + b^2 is a prime p in Z, then the quotient is the field F_p. I don't think the same thing is automatically true for prime powers. —Dan P.S. Is there a classification of finite commutative rings with multiplicative identity?
On May 14, 2016, at 10:05 AM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 14/05/2016 04:55, Keith F. Lynch wrote:
I'm not sure what you mean by modular arithmetic on Gaussian integers.
a = b mod m iff (a-b) = mk where k is an integer. In this case you'd interpret "integer" as "Gaussian integer".
* Modular in both real and imaginary, meaning that the complex plane is tessellated with rectangles of some integer side lengths, and the rectangles are all identified with each other? Equivalently, there's just one rectangle, and its opposite edges are connected, making a torus?
That's what you get if you work mod m in the Gaussian integers and m is a real integer. The rectangles are always squares. If, e.g., m = 2+3i then the squares will no longer be axis-aligned; one of the squares will have its corners at 0, 2+3i, -3+2i, -1+5i.
I think modular in both real and imaginary sounds most interesting. Especially since if you think of it as a tessellation of rectangles, you can generalize to offset rectangles, like bricks in a wall. And if you think of it as a torus, you can generalize to a Klein bottle.
I think that if you do either of those things you're no longer doing modular arithmetic. (Though you might be doing some other interesting thing.)
Long ago I noticed that modulo a number that's one more than a square, negative one has a real square root.
The precise condition is that all the odd prime factors of the modulus should be 1 mod 4, and that there should be at most one factor of 2 in the modulus. (If p|m and m=n^2+1 then -1 is a square mod p and therefore p equals 2 or is 1 mod 4; and you never have 4 | square+1.)
I never played with Eisenstein integers. Would you use a triangular tessellation for them when going modular?
It might be better to think of the fundamental domain as a rhombus (two of your triangles glued together).
Are those the only two interesting ways of generalizing integers to the complex plane?
No. Let D be any square-free positive integer > 1 and consider numbers of the form p+q.sqrt(d) where p,q are rational; this forms a field; the algebraic integers in the field are either the numbers of the form p+q.sqrt(d) where p,q are integers, or the numbers of the form p+q(1+sqrt(d))/2 where p,q are integers, and these are by any reasonable criteria an "interesting way of generalizing integers to the complex plane".
The Gaussian and Eisenstein integers are the nicest, though.
What about the vertices of a regular hexagonal tessellation?
I doubt there's any way to do arithmetic on those. Certainly not the "obvious" way -- they don't form a lattice.
-- g
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On 14/05/2016 19:11, Dan Asimov wrote:
So another way to phrase what Gareth is saying is that modular arithmetic is the ring of Gaussian integers modulo an ideal. Right?
Right. Just as "ordinary" modular arithmetic means working in the ring of integers modulo an ideal.
If the (an) element of Z[i] generating the ideal is a + bi, a, b in Z, then the ideal will have a^2 + b^2 elements.
Question: Is there a simple way to say which finite ring Z[i] / (a + bi) is isomorphic to? Like maybe it's always (isomorphic to) a product of rings of form Z/(n) ?
I don't know, but that is no evidence that others don't because to an excellent first approximation I don't know anything. I would guess the answer to your second question is no. There's some analysis of this here: http://home.wlu.edu/~dresdeng/papers/factorrings.pdf which appears to give a complete answer to your question. (And indeed you don't always get a product of Z/(n).)
P.S. Is there a classification of finite commutative rings with multiplicative identity?
There's some discussion of this here: http://mathoverflow.net/questions/7133 (it seems the answer is that no classification is known). -- g
Thanks for that reference, Gareth. Boy, things are a lot more complicated with factor rings of Z[i] than I would have guessed! But also very interesting. Now I wonder what's known in general about factor rings of Z[w_n] where w_n = exp(2pi*i/n) . That paper gives a lot of hints for when w_n is quadratic, i.e., its minimal polynomial is degree 2. —Dan
On May 14, 2016, at 2:55 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 14/05/2016 19:11, Dan Asimov wrote:
So another way to phrase what Gareth is saying is that modular arithmetic is the ring of Gaussian integers modulo an ideal. Right?
Right. Just as "ordinary" modular arithmetic means working in the ring of integers modulo an ideal.
If the (an) element of Z[i] generating the ideal is a + bi, a, b in Z, then the ideal will have a^2 + b^2 elements.
Question: Is there a simple way to say which finite ring Z[i] / (a + bi) is isomorphic to? Like maybe it's always (isomorphic to) a product of rings of form Z/(n) ?
I don't know, but that is no evidence that others don't because to an excellent first approximation I don't know anything. I would guess the answer to your second question is no.
There's some analysis of this here: http://home.wlu.edu/~dresdeng/papers/factorrings.pdf which appears to give a complete answer to your question. (And indeed you don't always get a product of Z/(n).)
P.S. Is there a classification of finite commutative rings with multiplicative identity?
There's some discussion of this here: http://mathoverflow.net/questions/7133 (it seems the answer is that no classification is known).
-- g
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* Keith F. Lynch <kfl@KeithLynch.net> [May 14. 2016 07:37]:
Dan Asimov <dasimov@earthlink.net> wrote:
P.S. Now I'd like to know the same thing for the Gaussian integers Z[i] and the Eisenstein integers Z[w] where w = exp(2pi*i/3). These apparently have some complicated factor rings.
I'm not sure what you mean by modular arithmetic on Gaussian integers.
[...]
Purely computationally, I find the following the most natural approach as it is the same for finite fields. Consider complex numbers as polynomials modulo the polynomial x^2 + 1 (Gaussian) or x^2 + x + 1 (Eisenstein). For Gaussian integers think i=x (a + i*b |--> a + b*x), for Eisenstein integers think w=x (a + w*b |--> a + b*x). An attempt to write up this idea is Section 39.12 "Complex modulus: the field GF(p^2)" of my (fxt)book. What is not in the fxtbook: For computations mod a given Gaussian integer A + i*B use the minimal polynomial P = x^2 - 2*A*x + (A^2 + B^2) as the modulus (is this correct?). For a given Eisenstein integer, OOPS, someone please fill me in (long ago that I looked at this, plus I claim lack of coffee, there also was a traffic jam, and an earthquake, somewhere)! Best regards, jj
participants (5)
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Dan Asimov -
Dan Asimov -
Gareth McCaughan -
Joerg Arndt -
Keith F. Lynch