We have a nonzero (onto) map from Z[i] to F_q, a finite field. This map carries the subring generated by 1 in Z[i] to the subring generated by 1 in F_q. In F_q, this subring is the subfield Z/p (characteristic p). So p is in the Kernel of the map.
This video took a while to shoot, but it is one of my favorites. This leads to quadratic reciprocity and characters, and all kinds of interesting number theory. It's definitely convinced me that I need to learn more about lattices in C.
@@MathDoctorBob ;) The real question is, have you learned how the fraud market works, how reserve currency works, and how lunatics have hijacked humanity and are destroying it? That is also math. (1.01)^t -> oo, (0.99)^t -> 0. Feedback loops, dynamical systems, AI, probability theory, etc... have you applied it to the world outside of formal mathematics yet?
Well, I am not sure if you can do this with any other number and its conjugate, you can take 5 and factor it into (2+i)(2-i), you can also do something else (if it is Z[i]/5) like take a Gaussian Integer such that is like (a+2ai) [or (2a+ai)] and multiply it by its conjugate you get something which will be 0 Modulo 5... you can also do this with (a+3ai)... but since I guess this isn't really a finite field, it is kinda hard to say that it is a flexible prime D:
I was thinking about how to apply the points at 15:50 to determine the primes of Z that stay prime in Z[w] (aka eisenstein integers). They're p=3k+2 (3 divides p-2). I understand that the divisibility by 3 arise from the fact that w^3=1 and therefore a homomorphism can only map phi(w) to a cyclic field with an element of order 3. However, 3|(p-2) implies that (Z[w]/p)^* will have p-2 elements. I can't see how a field of that size arises.
I realized where I was going wrong and can use your method to determine which primes in the integers remain prime in the eisenstein integers. They're the primes such that p^2-1 is divisible by 3. No need to answer my question posted earlier.
I keep learning from this video. Here comes another question on what you are discussing at 16:25 to 17:40. Is the following true? The fact that we know that Z[i]/p has order p when p is not maximal in Z[i] and Z[i]/p will have order p^2 when p is maximal (aka prime in in Z[i] since we are in an ED) in Z[i] is exactly why we have to extend the field Z/p of order p to the field Z/p/(x^2+1) of order p^2.
at 5:45 you state that p is in the Ker of phi= (a+bi). It's one of the pieces I'm missing to understanding the entire video. Do you have time to elaborate?
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I don´tunderstand much, but there is something very good and honest going on. Like a good day´s work.
We have a nonzero (onto) map from Z[i] to F_q, a finite field. This map carries the subring generated by 1 in Z[i] to the subring generated by 1 in F_q. In F_q, this subring is the subfield Z/p (characteristic p). So p is in the Kernel of the map.
This video took a while to shoot, but it is one of my favorites. This leads to quadratic reciprocity and characters, and all kinds of interesting number theory. It's definitely convinced me that I need to learn more about lattices in C.
So did you finally learn it?
@@MDNQ-ud1ty Yes.
@@MathDoctorBob ;) The real question is, have you learned how the fraud market works, how reserve currency works, and how lunatics have hijacked humanity and are destroying it? That is also math. (1.01)^t -> oo, (0.99)^t -> 0. Feedback loops, dynamical systems, AI, probability theory, etc... have you applied it to the world outside of formal mathematics yet?
Well, I am not sure if you can do this with any other number and its conjugate, you can take 5 and factor it into (2+i)(2-i), you can also do something else (if it is Z[i]/5) like take a Gaussian Integer such that is like (a+2ai) [or (2a+ai)] and multiply it by its conjugate you get something which will be 0 Modulo 5... you can also do this with (a+3ai)... but since I guess this isn't really a finite field, it is kinda hard to say that it is a flexible prime D:
i find that the gauss integer is a very interesting object.
I was thinking about how to apply the points at 15:50 to determine the primes of Z that stay prime in Z[w] (aka eisenstein integers). They're p=3k+2 (3 divides p-2). I understand that the divisibility by 3 arise from the fact that w^3=1 and therefore a homomorphism can only map phi(w) to a cyclic field with an element of order 3. However, 3|(p-2) implies that (Z[w]/p)^* will have p-2 elements. I can't see how a field of that size arises.
I realized where I was going wrong and can use your method to determine which primes in the integers remain prime in the eisenstein integers. They're the primes such that p^2-1 is divisible by 3. No need to answer my question posted earlier.
Beautiful. Thank you.
I keep learning from this video. Here comes another question on what you are discussing at 16:25 to 17:40. Is the following true?
The fact that we know that Z[i]/p has order p when p is not maximal in Z[i] and Z[i]/p will have order p^2 when p is maximal (aka prime in in Z[i] since we are in an ED) in Z[i] is exactly why we have to extend the field Z/p of order p to the field Z/p/(x^2+1) of order p^2.
Genial!
What defines a flexible prime?
at 5:45 you state that p is in the Ker of phi= (a+bi). It's one of the pieces I'm missing to understanding the entire video. Do you have time to elaborate?
Only thing I would add - check out Conway's book on Octonions and Quaternions. The first chapter has interesting properties of this lattice.
I just wanted to point out that 5 is a flexible prime... I am not sure if any more flexible primes exist.