Message from @YourFundamentalTheorum
Discord ID: 503831458004336660
That's neat. I wonder if it's simultaneously done in all areas of the animal.
Hey guys, I'm Nick, and I'm a ChemE undergraduate from NY, excited to talk with all of you
nic
*nice
@Nicholas1166 - NY roll'd
roll'd?
@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.
Ah, OK. Yeah, I'd be glad to lend a hand if I can, although I am just a junior right now
I'm graduated from a university with a degree in math
ask me math questions
if you need halp
I very well might, math is my weakest subject. Thank you for the offer.
I didn't know this sever had a role thing. Can I get a role and be alerted?
@micbwilli done!
@ThisIsChris hey do you have any good explanation on how to find isomorphisms between two polynomial factor groups?
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]
@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
@ThisIsChris quotient field
and F[x] is a field of polynomials in x with coefficients in Z_q
@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
On the right hand side I mean g times f2 divided by f1 in the normal sense for poly nomials
F[x]/f1(x) is usually represented by all the polynomials """"less than""""" f1(x)
a lesser degree than f1(x) that is
Hm, what happens in the three cases where f2 is higher, equal, or lesser degree than f1?
assume f1 and f2 are of the same degree
otherwise you don't have the same cardinalities
which means they can't be isomorphic
the Galois fields can't be isomorphic that is
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
yes you can prove that
it's quite easy actually
the degrees part htat is
we know the exact # of elements in F[x]/f1(x)
it's going to be q^n
i'm assuming q is prime btw
Sure since it's irreducible. Otherwise, hm...
If a polynomial has a non-prime degree, to guarantee reducibility you need to allow complex coefficients, right?
i don't think it matters what the degree of the polynomial is
like if it's prime or not
i think what matters are the coefficients