Message from @YourFundamentalTheorum

Discord ID: 503831587595616266


2018-09-23 19:12:00 UTC  

*nice

2018-09-24 01:38:13 UTC  
2018-09-24 02:36:03 UTC  

roll'd?

2018-09-24 03:21:31 UTC  

@Nicholas1166 - NY sorry, role'd not roll'd. I gave you the @`AE` role (academic expert) so you will be alerted when people ping @`AE` with academic questions you may be able to help with.

2018-09-24 03:37:27 UTC  

Ah, OK. Yeah, I'd be glad to lend a hand if I can, although I am just a junior right now

2018-09-24 04:23:03 UTC  

I'm graduated from a university with a degree in math

2018-09-24 04:23:05 UTC  

ask me math questions

2018-09-24 04:23:09 UTC  

if you need halp

2018-09-24 04:31:53 UTC  

I very well might, math is my weakest subject. Thank you for the offer.

2018-10-03 05:19:28 UTC  

I didn't know this sever had a role thing. Can I get a role and be alerted?

2018-10-03 16:45:13 UTC  
2018-10-22 06:55:35 UTC  

@ThisIsChris hey do you have any good explanation on how to find isomorphisms between two polynomial factor groups?

2018-10-22 06:56:33 UTC  

that is an isomorphism phi: F[x]/f1(x) -> F[x]/f2(x) where F[x] is the field of polynomials with coefficients in Z_q and f1(x) and f2(x) are irreducible polynomials in F[x]

2018-10-22 07:18:25 UTC  

@YourFundamentalTheorum by F[x]/f1(x) do you mean a quotient group? If so, what is the group you are doing F[x] modulo with? f1 doesn't generate a group unless I am missing something

2018-10-22 07:18:42 UTC  

@ThisIsChris quotient field

2018-10-22 07:19:06 UTC  

and F[x] is a field of polynomials in x with coefficients in Z_q

2018-10-22 07:24:17 UTC  

@YourFundamentalTheorum thanks. How about for an element of F[x]/f1 called G, take a representative g, phi(G) = the coset of g*f2/f1 . Need to prove that the coset is independent of the choice of representative g

2018-10-22 07:26:20 UTC  

the last part is a given.

2018-10-22 07:26:24 UTC  

On the right hand side I mean g times f2 divided by f1 in the normal sense for poly nomials

2018-10-22 07:26:44 UTC  

F[x]/f1(x) is usually represented by all the polynomials """"less than""""" f1(x)

2018-10-22 07:26:51 UTC  

a lesser degree than f1(x) that is

2018-10-22 07:27:59 UTC  

Hm, what happens in the three cases where f2 is higher, equal, or lesser degree than f1?

2018-10-22 07:28:10 UTC  

assume f1 and f2 are of the same degree

2018-10-22 07:28:16 UTC  

otherwise you don't have the same cardinalities

2018-10-22 07:28:20 UTC  

which means they can't be isomorphic

2018-10-22 07:28:42 UTC  

the Galois fields can't be isomorphic that is

2018-10-22 07:29:51 UTC  

Hm I think you can actually prove it if the degrees mismatch too. Since you assume the rep g has degree strictly less than f1, then g*f2/f1 should have degree strictly less than f2

2018-10-22 07:30:43 UTC  

yes you can prove that

2018-10-22 07:30:46 UTC  

it's quite easy actually

2018-10-22 07:31:14 UTC  

the degrees part htat is

2018-10-22 07:31:28 UTC  

we know the exact # of elements in F[x]/f1(x)

2018-10-22 07:31:35 UTC  

it's going to be q^n

2018-10-22 07:31:41 UTC  

i'm assuming q is prime btw

2018-10-22 07:32:18 UTC  

Sure since it's irreducible. Otherwise, hm...

2018-10-22 07:32:58 UTC  

If a polynomial has a non-prime degree, to guarantee reducibility you need to allow complex coefficients, right?

2018-10-22 07:33:48 UTC  

i don't think it matters what the degree of the polynomial is

2018-10-22 07:33:58 UTC  

like if it's prime or not

2018-10-22 07:34:06 UTC  

i think what matters are the coefficients

2018-10-22 07:34:20 UTC  

if you don't have prime coefficients, your polynomials stop being a field

2018-10-22 07:34:42 UTC  

the reason why F[x] is a field is because Z_q is a field when q is prime

2018-10-22 07:34:48 UTC  

if q is not prime, Z_q is not a field